CHAPTER 17 | Equilibrium in the Aqueous Phase
17.1. Collect and Organize
In Figure P17.1 four lines are shown to describe the possible dependence of percent ionization of acetic acid
with concentration. We are to choose the one that best represents the trend for this weak acid.
Analyze
The ionization of acetic acid is described by the following chemical reaction:
CH3COOH(aq)
CH3COO–(aq) + H+
The degree of ionization is the ratio of the quantity of a substance that is ionized to the concentration of the
substance before ionization.
Solve
According to Figure 17.8, the change in degree of ionization of a weak acid with concentration is not linear
and is best described by the red line in Figure P17.1. The degree of ionization increases with decreasing acetic
acid concentration.
Think about It
The percent ionization could be calculated for each concentration if we knew the equilibrium concentration of
the acetate ion in solution and the initial concentration of acetic acid dissolved.
 H + 
equilibrium
% ionization =
 100
 acetic acid initial
17.2. Collect and Organize
The bar graph in Figure P17.2 shows the percent ionization for three aqueous solutions of HOX where X = Cl,
Br, or I. From the graph, we can compare the Ka values for these three weak acids to determine which bar
represents the percent ionization of HOI.
Analyze
The Ka values for the acids (from Figure 17.12) are as follows: Ka = 2.9  10–8 for HOCl, Ka = 2.3  10–9 for
HOBr, Ka = 2.3  10–11 for HOI. From this we see that HOI is the weakest acid.
Solve
The weakest acid has the lowest percent ionization, so the third bar (shortest one) corresponds to HOI. Notice,
too, that as the electronegativity of X in HOX decreases, so too does the acidity and percent ionization.
Think about It
Notice that, even though all three acids are weak, the percent ionization differs greatly among them. The
strongest of these acids, HOCl, has a percent ionization of just under 0.5% while the weakest acid, HOI, has a
percent ionization of well under 0.1%, a more than fivefold difference.
17.3. Collect and Organize
From Figure P17.3 we are to choose which titration curve represents a strong acid and which represents a weak
acid, each of 1 M concentration at the start of the titration.
Analyze
A strong acid is completely ionized in solution and has a lower initial pH than the weak acid, which is only
partially ionized in solution.
Solve
The blue titration curve represents the titration of a 1 M solution of strong acid. The red titration curve
represents the titration of a 1 M solution of weak acid. This is because the pH of the strong acid is expected to
be much lower than that of the weak acid at the start of the titration (where no base has yet been added).
281
282 | Chapter 17
Think about It
Notice that the equivalence point of the titration of the strong acid (pH 7) does not equal that of the weak acid
(pH 10).
17.4. Collect and Organize
From the titration curve shown for a weak acid (red line in Figure P17.3), we can estimate the pKa. We can
then use that value to determine the pKb of the sodium salt of the weak acid to estimate the pH of a 0.5 M
solution of that sodium salt.
Analyze
At the midpoint, pKa = pH. From the plot, we see that the equivalence point is at pH 10. At the midpoint,
where half of the volume of titrant has been added, the pH is 6, so the pKa is 6 or Ka = 1  10– 6. The Kb then is
calculated using Kw:
K w 110–14
Kb 

 110–8
–6
Ka 110
We can then use this value of Kb with an ICE table to solve for the pH of the 0.5 M solution of A– (from the
sodium salt of the weak acid HA).
Solve
The pH of the sodium salt of HA is approximately equal to the pH at the equivalence point in the titration
curve of the weak acid (pH = 10). This makes sense because at the equivalence point we have the following
equilibrium:
A–(aq) + H2O ( )
HA(aq) + OH–(aq)
This is exactly the same as the reaction of NaA with water.
Think about It
We can solve for the pH of a 0.5 M A– solution where Kb = 1  10–8 using an ICE table:
[A–]
[HA]
[OH–]
Initial
0.5
0
0
Change
–x
+x
+x
Equilibrium
0.5 – x
x
x
(
x
)(
x
)
x2
K b  1  10 –8 

0.5 – x 0.5
x  7.1  10 –5
pOH  –log(7.1  10 –5 )  4.15
pH  14 – pOH  14 – 4.15  9.85
Our estimate of pH 10 was close!
17.5. Collect and Organize
For the red titration curve in Figure P17.3, we are to choose the indicator, according to its pKa, that would be
best for the titration.
Analyze
The best indicator is the one with a pKa that is nearest to the end point of the titration.
Solve
The end point for the red curve is at approximately pH 10. Therefore, the best indicator is the one with a pKa
of 9.0.
Equilibrium in the Aqueous Phase | 283
Think about It
The lower pKa indicators would show a color change before the end point of the titration was reached. Using
these would therefore underestimate the concentration of the weak acid in the original solution.
17.6. Collect and Organize
From the titration curve for the weak acid shown in Figure P17.3, we are to determine the pKa for the acid.
Analyze
In the titration curve the pH at the midpoint is equal to the pKa of the weak acid.
Solve
From the titration curve we see that when half of the volume of strong base has been added compared to that to
reach the equivalence point, the pH is 6. So the pKa for this weak acid is 6.
Think about It
At the midpoint [A–]  [HA] for the equilibrium reaction
HA(aq) + H2O ( )
A–(aq) + H3O+(aq)
This describes a buffer solution where [HA] = [A–]. By the Henderson–Hasselbalch equation,
[A – ]
pH = pK a  log
[HA]
and since log([A–]/[HA]) = log 1 = 0, then pH = pKa.
17.7. Collect and Organize
We are shown two titration curves in Figure P17.7. The blue curve has one equivalence point and the red curve
has two equivalence points. We are to assign each of these curves to either Na2CO3 or NaHCO3.
Analyze
Both of the bases being titrated are soluble sodium salts. The equation describing the titration of CO 32– (Na+ is
a spectator ion) shows CO32– to be “dibasic”; it reacts in two steps to form H2CO3.
CO32–(aq) + H+(aq)  HCO3–(aq)
HCO3–( aq) + H+(aq)  H2CO3(aq)
HCO3– , however, reacts with acid in one step; it is “monobasic”:
HCO3–(aq) + H+(aq)  H2CO3(aq)
Solve
The red titration curve represents the titration of Na 2CO3 because it shows two equivalence points. The blue
titration curve represents the titration of NaHCO3 because it shows one equivalence point.
Think about It
Notice that both titration curves start at high pH. This is due to the hydrolysis of CO 32– and HCO3– in water:
CO32–(aq) + H2O ( )  HCO3–(aq) + OH–(aq)
HCO3–(aq) + H2O ( )  H2CO3(aq) + OH–(aq)
17.8. Collect and Organize
Figure P17.8 shows three beakers: one containing a yellow solution, one containing a very light green solution,
and the last one containing a blue solution. Given that the bromthymol blue indicator in each beaker is yellow
in acidic solutions and blue in basic solutions, we are to match solutions of the dissolved salts NH 4Cl,
NH4C2H3O2, and NaC2H3O2 to the correct beaker.
Analyze
We must consider the hydrolysis of the constituent cation and anion in each salt. In NH 4Cl, the NH4+ ion reacts
with water to give an acidic solution:
NH4+(aq) + H2O ( )
NH3(aq) + H3O+(aq)
284 | Chapter 17
but Cl– does not. In NaC2H3O2 the Na+ ion does not hydrolyze, but the acetate ion does to give a basic
solution:
C2H3O2–(aq) + H2O ( )
HC2H3O2(aq) + OH–(aq)
In NH4C2H3O2 both cation and anion hydrolyze, giving a nearly neutral solution.
Solve
NH4Cl is dissolved in the yellow solution, NaC2H3O2 is dissolved in the blue solution, and NH4C2H3O2 is
dissolved in the light green solution.
Think about It
The relative magnitude of the Ka of NH4+ (5.7  10–10) compared to the Kb of C2H3O2– (5.7  10–10) shows that
both salts hydrolyze to the same extent and that we should expect a solution of NH 4C2H3O2 to be
approximately neutral.
17.9. Collect and Organize
For HBr(aq) we are to identify the Brønsted–Lowry acid and base.
Analyze
A Brønsted–Lowry acid is a proton donor. A Brønsted–Lowry base is a proton acceptor.
Solve
HBr is a strong acid in water. It acts as a Brønsted–Lowry acid, donating its proton to H2O, the Brønsted–
Lowry base:
HBr(aq) + H2O ( )  H3O+(aq) + Br–(aq)
Think about It
Hydrobromic acid is a strong acid in water. It completely dissociates in water to H 3O+ and Br–.
17.10. Collect and Organize
For HNO3(aq) we are to identify the Brønsted–Lowry acid and base.
Analyze
A Brønsted–Lowry acid is a proton donor. A Brønsted–Lowry base is a proton acceptor.
Solve
HNO3 is a strong acid in water. It acts as a Brønsted–Lowry acid, donating its proton to H2O, the Brønsted–
Lowry base:
HNO3(aq) + H2O ( )  NO3–(aq) + H3O+(aq)
Think about It
Nitric acid is a strong acid in water. It completely dissociates in water to H 3O+ and NO3–.
17.11. Collect and Organize
For NaOH(aq) we are to identify the Brønsted–Lowry acid and base.
Analyze
NaOH is a soluble salt that forms Na+ and OH– in water. Na+ does not react with water so it is a spectator ion.
We need then to consider the behavior of OH– in water. A Brønsted–Lowry acid is a proton donor. A
Brønsted–Lowry base is a proton acceptor.
Solve
OH– is a strong base in water. It acts as a Brønsted–Lowry base, removing a proton from H2O, the Brønsted–
Lowry acid:
OH–(aq) + H2O ( )
H2O ( ) + OH–(aq)
Equilibrium in the Aqueous Phase | 285
Think about It
In Problems 17.9 and 17.10, water acted as a Brønsted–Lowry base. In this problem, it acts as an acid. This
dual acid–base behavior makes water amphoteric.
17.12. Collect and Organize
We are to compare the neutralization capacity of equimolar solutions of NaOH and Ca(OH) 2.
Analyze
Both of these bases are strong and completely dissociate in water:
NaOH(s)  Na+(aq) + OH–(aq)
Ca(OH)2 (s)  Ca2+(aq) + 2 OH–(aq)
Solve
No, these two compounds do not have the same capacity to neutralize strong acids. As seen from the
dissolution equations, Ca(OH)2 produces two equivalents of OH– in solution and can neutralize twice as much
acid as the same molar concentration of NaOH.
Think about It
It is the case, however, that equimolar solutions of NaOH and CsOH have the same neutralizing capacity.
17.13. Collect and Organize
For three acid–base reactions we are to identify which reactant is the acid and which reactant is the base.
Analyze
For all of these reactions that involve the transfer of a proton between the acid and base, we can apply the
Brønsted–Lowry definitions of acid and base. A Brønsted–Lowry acid is a proton donor. A Brønsted–Lowry
base is a proton acceptor.
Solve
(a) HNO3 is the acid. It transfers H+ to the base NaOH.
(b) HCl is the acid. It transfers H+ to the base CaCO3.
(c) HCN is the acid. It transfers H+ to the base NH3.
Think about It
In reactions a and b, Na+, Ca2+, and Cl– do not get involved in the reaction. They are spectator ions. The net
ionic equations are
HNO3(aq) + OH–(aq)  NO3–(aq) + H2O ( )
CO32–(aq) + 2 H+(aq)  CO2(g) + H2O ( )
17.14. Collect and Organize
For three acid–base reactions we are to identify which reactant is the acid and which reactant is the base.
Analyze
For all of these reactions that involve the transfer of a proton between the acid and base, we can apply the
Brønsted–Lowry definitions of acid and base. A Brønsted–Lowry acid is a proton donor. A Brønsted–Lowry
base is a proton acceptor.
Solve
(a) H2O is the acid. It transfers a proton to the base, NH2–.
(b) HClO4 is the acid. It transfers a proton to the base, H2O.
(c) HSO4– is the acid. It transfers a proton to the base, CO32–.
286 | Chapter 17
Think about It
Notice that in reaction a water acts as an acid, but in reaction b water acts as a base. This behavior classifies
water as amphoteric. Also, NH2– is a strong base in aqueous solution and HClO4 is a strong acid in aqueous
solution. Both of these completely react with water to form NH 3 and ClO4–.
17.15. Collect and Organize
For each species listed we are to write the formula for the conjugate base.
Analyze
The conjugate base form of a species has H+ removed from its formula.
Solve
The conjugate base of HNO2 is NO2–.
The conjugate base of HOCl is OCl–.
The conjugate base of H3PO4 is H2PO4–.
The conjugate base of NH3 is NH2–.
Think about It
Be sure to account for the change in charge when H + is removed to form the conjugate base.
17.16. Collect and Organize
For each species listed we are to write the formula for the conjugate acid.
Analyze
The conjugate acid form of a species has an H+ added to its formula.
Solve
The conjugate acid of NH3 is NH4+.
The conjugate acid of ClO2– is HClO2
The conjugate acid of SO42– is HSO4–.
The conjugate acid of OH– is H2O.
Think about It
Be sure to account for the change in charge when the base adds H+ to form the conjugate acid.
17.17. Collect and Organize
Given that the concentration of a nitric acid solution is 1.50 M, we are to calculate the concentration of H +
ions in the solution.
Analyze
Nitric acid is a strong acid that completely dissociates in water, so the concentration of H + ions is
stoichiometrically related to the concentration of HNO3:
HNO3(aq)  H+(aq) + NO3–(aq)
Solve
[H+] = 1.50 M
Think about It
The concentration of OH– in strong base solutions, likewise, is the same as the concentration of the strong
base dissolved into the solution.
17.18. Collect and Organize
Given a solution of the strong acid HCl that is prepared by diluting 20.0 mL of 11.6 M HCl solution to
500 mL, we are to calculate the concentration of H+ ions in the final solution.
Equilibrium in the Aqueous Phase | 287
Analyze
For this problem we use the relationship from Chapter 4 for diluting solutions:
Vinitial  Minitial = Vfinal  Mfinal
where V is the volume of the solution and C is the concentration.
Solve
20.0 mL  11.6 M = 500 mL  Mfinal
Mfinal = 0.464 M
Think about It
Because this is a solution of a strong acid, we know that the acid is completely dissociated and so
[HCl] = [H +].
17.19. Collect and Organize
Given that a solution is 0.0800 M in the strong base Sr(OH)2, we are asked to calculate the concentration of
OH–.
Analyze
Sr(OH)2, being a strong base, completely dissociates according to the equation
Sr(OH)2(aq)  Sr2+(aq) + 2 OH–(aq)
Therefore, [OH–] = 2  [Sr(OH)2].
Solve
2  0.0800 M = 0.160 M = [OH–]
Think about It
Be sure to account for both OH– ions in this strong base.
17.20. Collect and Organize
Given that a solution is prepared by dissolving 5.0 g of NaOH in water and diluting to a final volume of
250 mL, we are asked to calculate the final concentration of OH – in the solution.
Analyze
Because NaOH is a strong base we know that it is completely ionized in solution:
NaOH(aq)  Na+(aq) + OH–(aq)
Therefore, we can relate the moles of NaOH dissolved to the moles of OH– in solution and then calculate the
molarity by dividing the moles of OH– by the final volume in liters.
Solve
5.0 g NaOH 
1 mol NaOH
1 mol OH –
1


 0.50 M
40.00 g NaOH 1 mol K NaOH 0.250 L
Think about It
Other strong bases used in the laboratory include KOH, LiOH, Mg(OH) 2, Ca(OH)2, Ba(OH)2, and Sr(OH)2.
17.21. Collect and Organize
We are asked how to prepare 2.50 L of 0.70 M OH– using NaOH(s).
Analyze
From the desired volume and concentration we first calculate the moles of OH – required for the solution.
Since 1 mol of OH– is produced for every 1 mol of NaOH dissolved, this is also the moles of NaOH required.
To calculate the mass of NaOH needed we multiply the number of moles by the molar mass of NaOH
(40.00 g/mol).
288 | Chapter 17
Solve
Mass of NaOH needed
0.70 mol OH – 1 mol NaOH 40.00 g NaOH


 70.0 g NaOH
L
1 mol NaOH
1 mol OH –
Dissolve 70.0 g of NaOH(s) in water and dilute to a total volume of 2.50 L.
Mass NaOH  2.50 L 
Think about It
Remember that volume times concentration for solutions gives us the moles of that substance in solution.
17.22. Collect and Organize
We are to calculate the volume of a 1.00 M NaOH solution that is needed to prepare 250 mL of a solution that
is 0.0200 M in OH– concentration.
Analyze
Since NaOH is a strong base, 1.00 M NaOH is 1.00 M in OH–. We can use the relationship from Chapter 4 for
dilutions to solve this problem for Vinitial:
Vinitial  Minitial = Vfinal  Mfinal
Solve
Vinitial  1.00 M = 250 mL  0.0200 M
Vinitial = 5.00 mL
Think about It
Be careful in this type of calculation if the strong base is one like Ba(OH)2. In that case, a 1.00 M Ba(OH)2
solution would be 2.00 M in OH– concentration.
17.23. Collect and Organize
We are to explain why the pH value decreases for solutions as acidity increases.
Analyze
The pH of a solution is calculated through
pH = –log[H+]
Solve
Because the pH function is a –log function, as [H+] increases, the value of –log[H+] decreases.
Think about It
The pH scale is typically 0 –14 for concentrations of H+ from 1 M to 1  10–14 M, but values of pH may be
negative or greater than 14.
17.24. Collect and Organize
We are to determine the difference in pH between two solutions when solution A is 100 times more acidic
than solution B.
Analyze
If solution A is 100 times more acidic than solution B, [H +]A = 100[H+]B and [H+]A/[H+]B = 100. Taking the
log of both sides of this equation gives
 [H + ] 
log  + A   log[H + ]A – log[H + ]B  log100
 [H ]B 
+
Because pHA = –log[H ]A and pHB = –log[H+]B then pHB – pHA = log 100.
Solve
The difference in pH between the two solutions is pHB – pHA = log 100 = 2.000.
Equilibrium in the Aqueous Phase | 289
Think about It
Because pH is a logarithmic scale, one unit change in pH means that there is a tenfold increase or decrease in
[H+].
17.25. Collect and Organize
We are asked under what conditions the pH of a solution may be negative.
Analyze
The pH scale is often seen as 0 –14. This occurs when [H+] is between 1 M and 1  10–14 M.
Solve
When [H+] is greater than 1 M, the pH of the solution is negative.
Think about It
For example, a 3.00 M solution of HCl has a pH of
pH = –log[H+] = –log(3.00 M) = – 0.48
17.26. Collect and Organize
For the autoionization of ethanol we are to explain why this ionization has a K value much smaller than the
autoionization constant of water.
Analyze
For water the autoionization is described by
2H2O ( )
H3O+(aq) + OH–(aq)
For ethanol the autoionization is described by
2 CH3CH2OH ( )
CH3CH2OH2+(ethanol) + CH3CH2O–(ethanol)
Solve
In both autoionizations an O — H bond breaks and a new O — H bond forms. The O — H bond in ethanol is
less polar compared to that of water (because it is a C — O — H bond rather than a H — O — H bond) and so
ethanol is a weaker base than water and the K value for the acid–base transfer of H+ for ethanol is smaller.
Think about It
Another species that might autoionize is ammonia:
2 NH3 ( )
NH4+(ammonia) + NH2–(ammonia)
17.27. Collect and Organize
Given either the [OH–] or [H+] for a solution, we are asked to calculate the pH and pOH and determine
whether the solution is acidic, basic, or neutral.
Analyze
To calculate the pH or the pOH from the [H+] or [OH–], respectively, we use
pH = –log[H+]
pOH = –log[OH–]
To find the pOH from the pH and vice versa we use the relationship
pH + pOH = 14
If the pH of a solution is less than 7, the solution is acidic. If the pH is equal to 7, the solution is neutral. If the
pH is greater than 7, the solution is basic.
Solve
(a) pH = –log(3.45  10–8) = 7.462
pOH = 14 – pH = 6.538
This solution is basic.
290 | Chapter 17
(b) pH = –log(2.0  10–5) = 4.70
pOH = 14 – pH = 9.30
This solution is acidic.
(c) pH = –log(7.0  10–8) = 7.15
pOH = 14 – pH = 6.85
This solution is basic.
(d) pOH = –log(8.56  10– 4) = 3.068
pH = 14 – pOH = 10.932
This solution is basic.
Think about It
When determining how many significant figures to include in your answers when computing the pH or pOH,
remember that the first number in the pH or pOH gives the location of the decimal point. The significant
digits, therefore, follow after the decimal point.
17.28. Collect and Organize
Given either the [OH–] or [H+] for a solution, we are asked to calculate the pH and pOH and determine
whether the solution is acidic, basic, or neutral.
Analyze
To calculate the pH or the pOH from the [H+] or [OH–], respectively, we use
pH = –log[H+]
pOH = –log[OH–]
To find the pOH from the pH and vice versa we use the relationship
pH + pOH = 14
If the pH of a solution is less than 7, the solution is acidic. If the pH is equal to 7, the solution is neutral. If the
pH is greater than 7, the solution is basic.
Solve
(a) pOH = –log(7.69  10–3) = 2.114
pH = 14 – pOH = 11.886
This solution is basic.
(b) pOH = –log(2.18  10–9) = 8.662
pH = 14 – pOH = 5.338
This solution is acidic.
(c) pH = –log(4.0  10–8) = 7.40
pOH = 14 – pH = 6.60
This solution is basic.
(d) pH = –log(3.56  10– 4) = 3.449
pOH = 14 – pH = 10.551
This solution is acidic.
Think about It
When determining how many significant figures to include in your answers when computing the pH or pOH,
remember that the first number in the pH or pOH gives the location of the decimal point. The significant
digits, therefore, follow after the decimal point.
17.29. Collect and Organize
Given that stomach acid is 0.155 M HCl, we are to calculate its pH.
Analyze
The pH is defined as pH = –log[H+]. Since HCl is a strong acid, we know that 0.155 M HCl is also
0.155 M H+.
Equilibrium in the Aqueous Phase | 291
Solve
pH = –log(0.155) = 0.810
Think about It
The acid in your stomach is fairly strong and the HCl is produced by parietal cells to break down food. Your
stomach is protected in turn by epithelial cells that secrete a bicarbonate solution that neutralizes the acid and
forms a coating to protect the stomach’s tissues.
17.30. Collect and Organize
Given a 0.00500 M HNO3 solution, we are to calculate its pH.
Analyze
The pH is defined as pH = –log[H+]. Since HNO3 is a strong acid, we know that 0.00500 M HNO3 is also
0.00500 M H+.
Solve
pH = –log(0.00500) = 2.301
Think about It
This solution is about as acidic as the juice of a lemon (Figure 17.6).
17.31. Collect and Organize
Given that the concentration of NaOH in a solution is 0.0450 M, we are to calculate its pH and pOH.
Analyze
Because NaOH is a strong base, we know that the [OH–] in this solution is 0.0450 M. The pOH is calculated
from pOH = –log[OH–], with the pH being found from the pOH through the relationship pH + pOH = 14.
Solve
pOH = –log (0.0450) = 1.347
pH = 14 – pOH = 12.653
Think about It
This solution is as alkaline (basic) as drain cleaner (Figure 17.6).
17.32. Collect and Organize
Given that the concentration of KOH in a solution is 0.160 M, we are to calculate its pH and pOH.
Analyze
Because KOH is a strong base, we know that the [OH–] in this solution is 0.160 M. The pOH is calculated
from pOH = –log[OH–], with the pH being found from the pOH through the relationship pH + pOH = 14.
Solve
pOH = –log (0.160) = 0.796
pH = 14 – pOH = 13.204
Think about It
This solution is as alkaline (basic) as drain cleaner (Figure 17.6).
17.33. Collect and Organize
Given a solution that is 1.33 M in HNO3, we are to calculate its pH.
Analyze
Since HNO3 is a strong acid, we know that for this solution the [H +] from the HNO3 is 1.33 M. The pH is
calculated from pH = –log[H+].
Solve
pH = –log (1.33) = –0.124
292 | Chapter 17
Think about It
This pH is negative because the concentration of acid is greater than 1.00 M, which corresponds to
pH = 0.000.
17.34. Collect and Organize
Given a solution that is 6.9  10–8 M in HBr, we are to calculate its pH.
Analyze
Since HBr is a strong acid, we know that for this solution the [H +] from the HBr is 6.9  10–8 M. The
calculated pH from this value, however, gives a pH of greater than 7, which would mean that this solution is
basic:
pH = –log(6.9  10–8) = 7.16
This cannot be since we have added an acid to water! This is a case where the autoionization of water (where
[H+] = 1  10–7 M) is important. We therefore have to derive an equation to solve for the pH of this solution.
In the solution of HBr in water we must have an overall charge of zero:
[H+] = [OH–] + [Br–] or [OH–] = [H+] – [Br–]
This relationship puts [OH–] in terms of [H+]. We also know that the concentration of Br – in solution is equal
to the concentration of HBr originally dissolved in the solution since HBr is a strong acid:
[Br–] = [HBr]initial
Combining the two equations we get
[OH–] = [H+] – [HBr]initial
In the Kw expression we can substitute in for [OH–]:
Kw = [H+]  [OH–]
Kw = [H+]  ([H+] – [HBr]initial)
Rearranging this equation we obtain the quadratic equation
[H+]2 – ( [H+]  [HBr]initial) – Kw = 0
where for this problem [HBr]initial = 6.9  10–8 M and Kw = 1.0  10–14.
Solve
[H+]2 – ([H+]  6.9  10–8) – 1.0  10–14 = 0
Solving by the quadratic equation gives [H+] = 1.403  10–7 M for a pH = –log 1.403  10–7 = 6.85.
Think about It
A similar formula can be derived for very dilute solutions of a strong base:
[H+]2 + ([H+]  [strong base]initial) – Kw = 0
17.35. Collect and Organize
For one-molar solutions of CH3COOH, HNO2, HClO, and HCl we are to rank these in order of decreasing
concentration of H+. For this we need the Ka values for each acid.
Analyze
The Ka values for these acids are listed in Tables 17.1 and 17.2. The greater the value of the Ka, the greater the
concentration of H+ in the solution of the acid. We note that HCl is a strong acid.
Solve
In order of largest Ka (strongest acid) to smallest Ka (weakest acid), HCl > HNO2 > CH3COOH > HClO.
Think about It
For the weak acids in this series, there is a wide range in their acidities (about 10,000-fold comparing their Ka
values).
Equilibrium in the Aqueous Phase | 293
17.36. Collect and Organize
Given the degree of ionization for four weak acids of 0.100 M concentration, we are to determine which acid
has the smallest value of its acid dissociation constant, Ka.
Analyze
The lower the degree of ionization, the weaker is the acid and the lower the value of its Ka.
Solve
Acetic acid, CH3CO2H, has the lowest degree of ionization and is the acid with the smallest Ka value.
Think about It
The acid with the largest degree of ionization, HF, has the highest Ka value.
17.37. Collect and Organize
We are to explain why the electrical conductivity of 1.0 M NaNO2 is much better than that of 1.0 M HNO2.
Analyze
Solutions with a larger concentration of dissolved ions conduct electricity better than those with lower
concentrations of dissolved ions.
Solve
NaNO2 is completely soluble in water, separating into Na + and NO2– ions, each in 1.0 M concentration for a
total ion concentration of 2.0 M. HNO2, however, only weakly dissociates in water:
HNO2(aq)
NO2–(aq) + H+(aq)
and so produces just slightly greater than 1.0 M ions in solution. NaNO2, therefore, with more dissolved ions
in solution, is a better conductor of electricity.
Think about It
Later in this chapter, you will learn that NO2– reacts with water to a small extent:
NO2–(aq) + H2O ( )
HNO2(aq) + OH–(aq)
However, this produces another ion, OH–, and so our analysis that a solution of 1.0 M NaNO2 is 2.0 M in ions
is still correct.
17.38. Collect and Organize
We are to explain why HCl, although a molecular species, when dissolved in water (also a molecular species)
is a good conductor of electricity.
Analyze
Solutions with dissolved ions are good conductors of electricity.
Solve
When HCl dissolves in water, it completely dissociates because it is a strong acid that produces 2 moles of
ions for every mole of HCl dissolved
HCl(aq)  H+(aq) + Cl–(aq)
This makes HCl(aq) a good electrical conductor.
Think about It
This is true for other typical strong acids and bases such as HNO3, HBr, HClO4, KOH, and NaOH.
17.39. Collect and Organize
The formula for hydrofluoric acid is HF. From this we can write the mass action expression for this weak
acid.
294 | Chapter 17
Analyze
The general form of the mass action expression for weak acids, based on HA
Ka 
+
A– + H+, is
–
[H ][A ]
[HA]
Solve
F–(aq) + H+(aq)
HF(aq)
Ka 
[H + ][F– ]
[HF]
Think about It
The hydrofluoric acid is transferring a proton to water in this equation, so an equivalent expression is
HF(aq) + H2O ( )
F–(aq) + H3O+(aq)
Ka 
[H3 O + ][F– ]
[HF]
17.40. Collect and Organize
Given the formula of formic acid, HCOOH, and the information that one H atom is ionizable, we can write the
mass action expression for this acid’s ionization.
Analyze
The general form of the mass action expression for weak acids, based on HA
A– + H+, is
[H + ][A – ]
[HA]
In formic acid, the H atom bonded to O (not the H bonded to C) is ionizable.
Ka 
Solve
HCOO–(aq) + H+(aq)
HCOOH(aq)
Ka 
[H + ][HCOO – ]
[HCOOH]
Think about It
The formic acid is transferring a proton to water in this equation, so an equivalent expression is
HCOOH(aq) + H2O ( )
HCOO–(aq) + H3O+(aq)
Ka 
[H3O+ ][HCOO – ]
[HCOOH]
17.41. Collect and Organize
Given that the Ka of alanine is less when it is dissolved in ethanol than it is when dissolved in water, we are to
determine which solvent ionizes alanine to the larger extent and which solvent is the stronger Brønsted–
Lowry base.
Analyze
The larger the value of Ka, the greater the extent to which a substance has ionized. The solvent in which an
acid ionizes the most must be the strongest Brønsted–Lowry base toward that acid.
Solve
(a) Because Ka for alanine is greater in water than in ethanol, alanine in water is ionized to a greater extent
than in ethanol.
(b) Water is the stronger base for alanine compared to ethanol because alanine is ionized to a greater extent in
water.
Equilibrium in the Aqueous Phase | 295
Think about It
This question demonstrates that acid–base strengths can depend on the basicity of the solvent.
17.42. Collect and Organize
Given the Ka values of proline in water, 28% aqueous ethanol, and aqueous formaldehyde, we are to
determine in which solvent proline is the strongest acid and rank the solvents in order of increasing basicity.
Analyze
The order of Ka values from weakest acid to strongest is as follows: proline in water (2.5  10–11) < proline in
28% ethanol (2.8  10–11) < proline in aqueous formaldehyde (1.66  10–8). The larger the value of the Ka of
proline in a solvent, the stronger is proline’s acidity.
Solve
(a) Proline is the strongest acid in formaldehyde.
(b) In order of increasing basicity toward proline, water < 28% ethanol < formaldehyde.
Think about It
This question demonstrates that acid–base strengths depends on the basicity of the solvent.
17.43. Collect and Organize
Given that CH3NH2 is slightly basic in water we can write the equation describing its reaction with water to
identify which species in the reaction is the Brønsted–Lowry acid and which is the base.
Analyze
Acting as a base, CH3NH2 accepts H+ from surrounding water molecules.
Solve
The reaction describing the basicity of CH 3NH2 is
CH3NH2(aq) + H2O ( )
CH3NH3+(aq) + OH–(aq)
In this reaction, H2O is the acid and CH3NH2 is the base.
Think about It
In another solvent, such as diethylamine, methylamine may act as an acid:
CH3NH2 + (CH3CH2)2NH
CH3NH– + (CH3CH2)2NH2+
This occurs because diethylamine (Kb = 8.6  10– 4) is a stronger base than methylamine (Kb = 4.4  10– 4).
17.44. Collect and Organize
Given that an aqueous solution of H2NCH2CH2NH2 is basic, we are to write the formula of the product of its
reaction with HCl.
Analyze
Because H2NCH2CH2NH2 is a base it will accept the H+ in solution from the dissolved HCl.
Solve
H2NCH2CH2NH3+ forms in the solution as described by the equation
H2NCH2CH2NH2(aq) + H+(aq)
H2NCH2CH2NH3+(aq)
If two molar equivalents of H +(aq) are present the second protonation occurs:
H2NCH2CH2NH3+(aq) + H+(aq)
H3NCH2CH2NH32+(aq)
The Cl– ions are the counterions in the salt: (H3NCH2CH2NH3)Cl2.
Think about It
Because this compound has two basic sites on the nitrogen atoms, it is called a dibasic compound.
296 | Chapter 17
17.45. Collect and Organize
Using the measured percent ionization of 1.00 M lactic acid of 2.94%, we are to calculate the Ka of this weak
acid.
Analyze
The equilibrium equation from Appendix 5 that describes the ionization of lactic acid is
CH3CHOHCOOH(aq)
H+(aq) + CH3CHOHCOO–(aq)
We can set up an ICE table to solve this problem, where x = 1.00 M  0.0294 = 0.0294. The Ka expression is
[H + ][CH3 CHOHCOO – ]
Ka 
[CH 3 CHOHCOOH]
Solve
Initial
Change
Equilibrium
[CH3CHOHCOOH]
1.00
–x = – 0.0294
0.9706
Ka 
[H+]
0
+x = + 0.0294
0.0294
[CH3CHOHCOO–]
0
+x = + 0.0294
0.0294
(0.0294)2
 8.9110– 4
0.9706
Think about It
Compare this with the Ka at 25˚C value of 1.4  10 – 4 listed in Appendix 5. The difference may be attributable
to a temperature difference because body temperature is 37˚C.
17.46. Collect and Organize
Using the measured percent ionization of 0.100 M butanoic acid of 1.23%, we are to calculate the Ka of this
weak acid.
Analyze
The equilibrium equation that describes the ionization of butanoic acid is
C3H7COOH(aq)
H+(aq) + C3H7COO–(aq)
We can set up an ICE table to solve this problem, where x = 0.100 M  0.0123 = 0.00123. The Ka expression
is
[H + ][C3 H 7 COO – ]
Ka 
[C3 H 7 COOH]
Solve
Initial
Change
Equilibrium
[C3H7COOH]
0.100
–x = – 0.00123
0.09877
Ka 
[H+]
0
+x = +0.00123
0.00123
[C3H7COO–]
0
+x = +0.00123
0.00123
(0.00123) 2
 1.53  10 –5
0.09877
Think about It
This value compares well with the Ka value listed in Appendix 5 of 1.5  10–5.
17.47. Collect and Organize
We are given the [H+] for an equilibrium solution of an unknown acid with an initial concentration of
0.250 M. From this information we can calculate the degree of ionization and Ka for this weak acid.
Equilibrium in the Aqueous Phase | 297
Analyze
The equilibrium expression for the unknown acid is
HA(aq)
H+(aq) + A–(aq)
with a Ka expression of
[H + ][A – ]
Ka 
[HA]
We can set up an ICE table to show how this weak acid ionizes; for this acid, [H +] at equilibrium is
4.07  10–3 M.
Solve
[HA]
Initial
0.250
Change
–x = –0.00407
Equilibrium
0.24593
(0.00407) 2
Ka 
0.24593
[H+ ]eq
Degree of ionization 

[HA]initial
[H+]
0
+x = +0.00407
0.00407
[A–]
0
+x = +0.00407
0.00407
 6.74  10 –5
0.00407 M
 0.0163 or 1.63%
0.250 M
Think about It
In this problem we can use the [H+] as equivalent to x in the ICE table, enabling us to calculate both the Ka
and the degree of ionization for the acid.
17.48. Collect and Organize
Because when HNO3 is very concentrated it is not completely ionized, we are able to calculate its Ka. To do
this we are given that only 33% of the HNO3 is ionized when its concentration is 7.5 M.
Analyze
The equilibrium expression for concentrated nitric acid is
HNO3(aq)
H+(aq) + NO3–(aq)
with a Ka expression of
[H + ][NO3 – ]
Ka 
[HNO3 ]
We can set up an ICE table to solve this expression when x = 7.5 M  0.33 = 2.475 M.
Solve
Initial
Change
Equilibrium
[H+]
0
+x = +2.475
2.475
2
(2.475)
Ka 
 1.2
5.025
[HNO3]
7.5
–x = –2.475
5.025
[NO3–]
0
+x = +2.475
2.475
Think about It
This Ka is still larger than 1, so the products (H+ and NO3–) are favored in this equilibrium.
17.49. Collect and Organize
Given that formic acid has Ka = 1.8  10– 4, we can calculate the pH of a 0.060 M aqueous solution of this
weak acid.
298 | Chapter 17
Analyze
To solve this problem we set up an ICE table where the initial concentration of formic acid, HCOOH, is
0.060 M. We let x be the amount of formic acid that ionizes. The equilibrium and Ka expressions are
HCOOH(aq)
H+(aq) + HCOO–(aq)
[H + ][HCOO – ]
 1.8  10 –4
[HCOOH]
The pH can be calculated through the equation pH = –log[H+].
Ka 
Solve
[H+]
[HCOO–]
Initial
0
0
Change
+x
+x
Equilibrium
x
x
2
2
x
x
1.8  10 –4 

0.060  x 0.060
x  3.29  10 –3
We should first check the simplifying assumption we made above before calculating the pH.
3.29  10 –3
 100  5.5%
0.060
This is a little over 5%, so technically we should solve this by the quadratic equation (which we do below).
However, if we allow this simplifying assumption to be valid, the pH of the solution would be calculated by
[HCOOH]
0.060
–x
0.060 – x
pH = –log 3.29  10–3 = 2.48
Now, solving by the quadratic equation gives
1.8  10– 4(0.060 – x) = x2
2
x + 1.8  10– 4x – 1.08  10–5 = 0
x = 3.20  10–3
pH = –log 3.20  10–3 = 2.49
Think about It
The difference between the pH values when making the simplifying assumption and solving the equation
exactly using the quadratic equation is 2.49 – 2.48 = 0.01, which is fairly small.
17.50. Collect and Organize
Given that uric acid has pKa = 3.89, we can calculate the pH of a 0.0150 M aqueous solution of this weak acid.
Analyze
To solve this problem we can set up an ICE table where the initial concentration of uric acid is 0.0150 M. We
let x be the amount of uric acid that ionizes. The equilibrium and Ka expressions are
Uric acid(aq)
H+(aq) + urate–(aq)
Ka 
[H + ][urate – ]
[uric acid]
From the pKa we can calculate the Ka:
Ka  110– pKa  110–3.89  1.29 10– 4
The pH can be calculated through the equation pH = –log[H+].
Solve
Initial
Change
Equilibrium
[Uric Acid]
0.0150
–x
0.0150 – x
[H+]
0
+x
x
[Urate–]
0
+x
x
Equilibrium in the Aqueous Phase | 299
x2
x2

0.0150  x 0.0150
x  1.39  10 –3
Checking the simplifying assumption shows that our simplifying assumption is not valid:
1.39  10 –3
 100  9.3%
0.0150
so we must solve this by the quadratic equation:
1.29  10– 4(0.0150 – x) = x2
x2 + 1.29  10– 4x – 1.935  10– 6 = 0
x = 1.33  10–3
pH = –log 1.33  10–3 = 2.876
1.29  10 – 4 
Think about It
Simplifying assumptions are more often valid when there is a large difference between a larger initial
concentration of the weak acid or base and a smaller value of Ka.
17.51. Collect and Organize
By comparing the pH of rain in a weather system in the Midwest of 5.02 with the pH of the rain in that same
system when it reached New England of 4.66, we can calculate how much more acidic the rain in New
England was.
Analyze
We want to find the ratio
[H + ]New England
[H + ]Midwest
Because pH = –log[H+], the [H+] = 1  10–pH. Therefore,
[H + ]New England 1  10–4.66

[H + ]Midwest
1  10–5.02
Solve
1  10–4.66 2.19  10–5 M

 2.3
[H+ ]Midwest
1  10–5.02 9.55  10–6 M
The rain in New England in this weather system was 2.3 times more acidic than the rain in the Midwest.
[H+ ]New England

Think about It
Among the causes of the acidity of the rain in New England are the coal-burning electricity power plants in
the Midwest, which expel SO2 and SO3 into the air which make H2SO3 and H2SO4.
17.52. Collect and Organize
We are given that the pH of a water sample in 1998 was 200 times lower than that of the lowest pH taken in
the preceding 23 years. We are to calculate what this difference is in pH units.
Analyze
The ratio is expressed mathematically as
 H + 
1
today

200
 H + 
previous
Solve
In pH units this difference is –log 200 = –2.301 pH units. So the measurement in 1998 was 2.301 pH units
more acidic than any reading in the previous 23 years.
300 | Chapter 17
Think about It
Since the pH scale is a log scale, a large change in [H+] gives just a small change in pH.
17.53. Collect and Organize
Given the Kb for dimethylamine of 5.9  10– 4, we are to find the pH of a solution of dimethylamine that is
1.20  10–3 M.
Analyze
The equilibrium and Kb expressions for dimethylamine are
(CH3)2NH(aq) + H2O ( )
(CH3)2NH2+(aq) + OH–(aq)
[OH – ][(CH 3 ) 2 NH 2  ]
[(CH 3 ) 2 NH]
We can set up an ICE table to solve this problem, where x is the amount of dimethylamine that ionizes. By
solving for x we can calculate [OH–], from which we can determine the pOH and pH:
pOH = –log[OH–]
pH = 14 – pOH
Kb 
Solve
Initial
Change
Equilibrium
[(CH3)2NH]
[OH–]
[(CH3)2NH2+]
1.20  10–3
–x
1.20  10–3 – x
0
+x
x
0
+x
x
x2
x2

–3
1.2  10  x 1.2  10 –3
x  8.41  10 –4
Checking the simplifying assumption shows that our simplifying assumption is not valid:
8.41  10 –4
 100  70%
1.20  10 –3
so we must solve this by the quadratic equation:
5.9  10– 4 (1.20  10–3 – x) = x2
x2 + 5.9  10– 4x – 7.08  10–7 = 0
x = 5.966  10– 4
pOH = –log 5.97  10– 4 = 3.22
pH = 14 – 3.22 = 10.77
5.9  10 –4 
Think about It
Our answer of a pH > 7 is consistent with dimethylamine’s behavior as a base in aqueous solution.
17.54. Collect and Organize
Given the pKb of morphine, a weak base, we are asked to calculate the pH of a 0.115 M solution.
Analyze
From the pKb we can calculate the Kb
Kb  110– pKb  110–5.79  1.62 10– 6
The equilibrium and Kb expressions for the ionization of morphine are
Morphine(aq) + H2O ( )
morphineH+(aq) + OH–(aq)
Kb 
[OH – ][morphineH + ]
 1.62  10 – 6
[morphine]
Equilibrium in the Aqueous Phase | 301
We can set up an ICE table to solve this problem, where x is the amount of morphine that ionizes. By solving
for x we can calculate [OH–], from which we can determine the pOH and pH:
pOH = –log[OH–]
pH = 14 – pOH
Solve
[OH–]
[MorphineH+]
Initial
0
0
Change
+x
+x
Equilibrium
x
x
2
2
x
x
1.62  10 –6 

0.115  x 0.115
x  4.32  10 –4
Checking the simplifying assumption shows that our simplifying assumption is valid:
4.32  10 –4
 100  0.38%
0.115
–
From [OH ] we can calculate the pH:
pOH = –log 4.32  10–4= 3.365
pH = 14 – 3.365 = 10.635
[Morphine]
0.115
–x
0.115 – x
Think about It
Our calculation of the pH of this solution as greater than 7 is consistent with morphine’s behavior as a base in
aqueous solution.
17.55. Collect and Organize
Given the pKa of the conjugate acid of codeine (pKa = 8.21), we are to calculate the pH of a 3.42  10– 4 M
solution of codeine, a weak base.
Analyze
We first need to determine the value of the Kb from the pKa for codeine:
Ka  1  10– pKa  1  10–8.21  6.17  10 –9
110–14
 1.62 10– 6
Ka 6.17 10–9
The equilibrium and Kb expressions for the ionization of codeine are
Codeine(aq) + H2O ( )
codeineH+(aq) + OH–(aq)
Kb 
Kw

[OH – ][codeineH + ]
 1.62 10– 6
[codeine]
We can set up an ICE table to solve this problem, where x is the amount of codeine that ionizes. By solving
for x we can calculate [OH–], from which we can determine the pOH and pH:
pOH = –log[OH–]
pH = 14 – pOH
Kb 
Solve
Initial
Change
Equilibrium
[Codeine]
3.42  10– 4
–x
3.42  10– 4 – x
1.62  10 – 6 
[OH–]
0
+x
x
x2
x2

–4
3.42  10  x 3.42  10 – 4
x  2.35  10 –5
[CodeineH+]
0
+x
x
302 | Chapter 17
Checking the simplifying assumption shows that our simplifying assumption is not valid:
2.35  10 –5
 100  6.9%
3.42  10 –4
so we must solve this by the quadratic equation:
1.62  10–6(3.42  10–4 – x) = x2
2
x + 1.62  10–6x – 5.54  10–10 = 0
x = 2.27  10–5
pOH = –log 2.27  10–5 = 4.644
pH = 14 – 4.644 = 9.356
Think about It
Comparing the pKb of codeine to that of morphine in Problem 17.54, we see that codeine and morphine are
identical in their basicity.
17.56. Collect and Organize
Given that the Kb of pyridine is 1.7  10–9 in Appendix 5, we are to calculate the pH of a 0.125 M solution of
this weak base.
Analyze
The equilibrium and Kb expressions for the ionization of codeine are
C5H5N(aq) + H2O ( )
C5H5NH+(aq) + OH–(aq)
[OH – ][C5 H 5 NH + ]
 1.7  10 –9
[C5 H 5 N]
We can set up an ICE table to solve this problem, where x is the amount of pyridine that ionizes. By solving
for x we can calculate [OH–], from which we can determine the pOH and pH:
pOH = –log[OH–]
pH = 14 – pOH
Kb 
Solve
[OH–]
[C5H5NH+]
Initial
0
0
Change
+x
+x
Equilibrium
x
x
2
2
x
x
1.7  10 –9 

0.125  x 0.125
x  1.46  10 –5
Checking the simplifying assumption shows that our simplifying assumption is valid:
1.46  10 –5
 100  0.01%
0.125
From [OH–] we can calculate the pH:
pOH = –log 1.46  10–5 = 4.84
pH = 14 – 4.84 = 9.16
[C5H5N]
0.125
–x
0.125 – x
Think about It
This solution of pyridine is more basic than a solution of baking soda.
17.57. Collect and Organize
We are to explain why for H3PO4 Ka1  Ka2  Ka3 .
Equilibrium in the Aqueous Phase | 303
Analyze
The equations describing these acid dissociation constants are as follows:
H3PO4(aq)
H2PO4–(aq) + H+(aq)
K a1
–
H2PO4 (aq)
HPO42–(aq) + H+(aq)
Ka2
HPO42–(aq)
PO43–(aq) + H+(aq)
K a3
Solve
With each successive ionization, it becomes more difficult to remove H + from a species that is negatively
charged. Therefore it is harder to remove H + from HPO42– than from H2PO4– than from H3PO4. This is
reflected in decreasing Ka values as H3PO4 is ionized.
Think about It
From Appendix 5 we can compare the Ka values for phosphoric acid: K a1 = 7.11  10–3, K a 2 = 6.32  10–8,
K a3 = 4.5  10–13. These span ten orders of magnitude.
17.58. Collect and Organize
We are to explain why we can ignore the second ionization of H2SO3 but not that in H2SO4 in calculating the
pH of 1.0 M aqueous solutions of these acids.
Analyze
The equations describing the ionization reactions of these two acids, with Ka values from Appendix 5, are as
follows:
H2SO3(aq)
HSO3–(aq) + H+(aq)
K a1 = 1.7  10–2
HSO3–(aq)
SO32–(aq) + H+(aq)
K a 2 = 6.2  10–8
H2SO4(aq)
HSO4–(aq)
HSO4–(aq) + H+(aq)
SO42–(aq) + H+(aq)
K a1 >> 1
K a 2 = 1.2  10–2
Solve
Because there is a large difference (six orders of magnitude) in K a 2 and K a1 for H2SO3, the ionization of
HSO3– occurs to only a very small extent compared to the ionization of H 2SO3, so the second ionization can
be ignored. For H2SO4, K a 2 is sufficiently large (10–2) that it cannot be ignored.
Think about It
For most calculations involving polyprotic acids the second ionization can be ignored.
17.59. Collect and Organize
We have to use the K a 2 of H2SO4 to calculate the pH of a solution of 0.300 M H2SO4.
Analyze
The first H+ is completely removed from the H2SO4 and the initial concentrations of the species in solution
before the second ionization are [H+] = 0.300 M, [H2SO4] = 0.0 M, and [HSO4–] = 0.300 M. The equation
describing the second ionization is
HSO4–(aq)
SO42–(aq) + H+(aq)
Ka 
[H + ][SO 4 2– ]
 1.2  10 –2
[HSO 4 – ]
Solve
Initial
Change
Equilibrium
[HSO4–]
0.300
–x
0.300 – x
[H+]
0.300
+x
0.300 + x
[SO42–]
0
+x
x
304 | Chapter 17
Plugging equilibrium concentrations into the Ka expression gives
x(0.300  x)
1.2  10–2 
0.300  x
Solving this by the quadratic equation gives
x2 + 0.312x – 0.0036 = 0
x = 0.0111
[H+] = 0.300 + 0.0111 = 0.311
pH = –log 0.311 = 0.51
Think about It
If we did not take into consideration the second ionization of H 2SO4 we would have underestimated the
acidity of the solution by 0.016 pH units.
17.60. Collect and Organize
Given a 0.150 M H2SO3 solution and K a1 = 1.7  10–2 and K a 2 = 6.2  10–8 for this weak acid, we are to
calculate the pH.
Analyze
Because K a 2 is so much smaller than K a1 for this acid, we can say that the second ionization of H2SO3
contributes little to the [H+] in the solution. Therefore, we can solve this by examining only the first
ionization.
Solve
Initial
Change
Equilibrium
[H2SO3]
0.150
–x
0.150 – x
[HSO3–]
0
+x
x
[H+]
0
+x
x
Substituting into the K a1 expression gives
1.7  10 –2 
x2
0.150  x
Solving this by the quadratic equation gives
x = 0.0427
pH = –log 0.0427 = 1.37
Think about It
A simplifying assumption would not be valid in this problem because the initial concentration of H 2SO3 is
close to the value of Ka1 .
17.61. Collect and Organize
Given a 0.250 M solution of ascorbic acid and the K a1 and K a 2 for this weak diprotic acid (1.0  10–5 and
5  10–12, respectively), we are to calculate the pH.
Analyze
Because the K a 2 is so much smaller than the K a1 for ascorbic acid, we can say that the second ionization
contributes little to the [H+] in the solution. Therefore, we can solve this by examining only the first ionization.
Solve
Initial
Change
Equilibrium
[Ascorbic acid]
0.250
–x
0.250 – x
[Ascorbate–]
0
+x
x
[H+]
0
+x
x
Equilibrium in the Aqueous Phase | 305
x2
x2

0.250  x 0.250
x  1.58  10 –3
Checking the simplifying assumption shows that it is valid:
1.58  10 –3
 100  0.63%
0.250
+
The pH is found from the [H ]:
pH = –log 1.58  10–3 = 2.80
1.0  10 –5 
Think about It
This solution is about as acidic as vinegar.
17.62. Collect and Organize
Given a 0.0288 M solution of oxalic acid (HO2CCO2H) and the pKa1 and pK a 2 (from Appendix 5) for this weak
diprotic acid (1.23 and 4.19, respectively), we are to calculate the pH.
Analyze
The K a1 and K a 2 are calculated from the pKa1 and pKa2 :
K a1 = 1  10–1.23 = 5.89  10–2
K a 2 = 1  10–4.19 = 6.46  10–5
Because the K a 2 is so much smaller than the K a1 for oxalic acid, we can say that the second ionization
contributes little to the [H+] in the solution. Therefore, we can solve this by examining only the first ionization.
Solve
[HO2CCO2–]
Initial
0
Change
+x
Equilibrium
x
2
x
5.89  10 –2 
0.0288  x
Solving this by the quadratic equation gives
x = 0.0212
pH = –log 0.0212 = 1.67
[HO2CCO2H]
0.0288
–x
0.0288 – x
[H+]
0
+x
x
Think about It
For a weak acid, oxalic acid with a K a1 of 5.89  10–2 is fairly strong. Uses for this acid include bleaches,
rustproofing chemicals, wood restorers, and rust removers.
17.63. Collect and Organize
Given a 0.00100 M solution of nicotine and the K b1 and K b2 (from Appendix 5) for this weak dibasic
compound, we are to calculate the pH.
Analyze
The K b2 (1.3  10–11) is so much smaller than the K b1 (1.0  10–6) for nicotine that we may ignore the
contribution of the second ionization of nicotine to the [OH –] in the solution. We can therefore solve this
problem by examining only the first ionization.
306 | Chapter 17
Solve
[NicotineH+]
Initial
0
Change
+x
Equilibrium
x
2
x
x2
1.0  10 – 6 

0.00100  x 0.00100
x  3.16  10 –5
Checking the simplifying assumption shows that it is valid:
3.16  10 –5
 100  3.2%
0.00100
[Nicotine]
0.00100
–x
0.00100 – x
[OH–]
0
+x
x
We can calculate the pOH and pH from the [OH–]:
pOH = –log 3.16  10–5 = 4.500
pH = 14 – 4.500 = 9.500
Think about It
The relatively high pH of this dilute solution of nicotine shows that this compound is a fairly strong weak
base.
17.64. Collect and Organize
Given a 2.50  10–4 M solution of H2NCH2CH2NH2 and the pKb2 and pK b1 for this weak dibasic compound
(3.29 and 6.44, respectively), we are to calculate the pH.
Analyze
The K b1 and K b2 values are calculated from the pK b1 and pKb2 :
K b1 = 1  10–3.29 = 5.13  10–4
K b2 = 1  10–6.44 = 3.63  10–7
Because K b2 is so much smaller than K b1 for H2NCH2CH2NH2, we can say that the second ionization
contributes little to the [OH–] in the solution. Therefore, we can solve this by examining only the first
ionization.
Solve
Initial
Change
Equilibrium
[H2NCH2CH2NH2]
[H2NCH2CH2NH3+]
[OH–]
2.50  10– 4
–x
2.50  10– 4 – x
0
+x
x
0
+x
x
5.13 10– 4 
x2
2.50 10 4  x
Solving this by the quadratic equation gives
x = 1.84  10– 4
pOH = –log 1.84  10– 4 = 3.735
pH = 14 – 3.735 = 10.265
Think about It
We must solve this problem using the quadratic equation because K b1 is about the same magnitude as the
initial concentration of the weak base. In these cases, it is often true that the usual simplifying assumption is
not valid.
Equilibrium in the Aqueous Phase | 307
17.65. Collect and Organize
Given a 0.01050 M solution of quinine and the K b1 and K b2 for this weak dibasic compound, we are to
calculate the pH.
Analyze
The K b2 (1.4  10–9) is so much smaller than K b1 (3.3  10–6) for quinine that we may ignore the contribution
of the second ionization of quinine to the [OH–] in the solution. We can therefore solve this problem by
examining only the first ionization.
Solve
Initial
Change
Equilibrium
[Quinine]
0.01050
–x
0.01050 – x
[QuinineH+]
0
+x
x
[OH–]
0
+x
x
x2
x2

0.01050  x 0.01050
x  1.86  10 – 4
Checking the simplifying assumption shows that it is valid:
1.86  10 – 4
 100  1.8%
0.01050
The pOH and pH can be calculated from the [OH–]:
pOH = –log 1.86  10– 4 = 3.730
pH = 14 – 3.730 = 10.270
3.3  10 – 6 
Think about It
Quinine has a very complicated molecular structure that includes aromatic rings, an alcohol, an amine, an
alkene, and an ether as functional groups.
17.66. Collect and Organize
Given a 0.0133 M solution of piperazine and the K b1 and K b2 for this weak dibasic compound, we are asked to
calculate the pH and then draw the structure of piperazine that would be present in 0.15 M HCl (stomach
acid).
Analyze
(a) The K b2 (2.15  10–9) is so much smaller than K b1 (5.38  10–5) for piperazine that we may ignore the
contribution of the second ionization of piperazine to the [OH –] in the solution. We can therefore solve this
problem by examining only the first ionization.
(b) Stomach acid is quite strong, so we would expect both nitrogen atoms to be protonated.
Solve
Initial
Change
Equilibrium
[Piperazine]
0.0133
–x
0.0133 – x
[PiperazineH+]
0
+x
x
[OH–]
0
+x
x
308 | Chapter 17
x2
x2

0.0133  x 0.0133
x  8.46  10 – 4
Checking shows that our simplifying assumption is not valid:
8.46 10– 4
100  6.4%
0.0133
so we must solve this by the quadratic equation:
x2 + 5.38  10–5x – 7.155  10–7 = 0
x = 8.19  10– 4
pOH = –log 8.19  10– 4 = 3.087
pH = 14 – 3.087 = 10.913
(b)
2+
5.38  10 –5 
H
H
N
N
H
H
Think about It
Piperazine is one of the stronger of the weak nitrogen-containing bases and a little more basic than ammonia
(Kb = 1.76  10–5).
17.67. Collect and Organize
We are asked to explain why H2SO4 is a stronger acid (greater K a1 ) than H2SeO4.
Analyze
The only difference in these acids is the central atom. Sulfur and selenium belong to the same group in the
periodic table. These elements differ in size and electronegativity.
Solve
Sulfur is more electronegative than selenium. This higher electronegativity on the sulfur atom stabilizes the
anion HSO4– more than the anion HSeO4–. Therefore, H2SO4 is a stronger acid.
Think about It
We would expect this trend to continue, so we predict that H 2TeO4 is a weaker acid than H2SeO4.
17.68. Collect and Organize
We are to explain why H2SO4 is a stronger acid (larger K a1 ) than H2SO3.
Analyze
The only difference between these acids is the extra oxygen atom bound to sulfur in H 2SO4.
Solve
The presence of the additional electron-withdrawing oxygen atom (because of its high electronegativity) on
H2SO4 compared to H2SO3 delocalizes the negative charge on HSO4– better than on HSO3–. This stabilization
of the anion makes H2SO4 a stronger acid.
Think about It
Even the second proton on H2SO4 is more acidic (Ka 2 = 1.2  10–2) than that of H2SO3 (Ka 2 = 6.2  10–8).
17.69. Collect and Organize
We are to predict which acid of a pair is stronger.
Equilibrium in the Aqueous Phase | 309
Analyze
The more oxygen atoms bound to the central atom and the more electronegative the central atom (X) in the
acid, the more acidic is the compound.
Solve
(a) H2SO3 is a stronger acid than H2SeO3.
(b) H2SeO4 is a stronger acid than H2SeO3.
Think about It
The presence of oxygen atoms bound to the central atom in an oxyacid can have a dramatic effect on acidity.
17.70. Collect and Organize
We are to predict which acid of a pair is stronger.
Analyze
The more oxygen atoms bound to the central atom and the more electronegative the central atom in the HOX
acid, the more acidic is the compound.
Solve
(a) HOBrO is more acidic than HOBr.
(b) HOCl is more acidic than HOBr.
Think about It
The presence of oxygen atoms bound to the central atom in an oxyacid can have a dramatic effect on acidity.
The range for Ka values for HClO to HClO4 is 2.9  10–8 to greater than 1, as shown in Figure 17.10.
17.71. Collect and Organize
Given the pKa values of conjugate acids of three pyridine derivatives where methyl groups are added, we are
to determine if more methyl groups increase or decrease the basicity of pyridine.
Analyze
The pKa is a measure of the conjugate acid’s acidity. From the equation
pKa = –log Ka
we see that the larger the pKa, the weaker is the acid. The weaker the acid, the stronger is the conjugate base.
Solve
As methyl groups are added the pKa increases so the acidity decreases. If the acidity decreases, the basicity of
the conjugate base increases. Therefore, more methyl groups on the parent pyridine increases the base
strength.
Think about It
Our prediction is true. The Kb values of the pyridine bases show that more methyl groups lead to increased
basicity.
N
Kb = 1.51  10–9
N
Kb = 9.77  10–8
N
Kb = 2.69  10–7
17.72. Collect and Organize
We are to explain why we do not need a table of Kb values if we already have the base’s conjugate acid Ka
values.
310 | Chapter 17
Analyze
The acid and conjugate base are related to each other. Their respective Ka and Kb values are related by Kw:
Kw = Ka  Kb
and their pKa and pKb values are related through pKw, which is 14 at 25˚C:
14 = pKa + pKb
Solve
If we have the Ka value of the conjugate acid we can easily compute the Kb value of the base:
Kw
Kb 
Ka
Think about It
Whether a species is listed in a Kb table or a Ka table usually depends on the neutral species’ dominant acid–
base character. For example, H2SO3 acts as an acid so it is listed in a Ka table, but NH3 acts as a base and so it
is found in a Kb table.
17.73. Collect and Organize
Of the three salts given, we are to determine which, when dissolved in water, gives an acidic solution.
Analyze
To give an acidic solution, the cation of the salt must donate a proton to water without the anion reacting with
water, or if the anion hydrolyzes, then the pKa of the cation must be lower than the pKb of the anion.
Solve
Both NH4+ and CH3COO– of ammonium acetate hydrolyze:
NH4+(aq) + H2O ( )
NH3(aq) + H3O+(aq)
–
pKa = 9.25
–
CH3COO (aq) + H2O ( )
CH3COOH(aq) + OH (aq)
pKb = 9.25
Because pKa = pKb, this salt’s solution is neutral.
Only NH4+ of ammonium nitrate hydrolyzes. This gives an acidic solution:
NH4+(aq) + H2O ( )
NH3(aq) + H3O+(aq)
–
Only HCOO of sodium formate hydrolyzes. This gives a basic solution:
HCOO–(aq) + H2O ( )
HCOOH(aq) + OH–(aq)
Therefore, of the three salts, only ammonium nitrate dissolves to give an acidic solution.
Think about It
Remember that neither Na+ nor NO3– hydrolyze because they would form either a strong base or a strong acid
which are always 100% ionized.
 NaOH(aq) + H+(aq)
Na+(aq) + H2O ( ) 
 HNO3(aq) + OH–(aq)
NO3–(aq) + H2O ( ) 
17.74. Collect and Organize
Of the three salts NaF, KCl, and NH4Cl we are to determine which would give a basic solution when
dissolved in water.
Analyze
All these salts are soluble so we need to examine how and whether the cation and anion of each salt react with
water (hydrolyze). If the hydrolysis forms a strong acid or base, then that cation or anion will not react with
water to change the pH of the solution.
Equilibrium in the Aqueous Phase | 311
Solve
For NaF
Na+(aq) + H2O ( )
NaOH(aq) + H+(aq)
F–(aq) + H2O ( )
HF(aq) + OH–(aq)
For KCl
K+(aq) + H2O ( )
KOH(aq) + H+(aq)
Cl–(aq) + H2O ( )
HCl(aq) + OH–(aq)
For NH4Cl
NH4+(aq) + H2O ( )
NH3(aq) + H+(aq)
–
Cl (aq) + H2O ( )
HCl(aq) + OH–(aq)
Therefore, only NaF produces a basic solution.
NaOH is a strong base so Na+ does not change the pH.
HF is a weak acid so F– does make this solution basic.
KOH is a strong base so K+ does not change the pH.
HCl is a strong acid so Cl– does not change the pH.
NH3 is a weak base so NH4+ does make this solution acidic.
HCl is a strong acid so Cl– does not change the pH.
Think about It
When both cation and anion hydrolyze, the relative values of the Ka and Kb for the hydrolysis reactions
determine whether the solution is acidic, basic, or neutral.
17.75. Collect and Organize
We consider why lemon juice is used to reduce the fishy odor due to the presence of (CH 3)3N in not-so-fresh
seafood.
Analyze
Trimethylamine is a weak base, and lemon juice contains citric acid that is a weak acid.
Solve
The citric acid in the lemon juice neutralizes the volatile trimethylamine to make a nonvolatile dissolved salt
that neutralizes the fishy odor:
HOC(CH2)2(COOH)3(aq) + (CH3)3N(aq)
HOC(CH2)2(COOH)2COO–(aq) + (CH3)3NH+(aq)
Think about It
Because the pKb of trimethylamine of 4.19 and the pKa of citric acid of 3.13 are lower than the pKa of
trimethylammonium (9.81) and the pKb of citrate (10.87), this equilibrium lies to the right favoring the
products.
17.76. Collect and Organize
We consider the acid–base properties of the calcium salt of malonic acid, Ca(OOCCH 2COO).
Analyze
The conjugate base of malonic acid, –OOCCH2COO–, reacts with water, but Ca2+ does not.
Solve
The hydrolysis of the anion of calcium malonate gives a basic solution as described by the equation
–
OOCCH2COO–(aq) + 2 H2O ( )
HOOCCH2COOH(aq) + 2 OH–(aq)
The pH of the beets is raised because of the presence of the calcium salt of malonic acid because of this
hydrolysis reaction.
Think about It
Ca2+ does not hydrolyze because if it did the strong base Ca(OH) 2 would be formed.
17.77. Collect and Organize
Given Ka = 2.1  10–11 for the conjugate acid of saccharin, we are asked to calculate the value of pKb for
saccharin.
312 | Chapter 17
Analyze
From the Ka we can calculate the pKa:
pKa = –log Ka
From the pKa we can calculate the pKb because pKw = 14.00 at 25˚C
pKb = 14.00 – pKa
Solve
pKa = –log(2.1  10–11) = 10.68
pKb = 14.00 – 10.68 = 3.32
Think about It
Alternatively, we could calculate the Kb from Ka using
Kb 
Kw
Ka
and then calculate pKb using
pKb = –log Kb
17.78. Collect and Organize
Given the K a1 and K a 2 values for H2CrO4 (0.16 and 3.2  10–7, respectively), we are to calculate the value of
K b1 and K b2 for CrO42–.
Analyze
The relevant equilibrium equations are
CrO42–(aq) + H2O ( )
HCrO4–(aq) + H2O ( )
HCrO4–(aq) + OH–(aq)
H2CrO4(aq) + OH–(aq)
Solve
The Kb for the first reaction is
K b1 
Kw
Ka 2

1  10–14
 3.1  10–8
3.2  10–7

1  10–14
 6.3  10–14
0.16
The Kb for the second reaction is
K b2 
Kw
Ka1
Think about It
The first equilibrium contributes the most to the basicity of a solution of CrO 42– since its Kb is much larger
than that of the second equilibrium equation.
17.79. Collect and Organize
From Appendix 5 we know that the Ka of HF is 6.8  10– 4. Using this we are to calculate the pH of a solution
that is 0.00339 M in NaF.
Analyze
When NaF dissolves in water the F– ion hydrolyzes to give a basic solution:
F–(aq) + H2O ( )
HF(aq) + OH–(aq)
The Kb for this reaction is
Kw
110–14
Kb 

 1.47 10–11
Ka 6.8 10– 4
We can solve for [OH–] using an ICE table and then compute the pH.
Equilibrium in the Aqueous Phase | 313
Solve
[F–]
0.00339
–x
0.00339 – x
[HF]
Initial
0
Change
+x
Equilibrium
x
2
x
x2
1.47  10 –11 

0.00339  x 0.00339
x  2.23  10 –7
Checking the simplifying assumption shows that it is valid:
2.23  10 –7
 100  0.0066%
0.00339
The pH is calculated from the [OH–]:
pOH = –log 2.23  10–7 = 6.65
pH = 14 – 6.65 = 7.35
[OH–]
0
+x
x
Think about It
Because HF is a moderately strong weak acid, F–, its conjugate base, is a fairly weak base.
17.80. Collect and Organize
Given the Kb of ephedrine (3.86), we are to calculate the pH of a 1.25  10–2 M solution of its hydrochloride
salt.
Analyze
When ephedrine  HCl dissolves in water it produces an acidic solution:
EphedrineH+(aq)
ephedrine(aq) + H+(aq)
The Ka for this reaction is
K w 1  10–14
Ka 

 7.24  10–11
Kb 1  10–3.86
We can solve for [H+] using an ICE table.
Solve
[EphedrineH+]
0.0125
–x
0.0125 – x
[Ephedrine]
Initial
0
Change
+x
Equilibrium
x
2
x
x2
7.24  10 –11 

0.0125  x 0.0125
x  9.51  10 –7
Checking the simplifying assumption shows that it is valid:
9.57  10 –7
 100  0.007%
0.0125
The pH is found from the [H+]:
pH = –log 9.51  10–7 = 6.02
[H+]
0
+x
x
Think about It
EphedrineH+ is quite a weak acid, and for this solution the percent ionization is only 0.007%.
17.81. Collect and Organize
We are to explain why a solution of CH3COOH with CH3COONa is a better pH buffer than a solution
containing NaCl and HCl.
314 | Chapter 17
Analyze
A buffer is composed of a weak acid and its conjugate base.
Solve
Because the weak acid CH3COOH is combined with its weak conjugate base, CH3COO–, this buffer can
absorb added H+ or OH–. The other mixture, HCl with Cl–, is a strong acid paired with its very, very weak
conjugate base. This pairing cannot absorb added H+ or OH–.
Think about It
It is a key idea that for the acid–base pair in a buffer system both conjugates be weak.
17.82. Collect and Organize
We consider why a solution composed of a weak base with its conjugate acid is a better buffer than a solution
of only the weak base.
Analyze
A buffer acts to absorb both OH– and H+ added to the solution.
Solve
The presence of both the base and the conjugate acid in a buffer means that the system can absorb both added
H+ and OH–. Because the concentration of the conjugate acid (present only because of a small degree of
hydrolysis of the base) is too small, a buffer composed only of a weak base cannot absorb much OH–.
Think about It
A good buffer has relatively high and roughly equal concentrations of the weak base and its conjugate acid.
17.83. Collect and Organize
For a buffer that is 0.244 M in acetic acid and 0.122 M in sodium acetate, we can use the Henderson–
Hasselbalch equation to calculate the pH of this buffer at 25˚C (Ka = 1.76  10–5) and at 0˚C (Ka = 1.64  10–5).
Analyze
The Henderson–Hasselbalch equation is
[base]
[acid]
For this problem, [base] = [acetate] = 0.122 M and [acid] = [acetic acid] = 0.244 M. Because the Ka values are
different for the two temperatures, the pH of these solutions will differ.
pH  pKa  log
Solve
0.122
 4.453
0.244
0.122
 4.484
At 0˚C, pH  –log(1.64  10–5 )  log
0.244
At 25˚C, pH  –log(1.76  10–5 )  log
Think about It
At the lower temperature the pH of this buffer is less acidic than at the higher temperature.
17.84. Collect and Organize
For a buffer that is 0.100 M in pyridine and 0.275 M in pyridinium chloride, we can use the Henderson–
Hasselbalch equation to calculate the pH of this buffer (pKb for pyridine = 8.8).
Analyze
We’re given the pKb for pyridine and from this we can calculate the pKa for use in the Henderson–Hasselbalch
equation at 25˚C.
pKa = 14 – pKb = 14 – 8.8 = 5.2
Equilibrium in the Aqueous Phase | 315
The Henderson–Hasselbalch equation is
[base]
[acid]
For this problem, [base] = [pyridine] = 0.100 M and [acid] = [pyridinium] = 0.275 M.
pH  pKa  log
Solve
pH  5.2  log
0.100
 4.8
0.275
Think about It
If we were to increase the concentration of base relative to the acid, the pH of the solution would increase
because the value of log([base]/[acid]) in the Henderson–Hasselbalch equation becomes more positive.
17.85. Collect and Organize
For a buffer that is 0.225 M in both HPO42– and PO43– (with the Ka of HPO42– equal to 4.5  10–13), we can use
the Henderson–Hasselbalch equation to calculate the pH of the buffer. From the pH we can also calculate the
pOH.
Analyze
The Henderson–Hasselbalch equation is
pH  pKa  log
[base]
[acid]
For this problem, [base] = [acid] = 0.225 M.
Solve
0.225M
 12.34
0.225M
pOH  14 – pH  14 –12.34  1.65
pH  –log(4.5  10 –13 )  log
Think about It
Notice that when the concentrations of the acid and its conjugate base are equal, pH = pKa because log 1 = 0.
17.86. Collect and Organize
For a buffer that is 0.0200 M in boric acid and 0.0250 M in sodium borate (with the pKa of boric acid equal to
9.27), we can use the Henderson–Hasselbalch equation to calculate the pH of the buffer. From the pH we can
also calculate the pOH.
Analyze
The Henderson–Hasselbalch equation is
pH  pKa  log
[base]
[acid]
For this problem, [base] = 0.0250 M and [acid] = 0.0200 M.
Solve
0.0250
 9.37
0.0200
pOH  14 – pH  14 – 9.37  4.63
pH  9.27  log
Think about It
Notice here that because the ratio [base]/[acid] is greater than 1, the calculated pH is greater than the pKa of
the acid.
316 | Chapter 17
17.87. Collect and Organize
Given the pH of an acetic acid–acetate buffer solution (pH = 3.56) where the Ka of acetic acid is 1.76  10–5,
we can use the Henderson–Hasselbalch equation to calculate the ratio of acetate ion to acetic acid in the
solution.
Analyze
Rearranging the Henderson–Hasselbalch equation to solve for the ratio gives
[acetate]
pH  pK a  log
[acetic acid]
[acetate]
log
 pH – pK a
[acetic acid]
[acetate]
( pH – pK a )
 110
[acetic acid]
Solve
The pKa = –log(1.76  10–5) = 4.754, so pH – pKa = 3.56 – 4.754 = –1.194 and the ratio of acetate to acetic
acid is
[acetate]
 1  10–1.194  0.064
[acetic acid]
Think about It
Because [acetic acid] > [acetate], the pH of this buffer is less than the pKa of acetic acid.
17.88. Collect and Organize
Given the pH of a lactic acid–lactate buffer solution (pH = 4.00) where the Ka of lactic acid is 1.4  10– 4, we
can use the Henderson–Hasselbalch equation to calculate the ratio of lactic acid to lactate in the solution.
Analyze
Rearranging the Henderson–Hasselbalch equation to solve for the ratio gives
[lactate]
pH  pK a  log
[lactic acid]
[lactate]
log
 pH – pK a
[lactic acid]
[lactic acid]
1

( pH – pK a )
[lactate]
1  10
Solve
The pKa = –log(1.4  10– 4) = 3.85, so pH – pKa = 4.00 –3.85 = 0.15 and the ratio of lactic acid to lactate is
[lactic acid]
1

 0.71
[lactate]
1  100.15
Think about It
Because [lactate] > [lactic acid] the pH of this buffer is higher than the pKa of lactic acid.
17.89. Collect and Organize
We can use the Henderson–Hasselbalch equation to determine the pH of a solution prepared by mixing a
volume of 0.05 M NH3 with an equal volume of 0.025 M HCl.
Equilibrium in the Aqueous Phase | 317
Analyze
Mixing equal volumes of solutions dilutes them both. Therefore, after mixing and before reaction, [NH 3] =
0.025 M and [HCl] = 0.0125 M in the combined solution. When HCl reacts with NH3, NH4+ is produced
according to the equation
NH3(aq) + H+(aq)  NH4+(aq)
Stoichiometrically, this would give [NH3] = 0.0125 M and [NH4+] = 0.0125 M after complete reaction with
H+. Because this is a solution of an acid (NH4+) and its conjugate base (NH3), we can use the Henderson–
Hasselbalch equation to calculate the pH of the solution. To do so we need the pKa of NH4+ (9.25) from
Appendix 5 (Table A5.1).
Solve
pH  pK a  log
[NH3 ]
0.0125
 9.25  log
 9.25

0.0125
[NH 4 ]
Think about It
Because the HCl added to the NH3 in this solution converts exactly half of the NH3 to NH4+, the ratio of the
concentrations equals 1 and the pH of the solution equals the pKa of NH4+.
17.90. Collect and Organize
We can use the Henderson–Hasselbalch equation to determine the pH of a solution prepared by mixing a
volume of 0.05 M acetic acid with an equal volume of 0.025 M NaOH.
Analyze
Mixing equal volumes of solutions dilutes them both. Therefore, after mixing and before reaction,
[CH3COOH] = 0.025 M and [NaOH] = 0.0125 M in the combined solution. When NaOH reacts with
CH3COOH, CH3COO– is produced according to the equation
CH3COOH(aq) + OH–(aq)  CH3COO–(aq) + H2O ( )
Stoichiometrically, this would give [CH3COOH] = 0.0125 M and [CH3COO–] = 0.0125 M after complete
reaction with OH–. Because this is a solution of an acid (CH3COOH) and its conjugate base (CH3COO–), we
can use the Henderson–Hasselbalch equation to calculate the pH of the solution. To do so we need the pKa of
CH3COOH from Appendix 5 (4.75).
Solve
pH  pK a  log
[CH 3 COO – ]
0.0125
 4.75  log
 4.75
[CH 3 COOH]
0.0125
Think about It
Because the NaOH added to the CH3COOH in this solution converts exactly half of the CH3COOH to
CH3COO–, the ratio of the concentrations equals 1 and the pH of the solution equals the pKa of CH3COOH.
17.91. Collect and Organize
We are to calculate how much of the strong acid HNO3 (10 M) we would add to a buffer solution that is
0.010 M acetic acid and 0.10 M sodium acetate to give a solution that is pH 5.00.
Analyze
We can use the Henderson–Hasselbalch equation to first calculate the ratio [acetate]/[acetic acid] needed to
give a solution of pH 5.00. We then use that ratio to determine how much HNO 3 to add in order to convert the
acetate in solution to acetic acid so as to give that ratio. The pKa of acetic acid is 4.75.
318 | Chapter 17
Solve
[acetate]
[acetic acid]
[acetate]
5.00  4.75  log
[acetic acid]
[acetate]
 1.78
[acetic acid]
When we add HNO3 to the solution, acetate is converted to acetic acid and the amounts of each are
[acetate] = 0.10 – x
[acetic acid] = 0.10 + x
The ratio of these is 1.78. Solving for x gives
[acetate]
0.10  x
 1.78 
[acetic acid]
0.10  x
0.10  x  0.0178  1.78 x
pH  pK a  log
x  0.0296
The value of x is the moles of HNO3 that we need to add to the solution. In mL of 10 M HNO3, then
1000 mL
0.0296 mol 
 3.0 mL for 1.00 liter of the buffer solution
10 mol
Think about It
The small volume addition of HNO3 does not appreciably change the concentration of the acid and the base,
so we do not need to account for it.
17.92. Collect and Organize
For this problem we are to calculate how much of the strong base NaOH (6.0 M) we would add to 0.500 L of
a buffer solution that is 0.0200 M acetic acid and 0.0250 M sodium acetate to give a solution that is pH 5.75.
Analyze
We can use the Henderson–Hasselbalch equation to first calculate the ratio of [acetate]/[acetic acid] needed to
give a solution of pH 5.75. We then use that ratio to determine how much NaOH to add in order to convert the
acetate in solution to acetic acid so as to give that ratio. The pKa of acetic acid is 4.75.
Solve
[acetate]
[acetic acid]
[acetate]
5.75  4.75  log
[acetic acid]
[acetate]
 10
[acetic acid]
When we add NaOH to the solution, acetic acid is converted to acetate and the amounts of each are
[acetate] = 0.0250 + x
[acetic acid] = 0.0200 – x
The ratio of these will be 10. Solving for x gives
[acetate]
0.0250  x
 10 
[acetic acid]
0.0200 – x
0.0250  x  0.200 –10 x
x  0.0159
pH  pKa  log
Equilibrium in the Aqueous Phase | 319
The value of x is the moles of NaOH that we need to add to one liter of the buffer solution. In milliliters of
6.0 M NaOH then
1000 mL
0.0159 mol 
 2.7 mL for 1 liter of the buffer solution
6.0 mol
For 500 mL of the buffer solution, we need only one-half of this amount, or 1.3 mL of 6 M NaOH solution.
Think about It
The small volume addition of NaOH does not appreciably change the concentration of the acid and the base,
so we do not need to account for it.
17.93. Collect and Organize
We are to compare the pH of 1.00 L of a buffer that is 0.120 M in HNO2 and 0.150 M in NaNO2 before and
after 1.00 mL of 12.0 M HCl is added.
Analyze
We can use the Henderson–Hasselbalch equation in both cases. After the addition of HCl, however, the
amounts (calculated in moles) of HNO2 and NO2– have to be adjusted before using the Henderson–
Hasselbalch equation. The pKa of HNO2 is 3.40.
Solve
Without added HCl the pH of the buffer solution is
0.150
 3.50
0.120
Because we have 1.00 L of the buffer solution, we originally have 0.120 mol HNO2 and 0.150 mol NO2– in
the solution. Adding 1.00 mL of 12.0 M HCl adds
12.0 mol
1.00 mL 
 0.0120 mol H+
1000 mL
This will increase the moles of HNO2 to 0.120 mol + 0.0120 mol = 0.132 mol and decrease the moles of NO 2–
to 0.150 mol – 0.0120 mol = 0.138 mol. Because the volume of the solution is 1.00 L, the concentrations of
these species are [HNO2] = 0.132 M and [NO2–] = 0.138 M. Using the Henderson–Hasselbalch equation to
calculate the pH gives
0.138
pH  3.40  log
 3.42
0.132
pH  3.40  log
Think about It
The pH of the buffer changed very little. The change in pH of 1.00 L of water after adding 0.0120 mol of H +
would be from 7.00 to 1.92.
17.94. Collect and Organize
In this question we are to compare the pH of 100.0 mL of a buffer that is 0.100 M in NH4Cl and 0.100 M in
NH3 before and after the addition of 1.0 mL of 6 M HNO3.
Analyze
We can use the Henderson–Hasselbalch equation in both cases. After the addition of HNO3, however, the
amounts (calculated in moles) of NH4+ and NH3 have to be adjusted before using the Henderson–Hasselbalch
equation. The pKa of NH4+ is 9.25.
Solve
Without added HNO3 the pH of the buffer solution is
pH  9.25  log
0.100
 9.25
0.100
320 | Chapter 17
Because we have 100.0 mL of the buffer solution, we originally have 0.0100 mol NH4+ and 0.0100 mol NH3
in the solution. Adding 1.0 mL of 6 M HNO3 adds
6.0 mol
1.00 mL 
 0.0060 mol H +
1000 mL
This will increase the amount of NH4+ to 0.0100 mol + 0.0060 mol = 0.0160 mol and decrease the amount of
NH3 to 0.0100 mol – 0.0060 mol = 0.0040 mol. Because the volume of the solution is 0.1000 L, the
concentrations of these species are [NH4+] = 0.160 M and [NH3] = 0.040 M. Using the Henderson–
Hasselbalch equation to calculate the pH gives
0.040
pH  9.25  log
 8.65
0.160
Think about It
Our answer shows that adding acid to the buffer lowers the pH (slightly), as we would expect.
17.95. Collect and Organize / Analyze
We compare the terms molar solubility and solubility product.
Solve
Molar solubility is the mole of a substance that dissolves in a solvent. The solubility product is the equilibrium
constant for the dissolution of a substance.
Think about It
Solubility has units of grams or moles per volume of solution, but, like other equilibrium constants, the
solubility product is unitless.
17.96. Collect and Organize
Using the definition of the common-ion effect we are to give an example of how addition of a common ion to
a solution saturated in a slighty soluble salt would limit that salt’s solubility.
Analyze
The common-ion effect says that the addition of an ion common to that of the sparingly soluble salt shifts the
equilibrium according to Le Châtelier’s principle to the left. This then decreases the amount of that salt that
dissolves.
Solve
For the sparingly soluble salt BaSO4 if we increase the concentration of sulfate anions in the solution by
adding soluble sodium sulfate (Na2SO4), some of the Ba2+ in solution will precipitate to form BaSO4.
Think about It
The Ksp expression for a sparingly soluble salt with singly-charged ions may be expressed as
MX(s)
M (aq)  X– (aq)
Ksp  [M ][X– ]
If either X– or M+ is added to the solution then Q > Ksp, so the equilibrium shifts to the left and less of the salt
dissolves.
17.97. Collect and Organize
By comparing the Ksp values of MgCO3, CaCO3, and SrCO3 we can identify which cation (Mg2+, Ca2+, or
Sr2+) precipitates first as carbonate mineral.
Analyze
From Appendix 5, the Ksp values are
MgCO3
Ksp  6.8  10–6
CaCO3
Ksp  5.0  10–9
SrCO3
Ksp  5.6  10–10
Equilibrium in the Aqueous Phase | 321
Solve
Because SrCO3 has the lowest Ksp, the cation Sr2+ precipitates first as a carbonate mineral.
Think about It
The order of solubility from least to most soluble for these carbonates is SrCO 3 < CaCO3 < MgCO3.
17.98. Collect and Organize
We are asked if an increase in solubility as the temperature increases means that the Ksp value increases or
decreases.
Analyze
As a substance becomes more soluble, the Ksp value increases.
Solve
Ksp increases for substances that have increased solubility as temperature increases.
Think about It
It is not always the case that a substance is more soluble at higher temperatures. For some substances
solubility decreases as the temperature is raised!
17.99. Collect and Organize
For the case of SrSO4 whose Ksp increases as the temperature increases, we are to determine whether the
dissolution is exothermic or endothermic.
Analyze
We can include heat as a reactant (for an endothermic reaction) or as a product (for an exothermic reaction)
and apply Le Châtelier’s principle:
SrSO4(s) + heat
Sr2+(aq) + SO42–(aq)
2+
SrSO4(s)
Sr (aq) + SO42–(aq) + heat
Solve
Applying Le Châtelier’s principle, we see that the reaction shifts to the right and more SrSO 4 dissolves as the
temperature is increased. The dissolution is endothermic.
Think about It
The opposite effect of temperature occurs for an exothermic dissolution: less solid dissolves as the
temperature is increased.
17.100. Collect and Organize
We are asked how the addition of concentrated NaOH affects the solubility of an Al 3+ salt.
Analyze
Let’s use AlPO4 as an example. The solubility equilibrium is
AlPO4(s)
Al3+(aq) + PO43–(aq)
The reaction of Al3+ with excess OH– has the overall reaction
Al3+(aq) + 4 OH–(aq)
Al(OH)4–(aq)
Solve
The solubility of an Al3+ salt increases in a concentrated solution of OH– because the formation of soluble
Al(OH)4– shifts the solubility equilibrium to the right.
Think about It
Likewise, because of the formation of Cr(OH)4–, “insoluble” salts like Cr2(SO4)3 are expected to dissolve in
concentrated OH–.
322 | Chapter 17
17.101. Collect and Organize
By writing the equation for the dissolution of hydroxyapatite, we can explain why acidic substances erode
tooth enamel.
Analyze
The solubility of hydroxyapatite is described by
Ca5(PO4)3OH(s)
OH–(aq) + 3 PO43–(aq) + 5 Ca2+(aq)
Solve
Acidic substances react with the OH – released upon dissolution of hydroxyapatite. The equilibrium is shifted
to the right, dissolving more hydroxyapatite.
Think about It
The equilibrium would be shifted in the opposite direction (to the left) in an alkaline environment.
17.102. Collect and Organize
Using the formula given for the product for the conversion of hydroxyapatite into fluorapatite, we can
explain why fluorapatite is less susceptible to erosion by acids.
Analyze
The dissolution of fluorapatite is described by the equation
Ca5(PO4)3F(s)
5 Ca2+(aq) + 3 PO43–(aq) + F–(aq)
Solve
The F– produced in the dissolution of fluorapatite, unlike the OH – produced in the dissolution of
hydroxyapatite, is a weak base. It does not react completely with H + and does not drive the solubility
equilibrium to the right.
Think about It
Also, the Ksp of fluorapatite is estimated to be about 10 times lower than the Ksp of hydroxyapatite.
17.103. Collect and Organize
Given the [Ba2+] in a saturated solution of BaSO 4 (1.04  10–5 M Ba2+), we are to calculate the value of Ksp
for BaSO4:
BaSO4 ( s)
Ba 2  (aq)  SO4 2  (aq)
Analyze
The Ksp expression is
Ksp = [Ba2+][SO42–]
Because for every mole of BaSO4 that dissolves we get 1 mol of Ba2+ and 1 mol of SO42–, the molarities of
Ba2+ and SO42– are the same for this saturated solution of BaSO4.
Solve
Ksp = (1.04  10–5)(1.04  10–5) = 1.08  10–10
Think about It
Because BaSO4 is quite insoluble, we can add SO42– to a solution of dissolved Ba2+ to quantitatively
precipitate the barium out of solution. After weighing the dried precipitate we can then calculate how much
Ba2+ was present in the original solution.
17.104. Collect and Organize
Given the [F–] in a saturated solution of BaF2 (1.5  10–2 M F–), we can calculate the value of Ksp for BaF2:
BaF2 ( s)
Ba 2  (aq)  2 F (aq)
Equilibrium in the Aqueous Phase | 323
Analyze
The Ksp expression is
Ksp = [Ba2+][F–]2
Because for every mole of BaF2 that dissolves we get 1 mol of Ba2+ and 2 mol of F–, the molarity of Ba2+ in
the saturated solution is half that of F–.
Solve
Ksp = (1.5  10–2/2)(1.5  10–2)2 = 1.7  10–6
Think about It
Don’t forget to square the concentration of F– to calculate the Ksp for BaF2.
17.105. Collect and Organize
Given that Ksp = 1.02  10–6, we are to calculate [Cu+] and [Cl–] for a saturated solution of CuCl.
Analyze
The solubility equation and Ksp expression for CuCl are
CuCl(s)
Cu  (aq)  Cl (aq)
Ksp [Cu + ][Cl – ]
The [Cu+] = [Cl–] in this solution because for every particle of CuCl that dissolves we get one particle of Cu +
and one particle of Cl–.
Solve
Let [Cu+] = [Cl–] = x. The Ksp expression becomes
Ksp = 1.02  10–6 = (x)(x)
x = 1.01  10–3
Therefore, [Cu+] = [Cl–] = 1.01  10–3 M.
Think about It
We do not need to know how much CuCl is originally placed into the solution because it does not appear in
the Ksp expression as a pure solid.
17.106. Collect and Organize
Given that Ksp = 3.2  10–8, we are to calculate [Pb2+] and [F–] for a saturated solution of PbF2.
Analyze
The solubility equation and Ksp expression for PbF2 are
PbF2 (s)
Pb2 (aq)  2 F (aq)
Ksp  [Pb2+ ][F– ]2
In this solution [F–] = 2  [Pb2+] because for every particle of PbF2 that dissolves we get one particle of Pb2+
and two particles of F–.
Solve
Let [Pb2+] = x and [F–] = 2x. The Ksp expression becomes
Ksp = 3.2  10–8 = (x)(2x)2
x = 2.0  10–3 M
Thus, [Pb2+] = x = 2.0  10–3 M and [F–] = 2x = 4.0  10–3 M.
Think about It
We do not need to know how much PbF2 is originally placed into the solution because it does not appear in
the Ksp expression as a pure solid.
17.107. Collect and Organize
Given the Ksp of CaCO3 (9.9  10–9), we are to calculate the solubility of this substance in units of grams per
milliliter.
324 | Chapter 17
Analyze
The solubility equation and Ksp expression for CaCO3 are
CaCO3 (s)
Ca 2 (aq)  CO32 (aq)
Ksp  [Ca 2+ ][CO32– ]
In this solution, [Ca2+] = [CO32–] because for every CaCO3 that dissolves one Ca2+ and one CO32– are
produced. We can then say that
Ksp = x2
and we can solve for x, which is the molar solubility (mol/L) of CaCO3. To convert this to grams per
milliliter we multiply by the molar mass of CaCO3 (100.09 g/mol) and divide by 1000 mL/L.
Solve
Ksp  9.9  10–9  x 2
x  9.95  10–5
9.95  10–5 mol 100.09 g
1L


 9.96  10–6 g/mL
L
1 mol
1000 mL
Think about It
The value of x that we calculate in the Ksp expression is the molar solubility of the solid because it is that
amount (“x”) that dissolves into the solution.
17.108. Collect and Organize
Given the Ksp of AgI (1.50  10–16), we are to calculate the solubility of this substance in units of grams per
milliliter.
Analyze
The solubility equation and Ksp expression for AgI are
AgI(s)
Ag (aq)  I (aq)
Ksp  [Ag + ][I  ]
In this solution, [Ag+] = [I–] because for every AgI that dissolves one Ag+ and one I– are produced. We can
then say that
Ksp = x2
and we can solve for x, which is the molar solubility (mol/L) of AgI. To convert this to grams per milliliter
we multiply by the molar mass of AgI (234.77 g/mol) and divide by 1000 mL/L.
Solve
Ksp  1.50  10–16  x 2
x  1.225  10–8
1.225  10–8 mol 234.77 g
1L


 2.88  10–9 g/mL
L
1 mol
1000 mL
Think about It
The value of x that we calculate in the Ksp expression is the molar solubility of the solid because it is that
amount (“x”) that dissolves into the solution.
17.109. Collect and Organize
Given the Ksp for the dissolution of AgOH (1.52  10–8) in Appendix 5, we are to calculate the pH of a
saturated solution.
Analyze
The solubility equation and Ksp expression for AgOH are
AgOH(s)
Ag (aq)  OH (aq)
Ksp  [Ag + ][OH ]
Equilibrium in the Aqueous Phase | 325
Letting [Ag+] = [OH–] = x (because the stoichiometry is 1:1), we can solve for x using the Ksp expression.
The pH of the solution will then be
pH = 14 – (–log x)
Solve
Ksp  1.52 10– 8  x 2
x  1.233 10– 4
pH  14 –  – log 1.233  10– 4   10.091
Think about It
Even though the Ksp of AgOH is not high, this solution is quite basic.
17.110. Collect and Organize
Using the Ksp for the dissolution of Mg(OH)2 from Appendix 5 (5.6  10–12), we can calculate the pH of a
saturated solution.
Analyze
The solubility equation and Ksp expression for MgOH are
Mg  OH2(s)
Mg2 (aq)  2 OH (aq)
Ksp  [Mg 2+ ][OH ]2
If we let [Mg2+] = x, then [OH–] = 2x at equilibrium. We can then solve for x using the Ksp expression and the
pH of the solution will be
pH = 14 – (–log 2x)
Solve
Ksp  5.6 10–12  x(2 x)2
x  1.12 10– 4
pH  14 – [– log(2 1.12 10– 4 )]  10.35
Think about It
Be careful to notice here that [OH–] = 2x and that the [OH–] must be squared in the Ksp expression.
17.111. Collect and Organize
Using the common-ion effect, we can determine in which 0.1 M solution (NaCl, Na2CO3, NaOH, or HCl) the
most CaCO3 dissolves.
Analyze
Whenever a common ion is already present in the solution, the CaCO 3 is less soluble. Any solution,
therefore, with Ca2+ or CO32– would have lower solubility of CaCO3 compared to water. We also should look
for solutions that might react with either Ca2+ or CO32– and shift the solubility equilibrium to the right.
Solve
(a) NaCl(aq) has neither an ion common to CaCO3 nor ions that react with either Ca2+ or CO32–. CaCO3 has
the same solubility in this NaCl solution as in water.
(b) The 0.1 M Na2CO3 solution is 0.1 M in CO32–. This decreases the solubility of CaCO3.
(c) NaOH(aq) has neither an ion common to CaCO3 nor reacts with either Ca2+ or CO32–. CaCO3 has the
same solubility in this NaOH solution as water.
(d) A solution of HCl reacts with CO32– to form H2CO3 which then decomposes to H2O and CO2. This
reaction shifts the solubility equilibrium to the right so more CaCO3 dissolves.
The solution of (d) 0.1 M HCl dissolves the most CaCO3.
Think about It
A higher concentration of acid dissolves even more CaCO3 as the equilibrium shifts to the right:
CaCO3(s) + 2 H+(aq)
Ca2+(aq) + H2CO3(aq)
326 | Chapter 17
17.112. Collect and Organize
Using the common-ion effect, we can determine in which solution, already containing either Ca 2+ or F–, the
solid CaF2 will be most soluble.
Analyze
The presence of either Ca2+ or F– in the solution causes less CaF2 to dissolve than in pure water. The CaF2 is
most soluble, then, in the solution that has the lowest concentration of either Ca 2+ or F–. For each solution,
the ion concentration is
(a) 0.010 M Ca2+
(b) 0.01 M F–
(c) 0.001 M F–
(d) 0.10 M Ca2+
Solve
The solution that has the highest solubility for CaF2 is (c) 0.001 M NaF.
Think about It
The solution that has the lowest solubility for CaF2 is the one highest in concentration of the common ions.
That is the case for (d) 0.10 M Ca2+.
17.113. Collect and Organize
Given the average concentrations of SO4–2 and Sr2+ in seawater (0.028 M and 9  10–5 M, respectively) and
the Ksp of SrSO4 (3.4 10–7we are to determine if the concentration of Sr 2+ is controlled by the relative
insolubility of SrSO4.
Analyze
The solubility equation and Ksp expression for SrSO4 are
SrSO4 (s)
Sr 2 (aq)  SO42– (aq)
Ksp  [Sr 2 ][SO42– ]
Solve
2–
K sp  3.4  10 –7  [Sr 2+ ][SO 4 ]
 [Sr 2+ ]  0.028 M 
[Sr 2  ]  1.21 10 –5 M
This is the expected concentration of Sr2+ in seawater with a known [SO42–] of 0.028 M. This [Sr2+] is lower
than the average [Sr2+] of 9 – M, so some other process must be controlling the [Sr2+].
Think about It
As the [SO42–] increases, the solubility of SrSO4 decreases because of the common-ion effect.
17.114. Collect and Organize
Given the Ksp for Fe(OH)3 (1.1  10–36), we are to calculate the [Fe3+] in seawater that has a pH of 8.
Analyze
The seawater has a [OH–] that is found from the pH
K
110–14
[OH – ]  w+ 
 1.26 10 – 6 M
[H ] 110 –8.1
This is the initial concentration of [OH–] before the Fe(OH)3 dissolves so the ICE table is
[Fe3+]
[OH–]
Initial
Change
Equilibrium
0
+x
x
1.26  10– 6
+3x
1.26  10– 6 + 3x
Because Ksp is so small, we may assume that 3x << 1.26  10– 6 in solving the problem.
Equilibrium in the Aqueous Phase | 327
Solve
Ksp = 1.1  10–36 = x(1.26  10– 6 + 3x)3  x(1.26  10– 6)3
x = 5.5  10–19
3+
–19
The maximum [Fe ] in seawater is 5.5  10 M. This is also the solubility (s) of Fe(OH)3 in seawater.
Think about It
The suppression of solubility by the initial presence of an ion also present in the species dissolving is called
the common-ion effect.
17.115. Collect and Organize
Given 125 mL solution that is 0.375M in Ca(NO3)2 we are asked whether CaF2 will precipitate when
245 mL of a 0.255 M NaF solution is added.
Analyze
The Ksp for CaF2 from Appendix 5 is 3.9  10–11. When the two solutions are mixed, the total volume is
370 mL and the [Ca2+] and [F –] in the mixed solution is
125 mL  0.375M  370 mL  [Ca 2  ]
[Ca 2  ]  0.127 M
245 mL  0.255M  370 mL  [F ]
[F ]  0.169 M
From the Ksp expression
Ksp = [Ca2+][F–]2 = 3.9  10–11
If the [Ca2+]initial  [F–]2initial > Ksp, then CaF2 will precipitate.
Solve
[Ca2+]initial  [F–]2initial = 0.127  (0.169)2 = 3.63  10–3. This is greater than the value of Ksp for CaF2, so it will
preciptate from the mixed solution.
Think about It
Because CaF2 has a small solubility product constant, we expect that most of the ions will precipitate as
CaF2. In this solution the Ca2+ ions are in excess (0.0469 mol) compared to that of F – (0.0625 mol) so the
final solution will have only a small amount of F– in solution.
17.116. Collect and Organize
Given 345 mL solution that is 0.0100 M in Sr2+(aq), we are asked whether SrSO4 will precipitate when
75 mL of a 0.175 M K2SO4 solution is added.
Analyze
The Ksp for SrSO4 from Appendix 5 is 3.4  10–7. When the two solutions are mixed, the total volume is
420 mL and the [Sr2+] and [SO42–] in the mixed solution is
345 mL  0.0100 M  420 mL  [Sr 2  ]
[Sr 2  ]  8.21 103 M
2
75 mL  0.175 M  420 mL  [SO 4 ]
2
[SO 4 ]  3.13  102 M
From the Ksp expression
Ksp = [Sr2+][SO42–] = 3.4  10–7
If the [Sr2+]initial  [SO22–]initial > Ksp, then SrSO4 will precipitate.
Solve
[Sr2+]initial  [SO22–]initial = (8.21  10–3 M)  (3.13  10–2 M) = 2.56  10– 4. This is greater than the value of Ksp
for SrSO4 so it will preciptate from the mixed solution.
328 | Chapter 17
Think about It
Because SrSO4 has a small solubility product constant, we expect that most of the ions will precipitate as
SrSO4. In this solution the SO42– ions are in excess (0.0131 mol) compared to that of Sr 2+ (0.00345 mol) so
the final solution will have only a small amount of Sr2+ in solution.
17.117. Collect and Organize
When a 0.250 M solution of Pb2+(aq) is added to a solution that is 0.010 M in Br– and SO42– we can use the
values of Ksp for PbBr2 and PbSO4 to determine which anion is the first to precipitate. We are then asked to
calculate the concentration of the anion that precipitates first at the moment that the second ion starts to
precipitate. This will be when the solution is saturated in the lead salt of the first anion to precipitate.
Analyze
The Ksp value shows that PbSO4 has a smaller solubility product constant (1.8  10–8) compared to that of
PbBr2 (6.6  10– 6).
Solve
(a) PbSO4 with a smaller solubility product constant will precipitate first from the solution, so the SO42–
anion will precipitate first.
(b) When Br– begins to precipitate, the maximum amount of Pb2+ that could be present that will not cause
Br– to precipitate is
K sp  6.6  106  [Pb 2+ ] [Br – ]2
[Pb 2+ ] 
6.6  106
 6.6  102 M
(0.010)2
The [SO42–] in the solution when [Pb2+] = 6.6  10–2 M is
K sp  1.8  108  [Pb 2+ ] [SO 4 2  ]
[Pb 2+ ] 
1.8  108
 2.7  107 M
6.6  102
Think about It
We do not need an ICE table to solve this problem.
17.118. Collect and Organize
Considering a solution (A) that is 0.0250 M Ag+ and 0.0250 M Pb2+ we are to determine how to separate
these ions using either a 0.500 M solution of NaCl (solution B) or a 0.500 M solution of NaBr (solution C).
Analyze
Here we must consider the values of Ksp for the possible salts formed in the reaction. From Appendix 5 these
are:
Ksp (AgCl) = 1.8  10–10
Ksp (AgBr) = 5.4  10–13
Ksp (PbCl2) = 1.6  10–5
Ksp (PbBr2) = 6.6  10–6
To separate Ag+ from Pb2+ we want to employ the anion solution (Cl – or Br–) with the greatest difference in
the Ksp for the Ag+ and Pb2+ salts of that anion.
Solve
(a) Because the solubility product constants for both Cl– and Br– salts of Ag+ are smaller than those of Pb2+,
Ag+ will precipitate first from the solution whether we add solution B or solution C. The difference in the
solubility product constant is approximately 10 7 for AgBr versus PbBr2, which is greater than the
approximately 105 difference in solubility product constants between AgC and PbCl 2. Therefore, we should
choose solution C (NaBr) to separate Ag+ from Pb2+ in the mixed solution.
Equilibrium in the Aqueous Phase | 329
(b) When Pb2+ begins to precipitate, the maximum amount of Br– that can be present to not cause Ag+ to
precipitate is
K sp  6.6  106  [Pb 2+ ] [Br – ]2
[Br  ] 
6.6  106
 1.62  10 2 M
0.0250
The [Ag+] in the solution when [Br–] is 1.62  10–2 M is
Ksp  5.4  1013  [Ag + ] [Br  ]
5.4  1013
 3.3  1011 M
1.62  102
Relative to the original concentration of Ag+ in the solution, this is
[Ag + ] 
3.3 ×10 11 M
 100  1.3  107 %
0.0250 M
This is less than 0.10% so the two cations are completely separated.
Think about It
We do not need an ICE table to solve this problem.
17.119. Collect and Organize
We are to describe the difference between a titration curve for a strong acid and that of a weak acid.
Analyze
A strong acid is completely ionized in aqueous solution whereas a weak acid is only partially hydrolyzed.
This affects the pH of the solution at the start of the titration for equal concentrations of the acids. The
equivalence or end point of the titration is where equal moles of OH – have been added to the acid. The
species formed at the end point for a strong acid and weak acid differ. This affects the pH of the solution at
the equivalence point.
Solve
The weak acid titration curve has an initial pH that is higher (less acidic) than that of an equimolar solution
of a strong acid (lower pH, more acidic). The pH at the equivalence point in the titration of a strong acid is
7.00 because the species formed in the titration are water and a nonhydrolyzing anion, such as Cl –:
HCl(aq) + OH–(aq)  H2O ( ) + Cl–(aq)
The pH at the equivalence point for a weak acid is basic because of the formation of a hydrolyzing anion,
such as NO2– in the titration of HNO2:
HNO2(aq) + OH–(aq)  NO2–(aq) + H2O ( )
NO2–(aq) + H2O ( )
HNO2(aq) + OH–(aq)
Think about It
Titration of a weak acid always gives an end point of pH > 7.
17.120. Collect and Organize
We are asked whether the pH at the equivalence point in all titrations of a strong base with a strong acid is
the same.
330 | Chapter 17
Analyze
A strong base such as OH– (e.g., from NaOH or KOH) when titrated with strong acid produces water plus a
dissolved salt.
Solve
Yes. The pH at the equivalence point of strong bases during titrations with strong acids is always pH 7.00, as
seen by the formation of H2O and salts that do not hydrolyze as shown in the following examples:
NaOH(aq) + HCl(aq)  H2O ( ) + Na+(aq) + Cl–(aq)
Ca(OH)2(aq) + 2 HNO3(aq)  2 H2O ( ) + Ca2+(aq) + 2 NO3–(aq)
Na2O(aq) + 2 HClO4(aq)  H2O ( ) + 2 Na+(aq) + 2 ClO4–(aq)
Think about It
Likewise, the equivalence point is at pH 7.00 for all titrations of strong acids with strong bases.
17.121. Collect and Organize
We are asked whether the pH at the equivalence point is the same for the titration of all weak acids with
strong base.
Analyze
When a weak acid is titrated, the species present in solution at the equivalence point is the conjugate base of
the weak acid. This conjugate base is a weak base and will hydrolyze in water according to the equation
A–(aq) + H2O ( )
HA(aq) + OH–(aq)
Solve
No. Because the extent to which A– (the conjugate base) hydrolyzes depends on the base strength of A –, the
pH at the equivalence point in the titration for weak acids is not expected to be the same.
Think about It
Likewise, the pH values at the equivalence point for weak bases differ based on the strength of the conjugate
acid.
17.122. Collect and Organize
We are to list the properties of an acid–base indicator.
Analyze
An indicator is a visual indicator of the pH of a solution.
Solve
An indicator must have a different color in its base from that in its acid form.
Think about It
In using an indicator in a titration we have to be careful to choose an indicator that changes color in the pH
range of the expected equivalence point.
17.123. Collect and Organize
We are to calculate the pH along various points of the titration curve when 25.0 mL of 0.100 M acetic acid
(Ka = 1.76  10–5) is titrated with 0.125 M NaOH.
Analyze
For each step of the titration we have to consider the moles of NaOH added that react with the moles of
CH3COOH initially present:
0.100 mol
25.0 mL 
 0.00250 mol CH3COOH
1000 mL
Equilibrium in the Aqueous Phase | 331
For each point in the titration curve the moles of OH– (from NaOH) added are as follows:
0.125 mol
10.0 mL 
 0.00125 mol OH –
1000 mL
0.125 mol
20.0 mL 
 0.00250 mol OH –
1000 mL
0.125 mol
30.0 mL 
 0.00375 mol OH –
1000 mL
The OH– reacts with the CH3COOH in solution to give CH3COO–, the conjugate base of acetic acid. Our
strategy for the problem is to first react as much of the added OH– with acetic acid as possible and then use
the equilibrium Ka expression to calculate the pH:
CH3COOH(aq)
CH3COO–(aq) + H+(aq)
[CH 3 COO – ][H  ]
[CH 3 COOH]
Or we can use the equivalent equilibrium Kb expression:
CH3COO–(aq) + H2O ( )
CH3COOH(aq) + OH–(aq)
Ka 
[CH 3 COOH][OH – ]
[CH 3 COO – ]
For that calculation we have to be careful to determine the molarity of the species in solution by remembering
that the volume in a titration increases through the addition of the titrant.
Kb 
Solve
When 10.0 mL of OH– are added, the 0.00125 mol of OH– reacts with the 0.00250 mol of CH3COOH to
produce 0.00125 mol of CH3COO– and leave 0.00125 mol of CH3COOH unreacted. Since the total volume
of the solution is now 25.0 + 10.0 = 35.0 mL the molarity of these species is
0.00125 mol
 0.0357 M
0.0350 L
Using an ICE table to calculate the pH of this solution gives the following:
[CH3COOH]
[CH3COO–]
[H+]
Initial
0.0357
0.0357
0
Change
–x
+x
+x
Equilibrium
0.0357 – x
0.0357 + x
x
x(0.0357  x) x(0.0357)

0.0357  x
0.0357
–5
x  1.76  10
The simplifying assumption is valid (0.05%) so
pH = –log 1.76  10–5 = 4.754
–
When 20.0 mL of OH are added we have added an equal number of moles of OH – as there are moles of
CH3COOH initially present. This reaction produces 0.00250 mol of CH 3COO– so it makes sense here to use
the Kb expression to calculate the pH of the solution. Since the total volume of the solution is now 45.0 mL,
the molarity of these species is
0.00250 mol
 0.0556 M
0.0450 L
Using an ICE table to calculate the pH of this solution gives the following:
[CH3COO–]
[CH3COOH]
[OH–]
Initial
0.0556
0
0
Change
–x
+x
+x
Equilibrium
0.0556 – x
x
x
1.76  10 –5 
332 | Chapter 17
Kb 
Kw
Ka

11014
x2
x2
–10

5.68

10


0.0556  x 0.0556
1.76 10–5
x  5.62 10– 6
The simplifying assumption is valid (0.01%) so
pOH = –log 5.62  10– 6 = 5.250
pH = 14 –5.250 = 8.750
When 30.0 mL of OH– are added we convert all of the CH3COOH to 0.00250 mol CH3COO– and have
0.00125 mol OH– remaining. Since the total volume of the solution is now 55.0 mL, the molarity of these
species is
0.00250 mol
 0.0455 M CH 3COO –
0.0550 L
0.00125 mol
 0.0227 M OH –
0.0550 L
Using an ICE table to calculate the pH of this solution gives the following:
[CH3COO–]
[CH3COOH]
[OH–]
Initial
0.0455
0
0.0227
Change
–x
+x
+x
Equilibrium
0.0455 – x
x
0.0227 + x
x(0.0227  x) x(0.0227)
5.68  10 –10 

0.0455  x
0.0455
x  1.14  10 –9
The simplifying assumption is valid so
pOH = –log 0.0227 = 1.644
pH = 14 –1.644 = 12.356
Think about It
When exactly half the moles of strong base are added in the titration of a weak acid (as in this problem
where 10 mL of the titrant were added), this point is the midpoint of the titration. Notice that at this point
pH = pKa of the weak acid.
17.124. Collect and Organize
We are to calculate the pH along various points of the titration curve when 25.0 mL of 0.100 M
trimethylamine (Kb = 6.46  10–5) is titrated with 0.125 M HCl.
Analyze
For each step of the titration we have to consider the moles of HCl added that react with the moles of
(CH3)3N initially present
0.100 mol
25.0 mL 
 0.00250 mol (CH3 )3 N
1000 mL
+
For each point in the titration curve the moles of H (from HCl) added are as follows:
0.125 mol
10.0 mL 
 0.00125 mol H +
1000 mL
0.125 mol
20.0 mL 
 0.00250 mol H +
1000 mL
0.125 mol
30.0 mL 
 0.00375 mol H +
1000 mL
The H+ reacts with the (CH3)3N in solution to give (CH3)3NH+, the conjugate acid of trimethylamine. Our
strategy for the problem is to first react as much of the added H + with trimethylamine as possible and then
use the equilibrium Kb expression to calculate the pH:
Equilibrium in the Aqueous Phase | 333
(CH3)3N(aq) + H2O ( )
(CH3)3NH+(aq) + OH–(aq)
[(CH3 )3 NH  ][OH – ]
[(CH3 )3 N]
Or we can use the equivalent equilibrium Ka expression:
(CH3)3NH+(aq)
(CH3)3N(aq) + H+(aq)
Kb 
[(CH 3 )3 N][H  ]
[(CH 3 )3 NH + ]
For that calculation we have to be careful to determine the molarity of the species in solution by
remembering that the volume in a titration increases through the addition of the titrant.
Ka 
Solve
When 10.0 mL of H+ are added, the 0.00125 mol H+ reacts with the 0.00250 mol of (CH3)3N to produce
0.00125 mol of (CH3)3NH+ and leave 0.00125 mol of (CH3)3N unreacted. Since the total volume of the
solution is now 25.0 + 10.0 = 35.0 mL the molarity of these species is
0.00125 mol
 0.0357 M
0.0350 L
Using an ICE table to calculate the pH of this solution gives
[(CH3)3N]
[(CH3)3NH+]
[OH–]
Initial
0.0357
0.0357
0
Change
–x
+x
+x
Equilibrium
0.0357 – x
0.0357 + x
x
x(0.0357  x) x(0.0357)
6.46  10–5 

0.0357  x
0.0357
–5
x  6.46  10
The simplifying assumption is valid (0.2%) so
pOH = –log 6.46  10–5 = 4.190
pH = 14 – 4.190 = 9.810
When 20.0 mL of H+ are added we have added an equal number of moles of H+ as there are moles of
(CH3)3NH initially present. This reaction produces 0.00250 mol (CH 3)3NH+, so it makes sense here to use
the Ka expression to calculate the pH of the solution. Since the total volume of the solution is now 45.0 mL,
the molarity of this species is
0.00250 mol
 0.0556 M
0.0450 L
Using an ICE table to calculate the pH of this solution gives
[(CH3)3NH+]
[(CH3)3N]
[H+]
Initial
0.0556
0
0
Change
–x
+x
+x
Equilibrium
0.0556 – x
x
x
14
2
2
Kw
110
x
x
Ka 

 1.55 10–10 

K b 6.46 10–5
0.0556  x 0.0556
x  2.94 10– 6
The simplifying assumption is valid (0.005%) so
pH = –log 2.94  10– 6 = 5.532
+
When 30.0 mL of H are added we convert all of the (CH 3)3N to 0.00250 mol (CH 3)3NH+ and have
0.00125 mol H+ remaining. Since the total volume of the solution is now 55.0 mL, the molarity of these
species is
0.00250 mol
 0.0455 M (CH 3 )3 NH +
0.0550 L
0.00125 mol
 0.0227 M H +
0.0550 L
334 | Chapter 17
Using an ICE table to calculate the pH of this solution
[(CH3)3NH+]
[(CH3)3N]
Initial
0.0455
0
Change
–x
+x
Equilibrium
0.0455 – x
x
[H+]
0.0227
+x
0.0227 + x
x(0.0227  x) x(0.0227)

0.0455  x
0.0455
–10
x  3.11  10
1.55  10–10 
The simplifying assumption is valid so
pH = –log 0.0227 = 1.644
Think about It
In the titration of a weak base with a strong acid, the pH at the midpoint is equal to the pKa of the
conjugate acid. In this problem, notice that we calculated the midpoint to be at pH = 9.810. This is equal to
–log Ka = –log 1.55  10–10.
17.125. Collect and Organize
In the titration of a 100.00 mL NH3 solution with 0.1145 M HCl, it takes 22.35 mL to reach the equivalence
point. From this information we are to calculate the concentration of ammonia in the solution.
Analyze
Because at the equivalence point the moles of HCl added as a titrant equal the moles of NH 3 in the solution,
we can calculate the amount of NH3 through
1 mol NH3
Moles NH3  mL HCl used as titrant  molarity of HCl solution 
1 mol HCl
Because we know the volume of the original solution of NH3, the molarity of the sample is
mol NH3
volume of sample in L
Solve
0.1145 mol 1 mol NH3

 2.559  10–3 mol
1000 mL
1 mol HCl
2.559  10 –3 mol
Molarity of NH 3 
 0.02559 M
0.100 L
Moles NH3  22.35 mL HCl 
Think about It
Remember that at the equivalence point, what is equal is the moles, not the volume nor the concentration of
acid and base.
17.126. Collect and Organize
For the titration of a 100.0 mL sample from a hot spring containing CO 32– and HCO3–, we are to calculate the
concentrations of each species knowing that the first equivalence point is reached when 2.56 mL of a
0.0355 M solution of HCl has been added and the second equivalence point is reached when 10.42 mL of the
HCl has been added.
Analyze
As the titration proceeds the CO32– ion is converted into the HCO3– ion:
CO32–(aq) + H+(aq)  HCO3–(aq)
This then adds to the concentration of HCO3– already present and titrated to reach the second equivalence
point:
HCO3–(aq) + H+(aq)  H2CO3(aq)
We therefore have to subtract the moles of CO32– found in the first equivalence point from the total moles of
HCO3– found from the second equivalence point to obtain the moles of HCO 3– present originally in the
sample.
Equilibrium in the Aqueous Phase | 335
Solve
From the first equivalence point,
0.0355 mol 1 mol CO32–

 9.09  10 –5 mol CO32–
1000 mL
1 mol H +
9.09 10–5 mol
CO32–  in sample =
 9.09 10– 4 M


0.100 L
From the second equivalence point,
0.0355 mol 1 mol HCO3–
10.42 mL HCl 

 3.70 10– 4 mol HCO3–
1000 mL
1 mol H+
Moles of HCO3– present in original sample = 3.70 10 – 4 mol – 9.09 10 –5 mol = 2.79 10 – 4 mol
2.56 mL HCl 
HCO3–  in sample 


2.79 10– 4 mol
 2.79 10–3 M
0.100 L
Think about It
In this sample the pH is such that [HCO3– ] > [CO32–]. At higher pH, less HCO3– would be expected.
17.127. Collect and Organize
We are to calculate the volume of 0.0100 M HCl required to titrate 250 mL of 0.0100 M Na2CO3 to the first
equivalence point.
Analyze
Before setting out to do a lot of calculations here, let’s stop and think. The concentration of the titrant (HCl)
is equal to the concentration of the base we are titrating! We don’t need, therefore, to do any calculations
because the same volume of HCl is needed to neutralize the base.
Solve
Titration of 250 mL of 0.0100 M Na2CO3 to the first equivalence point requires 250 mL of HCl.
Think about It
To reach the second end point for the Na2CO3 solution, 500 mL of the HCl titrant would be required.
17.128. Collect and Organize
We are to determine the volume of 0.0100 M HCl needed to titrate 250 mL of 0.0100 M Na2CO3 and 250 mL
of 0.0100 M HCO3–.
Analyze
Before setting out to do a lot of calculations here, let’s stop and think. The concentration of the titrant (HCl)
is equal to the concentration of the Na2CO3 we are titrating! We don’t need, therefore, to do any calculations
here. We do have to keep in mind that Na2CO3 is a dibasic compound and therefore two equivalents of acid
are needed to titrate it. HCO3– is monobasic and requires only one equivalent of HCl to reach the equivalence
point.
Solve
500 mL are needed to titrate the Na2CO3 and 250 mL are needed to titrate the HCO3–.
Think about It
To reach the third end point for 250 mL of 0.0100 M Na3PO4 solution, 750 mL of the 0.0100 M HCl titrant
are required.
17.129. Collect and Organize
In comparing the titration of a weak acid in which the amount of NaOH titrant needed to reach the
equivalence point is double that for another titration, we are asked what the pH halfway to the equivalence
point is for the second titration if the pH at that point for the first titration is 4.44.
336 | Chapter 17
Analyze
The midpoint is where half of the weak acid has been converted into its conjugate base. It is at this point
where [acid] = [conjugate base] that the pH = pKa.
Solve
The pH at the midpoint in the second titration would be the same as it is in the first titration, 4.44.
Think about It
Certainly the volume of the added titrant at which the midpoint is reached for the second titration would be
twice that as in the first titration, but the pH at those points would be the same.
17.130. Collect and Organize
Using the data of pH versus the volume of OH– added to a 125.0 mg sample of a weak acid dissolved in
100.0 mL of water, we can plot the titration curve to determine the Ka of this acid.
Analyze
From the titration curve we first determine the volume of OH– to reach the equivalence point. When one-half
of that amount was added, the midpoint is reached and at that point pH = pKa. By reading the pH off the
titration curve at this volume of OH–, we can calculate the Ka of the acid.
Solve
The equivalence point is at 22.5 mL; therefore, the midpoint in the titration is when 11.25 mL of OH– have
been added. The pH at this point is 4.2 and so the Ka = 6.31  10–5.
Think about It
Because we know the moles of OH– added to reach the equivalence point in this titration
0.050 mol OH –
22.5 mL 
 1.125  10–3 mol
1000 mL
we also know that there are 1.125  10–3 mol of the monoprotic acid present in the sample. Since the sample
was weighed, we can calculate the molar mass of the acid
0.125 g
 111.1 g/mol
1.125  10–3 mol
Equilibrium in the Aqueous Phase | 337
17.131. Collect and Organize
To sketch the titration curve for the titration of 50.0 mL of a solution of the weak acid HNO 2 with 1.00 M
NaOH, we must use the information provided to calculate the pH of the HNO 2 solution before any NaOH is
added and at the equivalence point.
Analyze
The initial pH of 0.250 M HNO2 is calculated from the value of Ka = 4.0  10– 4 after setting up an ICE table
where the initial concentrations of H+ and NO2– are 0.00 M. To calculate the pH of the solution at the
equivalence point we need to recognize that at that point all of the HNO 2 is converted to NO2–. We use Kb,
therefore, for the ICE table calculation. We also need to take into account the added volume of the solution
when NaOH titrant is added.
Solve
The initial pH of 0.250 M HNO2 is found as follows:
[HNO2]
Initial
0.250
Change
–x
Equilibrium
0.250 – x
[H+]
0.00
+x
x
[NO2–]
0.00
+x
x
x2
x2

0.250  x 0.250
x  0.0100
K a  4.0  10 – 4 
pH  –log 0.0100  2.00
At the equivalence point, moles of OH added = moles of HNO2 present in the initial solution:
0.250 mol
Moles HNO2  50.0 mL 
 0.0125 mol
1000 mL
Because all the HNO2 is converted to NO2– at the equivalence point, the amount of NO 2– at equivalence point
is initially 0.0125 mol. The NO2– produced hydrolyzes by the equation
–
NO2–(aq) + H2O ( )
HNO2(aq) + OH–(aq)
to give a slightly basic solution at the equivalence point.
We need to determine the volume of NaOH titrant added so that we can use the [NO 2–] in an ICE table to
calculate the pH at the equivalence point. The 0.0125 mol of HNO 2 requires 0.0125 mol of OH– to neutralize
it. The volume of NaOH needed to provide this 0.0125 mol OH – is
1 mol NaOH – 1000 mL
0.0125 mol OH – 

 12.5 mL
1.00 mol
1 mol OH –
The total volume of the solution at the equivalence point is 12.5 + 50.0 mL = 62.5 mL. The molarity of NO 2–
at the equivalence point is
0.0125 mol NO2
 0.200 M
0.0625 L
[NO2–]
[HNO2]
[OH–]
Initial
0.200
0
0
Change
–x
+x
+x
Equilibrium
0.200 – x
x
x
This is a Kb expression so we must calculate Kb from Ka:
Kw
11014
Kb 

 2.5 10–11
Ka 4.0 10– 4
338 | Chapter 17
Solving the equilibrium constant expression for x gives
x2
x2

0.200  x 0.200
x  2.24  10 6
pOH = –log 2.24 10– 6  5.65
pH = 14 – 5.65 = 8.35
K b  2.5  10 –11 
Think about It
Note in the titration curve that there is a relatively flat region before the equivalence point. Here the pH does
not change much despite the addition of more and more titrant. This is often called the buffer region.
17.132. Collect and Organize
To determine the color of the red cabbage juice at the equivalence point in the titration of 0.10 M acetic acid
with 0.10 M NaOH, we need to consider whether the pH at the equivalence point is acidic or alkaline (basic).
Analyze
We could calculate the pH of the solution at the equivalence point, but that is not necessary to answer the
question. At the equivalence point all of the acetic acid, CH3COOH, is converted to the acetate anion,
CH3COO–. The acetate anion reacts with water.
Solve
The equation describing acetate’s reaction with water shows that we have a basic (alkaline) solution at the
equivalence point:
CH3COO–(aq) + H2O ( )
CH3COOH(aq) + OH–(aq)
The color of red cabbage juice at the equivalence point, then, is yellow.
Think about It
To calculate the actual pH, solve for x in the following Kb expression:
x2
x2
K b  5.68 10 –10 

0.050  x 0.050
x  5.33  10 6
pOH  –log 5.33 10 6  5.27
pH  14 – 5.27  8.73
17.133. Collect and Organize
For the titration of quinine, a dibasic malaria drug, we need to calculate the initial pH and the position of
each of the equivalence points in order to sketch the titration curve.
Equilibrium in the Aqueous Phase | 339
Analyze
Because the second base ionization constant (K b2 = 1.35  10–9) is much smaller than the first (K b1 =
3.31  10 – 6), we can ignore it in calculating the initial pH of the solution. Because the concentration of the
quinine and HCl titrant are equal (both solutions are 0.100 M), we know that the first equivalence point will
be at the point where 40.0 mL of HCl have been added and the second equivalence point is at 80.0 mL of
HCl added.
Solve
The initial pH is calculated as follows:
Initial
Change
Equilibrium
[QuinineH+]
0
+x
x
[Quinine]
0.100
–x
0.100 – x
[OH–]
0
+x
x
x2
x2

0.100  x 0.100
x  5.75  10 4
K b  3.31 10 – 6 
1
pOH  –log 5.75 10 4  3.240
pH  14 – 3.240  10.760
The first equivalence point is where 40.0 mL of HCl have been added to give
0.100 mol quinine 1 mol quinineH +
40.0 mL HCl 

 4.00  10 –3 mol
1000 mL
1 mol quinine
The molarity of quinineH+ is
4.00  10 –3 mol
 0.0500 M
0.080 L
At this point quinineH+ can act both as an acid and as a base:
QuinineH+(aq) + H2O ( )
quinineH22+(aq) + OH–(aq)
K b1 =1.35  10–9
110–14
 3.02 10–9
K b2 3.3110– 6
This calculation is beyond the scope of general chemistry study, but we can see that Ka2  Kb1 and so we
expect this equivalence point to be slightly acidic. In fact, it is, at approximately pH 6.8.
The second equivalence point is where 80.0 mL of HCl have been added. This solution contains 4.00 
10–3 mol quinineH22+ with a molarity of
4.00  10 –3 mol
 0.0333 M
0.120 L
because the total volume is now 40.0 mL quinine solution + 80.0 mL HCl titrant. The pH at this equivalence
point is dependent on the equilibrium
Kw
110–14
QuinineH2+(aq)
quinineH+(aq) + H+(aq) Ka1 

 7.4110– 6
Kb2 1.35 10–9
The pH at this equivalence point is as follows:
[QuinineH2+]
[QuinineH+]
[H+]
Initial
0.0333
0
0
Change
–x
+x
+x
Equilibrium
0.0333– x
x
x
QuinineH+(aq)
quinine(aq) + H+(aq)
Ka 2 
Kw

340 | Chapter 17
x2
x2

0.0333  x 0.0333
x  4.97 10  4
pH  –log 4.97 10 4  3.304
K a  7.41 10 – 6 
1
Think about It
Notice that at the two midpoints (20 mL and 60 mL HCl added) that the pH equals pKa2 (8.52) and pKa1
(5.13).
17.134. Collect and Organize
For the titration of 100 mL of a solution of ascorbic acid, a diprotic acid, we need to calculate the initial pH
and the position of each of the equivalence points in order to sketch the titration curve. The molarities of the
ascorbic acid and NaOH solutions are 1.25  10–2 M and 1.00  10–2 M, respectively. We are also asked how
many equivalence points the curve should have and what color indicator(s) could be used.
Analyze
Because the second acid ionization constant ( Ka 2  1.6  10–12 ) is much smaller than the first ( Ka 1  8.0  10–5 ),
we can ignore it in calculating the initial pH of the solution.
Solve
The initial pH is
[Hascorbate–]
[H+]
Initial
0
0
Change
+x
+x
Equilibrium
x
x
2
x
K a 1  8.0  10 –5 
1.25  10 –2  x
Expanding and solving by the quadratic equation gives
x2  8.0 10–5 x 1.00 10–6  0
x  9.6110 4
pH  –log 9.6110 4  3.02
The first equivalence point is where moles of OH– added = moles ascorbic acid:
1.25  10–2 mol
Moles ascorbic acid  100 mL 
 1.25  10–3 mol
1000 mL
[H2ascorbic acid]
1.25  10–2
–x
1.25  10–2 – x
Equilibrium in the Aqueous Phase | 341
The volume of NaOH solution required to reach this equivalence point is
1000 mL
1.25  10–3 mol NaOH 
 125 mL
1.00  10–2 mol
At the first equivalence point ascorbate can act either as an acid or as a base
Ka2  1.6  10–12
Hascorbate–
ascorbate2– + H+
Hascorbate– + H2O
H2ascorbic acid + OH–
K b2 
Kw
Ka1

1  10–14
 1.25  10–10
8.0  10–5
This calculation is beyond the scope of general chemistry study, but we can see that Kb2  Ka2 so we expect
this equivalence point to be slightly basic.
The second equivalence point is where 250 mL of NaOH solution have been added. This solution contains
1.25  10–3 mol of ascorbate2– with a molarity of
1.25  10 –3 mol
 3.57  10 –3 M
0.350 L
because the total volume is now 100 mL ascorbic acid solution + 250 mL NaOH titrant. The pH at this
equivalence point is dependent on the equilibrium
Ascorbate2– (aq)  H 2 O( )
Hascorbate – (aq)  OH – (aq)
1  10–14
 6.25  10–3
Ka 2 1.6  10–12
The pH at this second equivalence point is calculated as follows:
[Ascorbate2–]
[Hascorbate–]
Initial
0
3.57  10–3
Change
–x
+x
Equilibrium
x
3.57  10–3 – x
2
x
6.25  10 –3 
3.57  10 –3  x
Expanding and solving by the quadratic equation gives
x2  6.25  10–3 x  2.23  10–5  0
x  2.54  103
pOH  –log 2.54  10 3  2.60
pH  14  2.60  11.40
K b1 
Kw

[OH–]
0
+x
x
There are two equivalence points in the titration of the diprotic acid ascorbic acid. The color indicators that
could be used (Figure 17.14) are bromthymol blue for the first equivalence point near pH 7 and Alizarin
yellow R for the second equivalence point near pH 11.
342 | Chapter 17
Think about It
In performing this experiment in the lab, we can use the pH at the half-equivalence points (62.5 mL and
187.7 mL) to determine pKa 1 and pK a 2 for this acid.
17.135. Collect and Organize
We are asked to describe the changes in bonding and intermolecular forces when the weak base CH 3NH2 is
dissolved in water.
Analyze
Methylamine is a gas and when dissolved in water the individual methylamine molecules are surrounded
with water. CH3NH2 also reacts with water according to the equation
CH3 NH 2 (aq)  H 2 O( )
CH 3 NH 3 ( aq)  OH – ( aq)
Solve
The hydrogen bonds between some of the water molecules must break and re-form around the species
CH3NH2. Also, the amine hydrolyzes and forms CH3NH3+ and OH–; thus, ion–dipole forces are added when
these ions are surrounded by water molecules.
Think about It
Depending on the strength of the forces formed versus those broken, this dissolution may be either
exothermic or endothermic.
17.136. Collect and Organize
High-sulfur fuels when burned produce acid rain that erodes marble. We are to describe those chemical reactions.
Analyze
Sulfur burned in air produces sulfur oxides such as SO 3. This nonmetal oxide is an acid anhydride. The
parent acid forms upon reaction with water in clouds to produce acid rain. The acid, then, erodes (dissolves)
CaCO3 in marble.
Solve
In burning S-containing fuels:
4 O2 ( g )
8 SO3(g)
S8(s) + 12 O2(g)  8 SO2(g) 
In the presence of water these produce a weak acid, H2SO3, and a strong acid, H2SO4:
SO2 ( g )  H2 O( )  H2SO3 (aq)
SO3 ( g )  H2 O( )  H2SO4 (aq)
These acids dissolve CaCO3:
2 H+(aq) + CaCO3(s)  H2CO3(aq) + Ca2+
Think about It
H2CO3 decomposes to H2O and CO2 according to the equilibrium expression
H2 CO3 (aq)
H2 O( )  CO2 ( g )
17.137. Collect and Organize
We are to compare the structures of H3PO3 and H3PO4 and explain why the K a values for these acids are similar.
1
Analyze
After drawing the Lewis structure and identifying which H atom is ionizable on H 3PO3, we can compare that
to the structure of H3PO4.
Equilibrium in the Aqueous Phase | 343
Solve
(a) and (b) Phosphorous acid has the Lewis structure
(c) Phosphoric acid has a similar structure with its ionizable H atoms also bonded to oxygen atoms, so it is
not surprising that these two acids have similar values for Ka1 .
Think about It
Notice that the H atom bonded to P in H3PO3 is not ionizable. The electronegativity difference between P
and H is not great enough to make this H atom acidic.
17.138. Collect and Organize
By comparing the Lewis structure of hypophosphorous acid, H 3PO2, with that of phosphoric acid, H3PO4, we
can explain why the K a1 values for the acids are nearly the same and identify the ionizable atoms in the
structure.
Analyze
The ionizable hydrogen atoms in both H3PO4 and H3PO2 are those that are bound to the oxygen atoms.
Solve
(a) and (b)
Think about It
Hypophosphorous acid has only one ionizable H atom and thus has only one Ka value. Phosphoric acid,
though, has three ionizable H atoms and therefore has three associated Ka values.
17.139. Collect and Organize
We are asked whether the pH of the solution changes when a cook adds more baking soda to water used in a
recipe and to explain why or why not.
Analyze
Baking soda is a soluble sodium salt with the formula NaHCO3. In solution this salt forms Na+(aq) + HCO3–.
Na+ does not react with water but HCO3– does, which changes the pH of the solution.
Solve
Yes, the pH of the solution increases due to the increase in hydrolysis of HCO 3– according to the equation
HCO3 – (aq)  H 2 O( )
H 2 CO3 (aq)  OH – ( aq)
344 | Chapter 17
Think about It
H2CO3 decomposes at baking temperatures to give CO2(g) and H 2 O( ).
17.140. Collect and Organize
For each of the ingredients in antacid tablets, we are to write a net ionic equation for its reaction with
HCl(aq) and then explain how substances such as MgCO3, CaCO3, and Mg(OH)2 can act as antacids even
though they are insoluble.
Analyze
Each of these substances acts as a base to absorb the added HCl. The Cl– anion and the Na+ cation of the
soluble salts are not involved in the net ionic reaction.
Solve
(a) HCO3–(aq) + H+(aq)  H2CO3(aq)
MgCO3(s) + 2 H+(aq)  Mg2+(aq) + H2CO3(aq)
CaCO3(s) + 2 H+(aq)  Ca2+(aq) + H2CO3(aq)
Mg(OH)2(s) + 2 H+(aq)  Mg2+(aq )  2 H2 O( )
(b) Insoluble antacids are effective because even as solids a small amount dissolves and reacts with H + to
neutralize the acid and raise the pH. Once that small amount is consumed in the reaction, more of the solid
dissolves to reestablish equilibrium.
Think about It
Carbonic acid, H2CO3, decomposes in aqueous solution to give H2O and CO2.
17.141. Collect and Organize
Given the change in pH of a lake from 6.1 to 4.7 when 400 gallons of 18 M H2SO4 were added to the lake,
we are to calculate the volume of the lake.
Analyze
To make this calculation easier, we can assume that we only ionize the first proton on H 2SO4. First, we must
calculate the moles of H2SO4 added to the lake and then determine the increase in the concentration of H + in
the lake in going from pH 6.1 to pH 4.7. Knowing that we simply divide the moles of acid added by the
molarity change of H+ in the lake, we can obtain the size of the lake.
Solve
The amount of H2SO4 added is
400 gal 
3.78 L 18 mol

 27, 216 mol
1 gal
L
The change in lake [H+] is
Initial [H+] = 1  10– 6.1 = 7.94  10–7 M
Final [H+] = 1  10– 4.7 = 2.00  10–5 M
∆[H+] = 1.92  10–5 M increase
The size of the lake is
27, 216 mol 
1L
 1.4  109 L
1.92  10–5 mol
Think about It
This volume is equivalent to 1.4  106 m3. If the lake were 10 m deep, it would cover an area of 1.4  105 m2.
If thought of as a square, that is 374 m on a side.
17.142. Collect and Organize
We are asked to explain why the pH of the water increases when Na 3PO4 is added.
Equilibrium in the Aqueous Phase | 345
Analyze
Na3PO4 is a soluble salt that forms Na + and PO43– in aqueous solution. The Na+ ion does not hydrolyze but
PO43– does.
Solve
The reaction with PO43– produces OH– as seen in the following equations, which increases the pH:
PO43–(aq) + H2 O( )
HPO42–(aq) + OH–(aq)
HPO42–(aq) + H2 O( )
H2PO4–(aq) + OH–(aq)
H2PO4–(aq) + H2 O( )
H3PO4(aq) + OH–(aq)
The first reaction contributes to nearly all of the increase in the pH of the water because K b2 (1.58  10–7) and
K b3 (1.41  10–12) are neglible compared to K b1 (2.2  10–2).
Think about It
The ultimate acid formed in this reaction is a weak acid, not strong, and therefore we write all of the acid
equilibrium species through to the parent acid, H3PO4.
17.143. Collect and Organize
For the drug Zoloft we are to use Figure P17.143 to determine which form is for the acid salt and whether
solutions of Zoloft are acidic or basic.
Analyze
It is important to remember that this drug is sold as the HCl salt.
Solve
(a) The acid salt form has H on the amine (R 2NH) moiety with Cl– as a counterion. This structure is shown
on the right of Figure P17.143.
(b) Because this drug is sold as the HCl salt, solutions of this drug are acidic.
Think about It
Many drugs are sold as HCl salts to render the drugs more soluble in aqueous solution.
17.144. Collect and Organize
Given the structure of Prozac in Figure P17.144, we are asked about its acid–base character and the
solubility of its HCl salt.
Analyze
The functional group that is likely to show acid–base character is the amine group (–NHCH3) on one end of
the Prozac molecule.
Solve
(a) In water, amines show basic character. They pick up a proton from water to form ammonium cations and
OH–. Therefore, dissolving Prozac in water gives a slightly basic solution.
(b) The N atom on Prozac is more likely to react with HCl than the O atom.
(c) The HCl salt of Prozac is more soluble because it is charged and water molecules form stronger ion–
dipole forces around the molecule compared to the dipole–induced dipole forces between the neutral
molecule and water.
Think about It
Being soluble in water also helps to deliver the drug to the body.
17.145. Collect and Organize
By examining the equilibrium reactions of HF in water and aqueous F–, we are to determine the major
species present at pH 7.00, calculate the equilibrium constant for the combination of the two equilibrium
equations, and, finally, calculate the pH and [HF2–]eq when [HF] is 0.150 M.
346 | Chapter 17
Analyze
(a) By examining the values of the two equilibrium constants, and considering the concentrations of species
in solution, we can predict which species is more likely to be present at pH 7.00, F – or HF2–.
(b) The overall equation is the sum of the two equilibrium reactions, so the K for the combined reaction is
the product of the two K values for the individual reactions.
(c) We can use an ICE table and the value of the overall K calculated in part b to determine the pH and [HF2–]eq.
Solve
(a) Because the equilibrium constant of the reaction of F– with HF is larger than the dissociation of HF, we
might expect the most likely species to be HF2–. However, because HF is weak, the [F–] is low compared to
that of water and so HF reacts with H2O to form F– as the major anionic species.
(b) Koverall = Ka  K = (1.1  10–3)  (2.6  10–1) = 2.86  10– 4
(c) We must tackle this problem in two steps. First, we consider the hydrolysis of HF (aq).
[HF]
[H3O+]
[F–]
Initial
0.150
0
0
Change
–x
+x
+x
Equilibrium
0.150 – x
x
x
1.1 10 –3 
x2
 0.150  x 
x 2  1.1 10 –3 x  1.65  10 – 4  0
x  0.0123
–
So [F ] = 0.0123 M and [HF] = 0.138 M
Now, we consider the second equilibrium
[F–]
Initial
0.0123
Change
–x
Equilibrium
0.0123 – x
0.26 
[HF]
0.138
–x
0.138 – x
[HF2–]
0
+x
x
x
x

(0.0123  x)(0.138  x) (0.0123)(0.138)
x  4.4 10 – 4
Therefore,
pH = –log (0.0123) = 1.91
[HF2–] = 4.4  10– 4 M
Think about It
Be careful in making simplifying assumptions. For the first equilibrium, we must solve using the quadratic
equation.
17.146. Collect and Organize
We are to explain why C5F5H is such a strong acid compared to other organic acids, after we draw the
structure of its conjugate base.
Analyze
The presence of many nearby electronegative atoms increases the acidity of a proton.
Equilibrium in the Aqueous Phase | 347
Solve
(a)
4n + 2 = 6 electrons, so this ring is aromatic
(b) C5F5H is very acidic because the presence of five very electronegative F atoms on the carbon ring
stabilizes the anion formed when the proton is lost.
Think about It
The formation of a stable aromatic ring upon deprotonation enhances the stability of the conjugate base
above the stability imparted by the presence of F atoms on the ring.
17.147. Collect and Organize
Given the structure of Naproxen (Figure P17.147), we are to draw the structure of the sodium salt, explain
whether a solution of the salt is acidic or basic, and explain why the salt is more soluble than Naproxen itself.
Analyze
The ionizable functional group in Naproxen is the carboxylic acid (– COOH) group.
Solve
(a)
(b) A solution of the salt of Naproxen is basic because the ionized – COO– group reacts with water, giving
– COOH + OH–:
(c) The salt is more soluble because it is charged and water molecules form stronger ion – dipole forces
around the molecule compared to the dipole–induced dipole forces between the neutral molecule and water.
Think about It
Being soluble in water also helps to deliver the drug to the body.
17.148. Collect and Organize
For a drop in the pH of the oceans by 0.77 pH units due to increased CO 2 in the atmosphere, we are asked to
explain with chemical equations how CO2 might cause this decrease in pH. We are also to calculate the
extent of the increase in acidity this change would cause. Finally, we are to explain how coral reef survival
could be affected by the decrease in pH.
Analyze
Carbon dioxide gas dissolves in water to form carbonic acid, a weak diprotic acid. Coral reefs are composed
of CaCO3 that react with acids.
348 | Chapter 17
Solve
(a) By increasing the partial pressure of CO2 in the atmosphere the following equilibria are shifted to the
right (Le Châtelier’s principle):
CO2(g)
CO2(aq)
CO2(aq) + H2 O( )
H2CO3(aq)
H2CO3(aq)
H+(aq) + HCO3–(aq)
–
HCO3 (aq)
H+(aq) + CO32–(aq)
The acidity of the oceans increases (lower pH) because of increased ionization of the weak acid carbonic
acid.
(b) pH f – pH i  –0.77
– log[H + ]f –  –log[H + ]i   –0.77
log[H  ]f  log[H  ]i  0.77
log
[H  ]f
 0.77
[H + ]i
[H  ]f
 100.77  5.9
[H  ]i
[H  ]f  5.9[H  ]i
The acidity of the ocean will be about 6 times greater.
(c) Because coral reefs are composed of CaCO3 they are likely to dissolve to a greater extent as the pH
decreases.
CaCO3(s) + 2 H+(aq)  Ca2+(aq) + H2CO3(aq)
Think about It
Our answer to part b is consistent with our knowledge that one pH unit represents a tenfold difference in
[H+].
17.149. Collect and Organize
For three reactions of nitrogen and sulfur compounds, we are to complete the equations.
Analyze
All three reactants are covalent compounds. Reactions of covalent compounds are discussed on page 865 of
Chapter 17.
Solve
(a) SO3(g) + H2 O( )  H2SO4 ( )
(b) 3 NO2(g) + H2 O( )  2 HNO3 ( ) + NO(g)
(c) 4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g)
Think about It
The first two reactions show how acids are produced from nonmetal oxides.
17.150. Collect and Organize
We need to consider the steps of the Ostwald synthesis of nitric acid to determine which steps would have a
higher yield at higher pressure.
Analyze
By Le Châtelier’s principle, the yield of a reaction increases as pressure increases when the moles of gaseous
products are less than the moles of gaseous reactants in the balanced equation.
Equilibrium in the Aqueous Phase | 349
Solve
The steps of the Ostwald process are
4 NH3(g) + 5 O2(g)
4 NO(g) + 6 H2O(g)
2 NO(g) + O2(g)
2 NO2(g)
3 NO2(g) + H2 O( )
2 HNO3 ( ) + NO(g)
In the first step 9 moles of gaseous reactants are made into 10 moles of gaseous products. In the second step,
3 moles of gaseous reactants are made into two moles of gaseous product. In the third step, 3 moles of
gaseous reactant are made into one mole of gaseous product. Therefore, the yield of both steps 2 and 3
increases with increasing pressure.
Think about It
Increasing the pressure also likely increases the rate of the reaction because the concentration of the reactants
increases at high pressure.
17.151. Collect and Organize
We are asked which steps of the Ostwald synthesis of nitric acid would have a higher yield at higher
temperature.
Analyze
By Le Châtelier’s principle the yield of a reaction increases as temperature increases for endothermic
reactions.
Solve
The steps in the Ostwald process with their ∆H values calculated from the data in Appendix 4 are as follows:
4 NH3(g) + 5 O2(g)
4 NO(g) + 6 H2O(g)
H = [(4  90.3) + (6  –241.8)] – [(4  – 46.1) + (5  0)]
= –905.2 kJ
2 NO(g) + O2(g)
2 NO2(g)
H = (2  33.2) – [(2  90.3) + (1  0)]
= –114.2 kJ
3 NO2(g) + H2 O( )
2 HNO3 ( ) + NO(g)
H = [(2  –174.1) + (1  90.3)] – [(3  33.2) + (1  –285.8)]
= –71.7 kJ
All of these reactions are exothermic, so none of the steps has a higher yield at a higher temperature.
Think about It
The reactions, however, are run at 900˚C, which enhances the rate of the reaction.
17.152. Collect and Organize
We are asked to calculate H rxn for the reaction
SO2 ( g )  12 O2 ( g )  SO3 ( g )
Analyze
The H rxn equation is

 
 

H rxn  mol SO3  H f,SO3   mol SO 2  H f,SO2  mol O2  H f,O2 
We are given values for H fo for SO2(g) and SO3(g) of –296.8 kJ/mol and –395.7 kJ/mol, respectively.
350 | Chapter 17
Solve
H rxn  1 –395.7   1 –296.8   12  0 
 –98.9 kJ
Think about It
This reaction is exothermic.
17.153. Collect and Organize
After writing the equations corresponding to H f,SO2 and H f,SO3 , we are to apply Hess’s law to show how to
calculate H rxn for
2 SO2(g) + O2(g)  2 SO3(g)
Analyze
The chemical reactions corresponding to H f involve preparing the compound from the elements.
Solve
The balanced formation reactions for SO2 and SO3 are
1
8 S8 ( s )  O 2 (g )  SO 2 (g )
1
8
S8 ( s)  O 2 (g )  SO3 (g )
3
2
H f,SO2
H f,SO3
Reversing the first reaction, multiplying it by 2, and adding to the second reaction (also multiplied by 2)
allow us to calculate H rxn .
2 SO2(g)  14 S8(s) + 2 O2(g)
–2 H f,SO2
1
4
S8(g) + 3 O2(g)  2 SO3(g)
2 SO2(g) + O2(g)  2 SO3(g)
2 H f,SO2
2 H f,SO3 – 2 H f,SO2  H rxn
Think about It
Remember that Hess’s law is applicable to other state functions, such as S and G.
17.154. Collect and Organize
We are to consider the relative strengths of the hypothetical acid H 3NO4 and the strong acid HNO3.
Analyze
We are given that H3NO4 would have as part of its structure NO43–:
Solve
The H atoms on H3NO4 must be on the oxygen atoms so that the N atom maintains an octet in its Lewis
structure. In HNO3 the H atom is also bonded to an O atom. The difference between these compounds is the
number of oxygen atoms bound to the N atom. As shown in Figure 17.9 for sulfur oxyacids, we expect
H3NO4 to be a stronger acid because the negative charge is delocalized over more atoms in H 2NO4–
compared to NO3–.
Think about It
This is consistent with the greater acidity of H2SO4 compared to H2SO3.
Equilibrium in the Aqueous Phase | 351
17.155. Collect and Organize
We are asked to compare the solubility in water of SO 3 and NO2 with CO2 and whether the Henry’s law
constants for SO3 and NO3 are greater than that for CO2 (which is 3.5  10–2 M/atm).
Analyze
Carbon dioxide is a linear nonpolar molecule that forms dipole–induced dipole intermolecular interactions
when dissolved in water. To determine whether SO3 and NO2 are more or less soluble in water than CO2 we
must draw their Lewis structures and determine their molecular polarity.
Solve
SO3 is trigonal planar and therefore nonpolar like CO2:
SO3, like CO2, is nonpolar and forms dipole–induced
dipole interactions with water. These intermolecular
forces, however, are stronger for SO3 because SO3 is a
larger molecule. SO3, therefore, is slightly more soluble in
water than CO2.
NO2 is bent and therefore polar.
NO2 forms dipole – dipole interactions with water. These
are stronger than the water – CO2 intermolecular forces, so
NO2 is significantly more soluble in water compared to
CO2.
Yes, the corresponding constants for SO3 and NO2 are expected to be greater than the Henry’s law constant
for CO2.
Think about It
Of the three substances, NO2 is the most soluble in water.
17.156. Collect and Organize
For two reactions of sulfuric acid and one of nitric acid, we are to write balanced equations.
Analyze
All of these reactions are acid–base reactions involving the transfer of a proton from the acid to the base.
Solve
(a) HNO3(aq) + NH3(aq)  NO3–(aq) + NH4+(aq)
(b) H2SO4(aq) + NH3(aq)  NH4+(aq) + HSO4–(aq)
(c) H2SO4(aq) + H2 O( )  HSO4–(aq) + H3O+(aq)
HSO4–(aq) + H2 O( )
H3O+(aq) + SO42–(aq)
Think about It
Notice in reaction c that we indicate the first H on sulfuric acid is completely ionized by drawing a singleheaded arrow. The second H, however, is not 100% dissociated, which we indicate in the reaction with a
double-headed equilibrium arrow.
17.157. Collect and Organize
After drawing the Lewis structure for H2S2O3, we are to consider the acid properties of thiosulfuric acid
compared to H2SO4.
Analyze
We are given that H2S2O3 is isostructural with H2SO4. This means that they have their atoms arranged in the
same way with an S atom in H2S2O3 taking the place of one of the O atoms in H2SO4.
352 | Chapter 17
Solve
(a)
(b) When a less electronegative sulfur atom replaces an oxygen atom in the acid, the acidity decreases.
Therefore, H2S2O3 is less acidic than H2SO4.
Think about It
Thiosulfuric acid is indeed less acidic than sulfuric acid. H2S2O3 has pKa1 = 0.6 and pKa2 = 1.6 whereas
H2SO4 has pKa1 < 0 and pKa2 = 1.92. While pKa2 for H2S2O3 is higher than that of H2SO4, remember that the
first ionization constant in diprotic acids dominates, so H2SO4 is stronger as an acid than H2S2O3.
17.158. Collect and Organize
For the reaction of nitric acid with sulfuric acid to produce NO2, H3O+, and HSO4–, we are to determine
whether the reaction is a redox reaction and identify the acid and base and their conjugates by drawing the
Lewis structures of the reactants and products.
Analyze
(a) If the reaction is a redox reaction one reactant would be oxidized and one would be reduced. We can
determine this by looking at the oxidation states of the atoms in the reactants and products.
(b) The acid differs from its conjugate base by only one proton.
Solve
(a)
HNO3 is reduced.
(b)
Acid = H2SO4, conjugate base = HSO4– Base = HNO3 (the OH group), conjugate acid = H3O+
Think about It
Because we see a transfer of a proton from H2SO4 to HNO3, this reaction indicates that as pure substances
(not in water) H2SO4 is a stronger acid than HNO3.