Tutorial 4: Equilibrium reactions
Some of the problems below require one or more equilibrium constants or standard
state potentials. The following appendices, from the prescribed/recommended books,
will be of interest:
‐ Solubility Products
‐ Acid Dissociation Constants
‐ Standard Reduction Potentials
Equilibrium constants and thermodynamic data
Problem #1: At 25 °C the enthalpy change, ΔH°, for the ionization of trichloroacetic acid
is +6.3 kJ/mol and the entropy change, ΔS°, is +0.0084 kJ/mol‐K. What is the pKa of
trichloroacetic acid?
Answer: pKa = 0.664
Problem #2. Write equilibrium constant expressions for the following reactions.
a. NH3(aq) + HCl(aq) ↔ NH4+(aq) + Cl‐(aq)
b. PbI2(s) + S2‐(aq) ↔ PbS(s) + 2I‐(aq)
c Cd(EDTA)2‐(aq) + 4CN‐(aq) ↔ Cd(CN)42‐(aq) + EDTA4−(aq)
d. AgCl(s) + 2NH3(aq) ↔ Ag(NH3)2+(aq) + Cl‐(aq)
e. BaCO3(s) + H3O+(aq) ↔ Ba2+(aq) + H2CO3(aq) + H2O(l)
Balancing half‐reactions in acidic solution
Reminder: a half‐reaction MUST be balanced both for atoms and charge in order to be
correct. It is VERY easy to balance for atoms only, forgetting to check the charge.
DON'T FORGET TO CHECK THE CHARGE.
Here is the half‐reaction to be considered:
MnO4ˉ → Mn2+
It is to be balanced in acid solution.
Step One: Balance the atom being reduced/oxidized. In our example, there is already
one Mn on each side of the arrow, so this step is already done.
MnO4ˉ → Mn2+
Step Two: Balance the oxygens. Do this by adding water molecules (as many as are
needed) to the side needing oxygen. In our case, the left side has 4 oxygens, while the
right side has none, so:
MnO4ˉ → Mn2+ + 4H2O
Notice that, when the water is added, hydrogens also come along. There is nothing that
can be done about this; we'll take care of it in the next step. A common question is:
"Why can't I just add 4 oxygen atoms to the right side?" Quick answer: don't do it, it's
wrong. The "why" will be left to another day.
Step Three: Balance the hydrogens. Do this by adding hydrogen ions (as many as are
needed) to the side needing hydrogen. In our example, we need 8 (notice the water
molecule's formula, then consider 4 x 2 = 8).
8H+ + MnO4ˉ → Mn2+ + 4H2O
Step Four: Balance the total charge. This will be done using electrons. It is ALWAYS the
last step.
First, a comment: you do not need to look at the oxidation number for each atom. You
only need to look at the charge on the ion or molecule, then sum those up.
Left side of the reaction, total charge is +7. There are 8 H+, giving 8 x +1 = +8 and a minus
one from the permanganate. A very typical wrong answer for the left side is zero. The
person sees only the +1 and the ‐1, they forget the 8.
Right side of the reaction, total charge is +2. The water molecule is neutral (zero charge)
and the single Mn is +2.
5eˉ + 8H+ + MnO4ˉ → Mn2+ + 4H2O
Five electrons reduces the +7 to a +2 and the two sides are EQUAL in total charge. The
half‐reaction is now correctly balanced.
Practice Problems
Acidic solution
1) Re ‐‐‐> ReO2
Step One: Balance the atom being reduced/oxidized: Re ‐‐‐> ReO2
Step Two: Balance the oxygens: 2H2O + Re ‐‐‐> ReO2
Step Three: Balance the hydrogens: 2H2O + Re ‐‐‐> ReO2 + 4H+
Step Four: Balance the total charge: 2H2O + Re ‐‐‐> ReO2 + 4H+ + 4eˉ
2) Cl2 ‐‐‐> HClO
Cl2 ‐‐‐> 2HClO (chlorine is balanced)
2H2O + Cl2 ‐‐‐> 2HClO (now there are 2 oxygens on each side)
2H2O + Cl2 ‐‐‐> 2HClO + 2H+ (2H in HClO plus 2H+ makes 4 hydrogens)
2H2O + Cl2 ‐‐‐> 2HClO + 2H+ + 2eˉ (zero charge on the left; +2 from the hydrogen ions so
2 electrons gives the ‐2 charge required to make zero on the right)
3) NO3ˉ ‐‐‐> HNO2
4) H2GeO3 ‐‐‐> Ge
5) H2SeO3 ‐‐‐> SeO42ˉ
6) Au ‐‐‐> Au(OH)3
7) H3AsO4 ‐‐‐> AsH3
8) H2MoO4 ‐‐‐> Mo
9) NO ‐‐‐> NO3ˉ
10) H2O2 ‐‐‐> H2O
Balancing half‐reactions in basic solution
Here is the half‐reaction to be considered:
PbO2 ‐‐‐> PbO
Step One to Four: Balance the half‐reaction AS IF it were in acid solution. I hope you
got that. The half‐reaction is actually in basic solution, but we are going to start out as if
it were in acid solution. Here are the 4 acid steps:
1) Balance the atom being reduced/oxidized.
2) Balance the oxygens (using H2O).
3) Balance the hydrogens (using H+).
4) Balance the charge.
When you do that to the above half‐reaction, you get:
2eˉ + 2H+ + PbO2 ‐‐‐> PbO + H2O
+
Step Five: Convert all H to H2O. Do this by adding OHˉ ions to both sides. The side with
the H+ will determine how many hydroxides to add. In our case, the left side has 2
hydrogen ions, while the right side has none, so:
2eˉ + 2H2O + PbO2 ‐‐‐> PbO + H2O + 2OHˉ
Notice that, when the two hydroxide ions on the left were added, they immediately
reacted with the hydrogen ion present. The reaction is:
H+ + OHˉ ‐‐‐> H2O
Step Six: Remove any duplicate molecules or ions. In our example, there are two water
molecules on the left and one on the right. This means one water molecule may be
removed from each side, giving:
2eˉ + H2O + PbO2 ‐‐‐> PbO + 2OHˉ
The half‐reaction is now correctly balanced.
By the way, notice the 2OHˉ. Be careful to read that as two hydroxide ions (2 OHˉ) and
NOT twenty hydride ions (2O Hˉ). People have been known to do that.
Practice problems
Base solution
1) NiO2 ‐‐‐> Ni(OH)2
Step One: Balance the half‐reaction AS IF it were in acid solution: 2eˉ + 2H+ + NiO2 ‐‐‐>
Ni(OH)2
Step Two: Convert all H+ to H2O: 2eˉ + 2H2O + NiO2 ‐‐‐> Ni(OH)2 + 2OHˉ
Step Three: Remove any duplicate molecules or ions: none to be removed in this
example. Other problems below will have duplicates.
2) BrO4ˉ ‐‐‐> Brˉ
Step One: Balance the half‐reaction AS IF it were in acid solution: 8eˉ + 8H+ + BrO4ˉ ‐‐‐>
Brˉ + 4H2O
Step Two: Convert all H+ to H2O: 8eˉ + 8H2O + BrO4ˉ ‐‐‐> Brˉ + 4H2O + 8OHˉ
Step Three: Remove any duplicate molecules or ions: 8eˉ + 4H2O + BrO4ˉ ‐‐‐> Brˉ +
8OHˉ
3) SbO3ˉ ‐‐‐> SbO2ˉ
4) Cu2O ‐‐‐> Cu
5) S2O32ˉ ‐‐‐> SO32ˉ
6) Tl+ ‐‐‐> Tl2O3
7) Al ‐‐‐> AlO2ˉ
8) Sn ‐‐‐> HSnO2ˉ
9) CrO42ˉ ‐‐‐> Cr(OH)3
10) HfO(OH)2 ‐‐‐> Hf
Balancing redox reactions in acidic solution
Points to remember:
1) Electrons NEVER appear in a correct, final answer. In order to get the electrons in
each half‐reaction equal, one or both of the balanced half‐reactions will be multiplied by
a factor.
2) Duplicate items are always removed. These items are usually the electrons, water and
hydrogen ion.
Example #1: ClO3ˉ + SO2 ‐‐‐> SO42ˉ + Clˉ
Solution:
1) Split into unbalanced half‐reactions (oxidation and reduction):
ClO3ˉ ‐‐‐> Clˉ (Reduction)
SO2 ‐‐‐> SO42ˉ (Oxidation)
2) Balance the half‐reactions:
6eˉ + 6H+ + ClO3ˉ ‐‐‐> Clˉ + 3H2O
2H2O + SO2 ‐‐‐> SO42ˉ + 4H+ + 2eˉ
3) Make the number of electrons equal:
6eˉ + 6H+ + ClO3ˉ ‐‐‐> Clˉ + 3H2O
6H2O + 3SO2 ‐‐‐> 3SO42ˉ + 12H+ + 6eˉ <‐‐‐ multiplied through by a factor of 3
4) Add the two half‐reactions for the final answer:
ClO3ˉ + 3H2O + 3SO2 ‐‐‐> 3SO42ˉ + Clˉ + 6H+
Note that items duplicated on each side were cancelled out. The duplicates are 6eˉ,
3H2O, and 6H+
Example #2: H2S + NO3ˉ ‐‐‐> S8 + NO
Solution:
1) The unbalanced half‐reactions:
H2S ‐‐‐> S8
NO3ˉ ‐‐‐> NO
2) balance each half‐reaction:
8H2S ‐‐‐> S8 + 16H+ + 16eˉ
3eˉ + 4H+ + NO3ˉ ‐‐‐> NO + 2H2O
3) Make the number of electrons equal:
24H2S ‐‐‐> 3S8 + 48H+ + 48eˉ <‐‐‐ multiplied by a factor of 3
48eˉ + 64H+ + 16NO3ˉ ‐‐‐> 16NO + 32H2O <‐‐‐ multiplied by a factor of 16
Note that 16 and 3 have no common factors except 1, so both 16 and 3 had to be used
to obtain the lowest common multiple of 48 for the number of electrons.
4) Add:
24H2S + 16H+ + 16NO3ˉ ‐‐‐> 3S8 + 16NO + 32H2O
Comment: removing a factor of 8 does look tempting, doesn't it? However, the three in
front of the S8 (or the five in the next problem) makes it impossible. Also, note that
duplicates of 48 electrons and 48 hydrogen ions were removed.
Example #3: MnO4ˉ + H2S ‐‐‐> Mn2+ + S8
Example #4: Cu + SO42ˉ ‐‐‐> Cu2+ + SO2
Example #5: MnO4ˉ + CH3OH ‐‐‐> HCOOH + Mn2+
Balancing redox reactions in basic solution
Example #1: NH3 + ClOˉ ‐‐‐> N2H4 + Clˉ
Solution:
1) The two half‐reactions, balanced as if in acidic solution:
2NH3 ‐‐‐> N2H4 + 2H+ + 2eˉ
2eˉ + 2H+ + ClOˉ ‐‐‐> Clˉ + H2O
2) Electrons already equal, convert to basic solution:
2OHˉ + 2NH3 ‐‐‐> N2H4 + 2H2O + 2eˉ
2eˉ + 2H2O + ClOˉ ‐‐‐> Clˉ + H2O + 2OHˉ
3) The final answer:
2HN3 + ClOˉ ‐‐‐> N2H4 + Clˉ + H2O
Notice that no hydroxide appears in the final answer. That means this is a base‐
catalyzed reaction. For the reaction to occur the solution must be basic and hydroxide IS
consumed. It is just regenerated in the exact same amount, so it cancels out in the final
answer.
Example #2: Au + O2 + CNˉ ‐‐‐> Au(CN)2ˉ + H2O2
Solution:
1) the two half‐reactions, balanced as if in acidic solution:
2CNˉ + Au ‐‐‐> Au(CN)2ˉ + eˉ
2eˉ + 2H+ + O2 ‐‐‐> H2O2
2) Make electrons equal, convert to basic solution:
4CNˉ + 2Au ‐‐‐> 2Au(CN)2ˉ + 2eˉ <‐‐‐ multiplied by a factor of 2
2eˉ + 2H2O + O2 ‐‐‐> H2O2 + 2OHˉ
3) The final answer:
4CNˉ + 2Au + 2H2O + O2 ‐‐‐> 2Au(CN)2ˉ + H2O2 + 2OHˉ
Comment: the CNˉ is neither reduced nor oxidized, but it is necessary for the reaction.
For example, you might see this way of writing the problem:
Au + O2 ‐‐‐> Au(CN)2ˉ + H2O2
Notice that CNˉ does not appear on the left side, but does so on the right. Since you
MUST balance the equation, that means you are allowed to use CNˉ in your balancing.
An important point here is that you know the cyanide polyatomic ion has a negative one
charge.
Example #3: Brˉ + MnO4ˉ ‐‐‐> MnO2 + BrO3ˉ
Example #4: AlH4ˉ + H2CO ‐‐‐> Al3+ + CH3OH
Example #5: Se + Cr(OH)3 ‐‐‐> Cr + SeO32ˉ
How do you balance the reaction if there is only one reactant or product?
The key point to remember: repeat the single substance when you write the two half‐
reactions needed to balance the reaction. Then balance the half‐reactions as you
normally would. When you add the two half‐reactions together at the end, remember to
add items that are the same (for example, the P4 in problem three).
Here are five examples:
1) ClOˉ ‐‐‐> Clˉ + ClO3ˉ [basic solution]
2) IO3ˉ + Iˉ ‐‐‐> I2 [acidic solution]
3) P4 ‐‐‐> HPO32ˉ + PH3 [basic solution]
4) NO2 ‐‐‐> NO3ˉ + NO [acid solution]
5) Cl2 ‐‐‐> Clˉ + ClO3ˉ [basic solution]
Partial Solution to Example One:
1) Half‐reactions:
ClOˉ ‐‐‐> Clˉ
ClOˉ ‐‐‐> ClO3ˉ
2) Balance:
2eˉ + H2O + ClOˉ ‐‐‐> Clˉ + 2OHˉ
4OHˉ + ClOˉ ‐‐‐> ClO3ˉ + 2H2O + 4eˉ
Partial Solution to Example Two:
1) Half‐reactions:
IO3ˉ ‐‐‐> I2
Iˉ ‐‐‐> I2
2) Balance:
10eˉ + 12H+ + 2IO3ˉ ‐‐‐> I2 + 6H2O
2Iˉ ‐‐‐> I2 + 2eˉ
More Redox Practice Questions
Problem #1. Balance the following redox equations, supplying H+, OH‐, and H2O as
required:
(a)
V2+ + V(OH)4+ = VO2+
(basic medium)


2
2
(i) V  2OH  VO  H 2 O  2e
(ii) 2V (OH ) 4  2e   2VO 2  4OH   2 H 2 O
(i) + (ii) V 2  2V (OH ) 4  3VO 2  2OH   3H 2 O
(b)
Zn + H2MoO4 = Zn2+ + Mo3+
(i) Zn  Zn 2   2e 
(ii) MoO42  8H   3e   Mo 3  4 H 2 O
(acid medium)
3(i) + 2(ii) 3Zn  2MoO42  16 H   3Zn 2  2Mo 3  8H 2 O
(c)
P4 + HOCl
= H3PO4 + Cl‐
(i) P4  16 H 2 O  4 H 3 PO4  20 H   20e 
(acid medium)
(ii) HOCl  H   2e   Cl   H 2 O
(i) + 10(ii) P4  6 H 2 O  10 HOCl  4 H 3 PO4  10 H   10Cl 
(d)
Cr(OH)4‐ + BrO‐ = CrO42‐ + Br‐
(i) BrO   H 2 O  2e   Br   2OH 
(basic medium)
(ii) Cr (OH ) 4  4OH   CrO42  4 H 2 O  3e 
3(i) + 2(ii) 3BrO   2Cr (OH ) 4  2OH   3Br   2CrO42  5H 2 O
(e)
C2O42‐ + MnO4‐ = CO2 + Mn2+
(i) MnO4  8 H   5e   Mn 2  4 H 2 O
(acid medium)
(ii) C 2 O42  2CO2  2e 
2(i) + 5(ii) 2 MnO4  16 H   5C 2 O42  2 Mn 2  10CO2  8H 2 O
Problem #2. Calculate the potential for the following redox reaction:
2Fe3+(aq) + Sn2+(aq) ↔ Sn4+(aq) + 2Fe2+(aq)
for a solution in which [Fe3+] = 0.050 M, [Fe2+] = 0.030 M, [Sn2+] = 0.015 M and [Sn4+] =
0.020 M. Answer: E = 0.629 V
Problem # 3. Calculate the standard state potential (E0) and the equilibrium constant (K)
for each of the following redox reactions. Assume that [H3O+] is 1.0 M for acidic
solutions, and that [OH–] is 1.0 M for basic solutions.
a. MnO4‐(aq)+ H2SO3(aq) ↔ Mn2+(aq) + SO42‐(aq) Acidic solution
b. IO3‐(aq) + I‐(aq) ↔ I2(s) acidic solution
c. ClO‐(aq) + I‐(aq) ↔ IO3‐(aq) + Cl‐(aq) basic solution
Notes: These redox reactions are not balanced. You will need to balance them before
calculating their standard state potentials.
Problem #4. A solution contains 0.100 M Ce3+, 1.00 * 10‐4 M Ce4+, 1.00 * 10‐4 M Mn2+,
0.100 M MnO4‐, and 1.00 M HClO4.
(a)
Write a balanced net reaction that can occur between the species in this
solution.
o
o
E Ce
 1.70V , and E MnO
 1.507V
4

/ Ce 3 
/ Mn 2 
4
(b)
Calculate ΔGo and K for the reaction.
Answer: ΔGo = ‐93 kJ; K = 2 x 1016
(c)
Calculate E for the conditions given above.
Answer: E = ‐0.02 V
(d)
Calculate ΔG for the conditions given above.
At what pH would the concentrations of Ce4+, Ce3+, Mn2+, and MnO4‐ listed above be in
equilibrium at 298 K?
Answer: ΔG = + 10 kJ; pH = 0.21
Problem #5. Use the following standard‐state cell potentials to calculate the complex
formation equilibrium constant for the Zn(NH3)42+ complex ion.
Zn(NH3)42+ + 2e‐ ⇌ Zn + 4NH3
Eored = ‐1.04 V
Zn2+ + 2e‐ ⇌ Zn
Eored = ‐0.7628 V
Answer: K = 2.92 x 109
Problem #6. One analytical method for determining the concentration of sulfur is to
oxidize it to SO42‐ and then precipitate it as BaSO4 by adding BaCl2. The mass of the
resulting precipitate is proportional to the amount of sulfur in the original sample. The
accuracy of this method depends on the solubility of BaSO4, the reaction for which is
shown here.
BaSO4(s) ↔ Ba2+(aq) + SO42‐(aq)
How do the following affect the solubility of BaSO4 and, therefore, the accuracy of the
analytical method?
a. decreasing the solution’s pH
b. adding more BaCl2
c. increasing the volume of the solution by adding H2O
Problem #7. Write a charge balance equation and mass balance equations for the fol‐
lowing solutions. Some solutions may have more than one mass balance equation
(Please see examples in DC Harris, Quantitative Chemical Analysis, 8th ed, pp 150‐152
or the corresponding section from any earlier/later edition of this book)
a. 0.10 M NaCl
b. 0.10 M HCl
c. 0.10 M HF
d. 0.10 M NaH2PO4
e. MgCO3 (saturated solution)
f. 0.10 M Ag(CN)2–
g. 0.10 M HCl and 0.050 M NaNO2
Calculate the pH of a Saturated Solution When Given the Ksp
These problems do not require a concentration to be given. Knowing the Ksp and That
the solution is saturated is sufficient.
Warning: you need to know about Ksp AND acid base ideas to solve this problem type. If
you lack one or the other of these skills, just be aware you just might struggle a little in
your understanding of this problem type.
To solve the problem, we must first calculate the [OHˉ]. To do this, we will use the Ksp
expression and then, at the end, we will use acid base concepts to get the pH.
Final Note: Ksp are almost always given at 25.0°C in reference sources. All problems in
this tutorial are taken to be at 25.0°C. If you were to see a problem where the specified
temperature was different, it's probably just that the reference source gave a Ksp that,
for whatever reason, was not at 25.0°C.
Problem #1: Calculate the pH of a saturated solution of AgOH, Ksp = 2.0 x 10ˉ8
Solution:
AgOH ⇌ Ag+ + OHˉ
Ksp = [Ag+] [OHˉ]
2.0 x 10ˉ8 = (s) (s)
s = 1.4 x 10ˉ4 M (this is the [OHˉ])
pOH = ‐ log [OHˉ] = ‐ log 1.4 x 10ˉ4 = 3.85 (I actually took the negative log of 1.414 . . . x
10ˉ4; I did not use just 1.4)
Because pH + pOH = 14, we have pH = 14 ‐ 3.85 = 10.15
Problem #2: Calculate the pH of a saturated solution of Cu(OH)2, Ksp = 1.6 x 10ˉ19
Answer: pH = 7.835
Problem #3: Calculate the pH of a saturated solution of Mg(OH)2, Ksp = 5.61 x 10ˉ12
Answer: pH = 10.350
Problem #4: Calculate the pH of a saturated solution of Ba(OH)2, Ksp = 5.0 x 10ˉ3.
Answer: pH = 13.33
The following three problems are all of the form X(OH)2. These are the ones most
commonly asked on tests and in worksheets. Calculate the pH of a saturated solution of:
Problem #5: Ca(OH)2, Ksp = 7.9 x 10ˉ6 (pH = 12.10)
Problem #6: Mn(OH)2, Ksp = 4.6 x 10ˉ14 (no answer provided)
Problem #7: Ni(OH)2, Ksp = 2.8 x 10ˉ16 (no answer provided)
Calculate the Ksp of a Saturated Solution When Given the pH
These problems do not require a concentration to be given. Knowing the pH and the fact
that the solution is saturated is sufficient.
To solve the problem, we must first calculate the [OHˉ]. To do this, we will use the pH
and acid base concepts. Then, we will use the Ksp expression to calculate the Ksp.
Example #1: A saturated solution of Mg(OH)2 is prepared. The pH of the solution is
10.17. What is the Ksp for this compound?
Solution:
1) The chemical equation:
Mg(OH)2 ⇌ Mg2+ + 2OHˉ
2) The Ksp expression:
Ksp = [Mg2+] [OHˉ]2
3) Use the pH to get the pOH:
14.00 ‐ 10.17 = 3.83
4) Use the pOH to get the [OHˉ]:
[OHˉ] = 10ˉpOH = 10ˉ3.83 = 1.479 x 10ˉ4 M
5) From the chemical equation, we note that the [Mg2+] is half the value of the [OHˉ],
therefore:
[Mg2+] = 1.479 x 10ˉ4 M divided by 2 = 7.395 x 10ˉ5 M
6) We now have the necessary values to put into the Ksp expression:
Ksp = (7.395 x 10ˉ5) (1.479 x 10ˉ4)2
Ksp = 1.62 x 10ˉ12
Note: the book value for the Ksp of Mg(OH)2 is 5.61 x 10ˉ12. The reason for the difference
is that teachers like to change the values slightly so as to not allow a student to simply
look up the correct answer and claim they did all their work on the calculator.
What is the minimum pH required for precipitation?
Problem #1: What is the minimum pH at which Cr(OH)3 will precipitate? Ksp of Cr(OH)3 is
6.70 x 10ˉ31
Solution:
1) Write the dissociation equation:
Cr(OH)3 ⇌ Cr3+ + 3OHˉ
2) Write the Ksp expression:
Ksp = [Cr3+] [OHˉ]3
3) Plug into the Ksp expression:
6.70 x 10ˉ31 = (s) (3s)3
4) Solve for [OHˉ]:
[Cr3+] = s = 1.255 x 10ˉ8 M (I kept some guard digits)
[OHˉ] = 3s = 3.765 x 10ˉ8 M
5) Calculate the pH given the [OHˉ]:
pOH = ‐log 3.765 x 10ˉ8 = 7.424
pH = 14 ‐ pOH = 14 ‐ 7.424 = 6.576
Note that even in a slightly acidic environment, Cr(OH)3 will start to precipitate.
Problem #2: What is the minimum pH at which Cr(OH)3 will precipitate if the solution
has [Cr3+] = 0.0670 M? Ksp of Cr(OH)3 is 6.70 x 10ˉ31
Answer: pH = 4.333
Problem #3: At what pH will Al(OH)3(s) begin to precipitate from 0.10 M AlCl3? The Ksp of
Al(OH)3 is 1.90 x 10ˉ33
Answer: pH = 3.426
Buffers
Problem #1: If an acetate buffer solution was going to be prepared by neutralizing
HC2H3O2 with 0.10 M NaOH, what volume (in mL) of 0.10 M NaOH would need to be
added to 10.0 mL of 0.10 M HC2H3O2 to prepare a solution with pH = 5.50?
Solution:
Comment: In doing the salt (sodium acetate) and the acid (acetic acid), I'm going to use
moles rather than molarity. Since everything occurs in the same volume of solution, the
ratio of salt moles to acid moles is the same as the ratio of molarities. Besides, we don't
know the final molarities since we are adding an unknown volume of NaOH solution.
1) We need to know the initial moles of acetic acid in the solution:
(0.010 L) (0.1 mol/L) = 0.001 mol
2) Let us insert values into the H‐H equation:
pH = pKa + log (salt/acid)
5.50 = 4.752 + log (x / (0.001 ‐ x))
4.752 is the pKa of acetic acid
x is the moles of sodium acetate produced by the NaOH reacting
0.001 ‐ x is the amount of acetic acid remaining in solution.
The moles of acetate will give us moles of NaOH since there is a 1:1 molar ratio between
the two.
3) Continue solving:
log (x / 0.001 ‐ x) = 0.748
(x / 0.001 ‐ x) = 100.748
(x / 0.001 ‐ x) = 5.598
x = 5.598 x 10‐3 ‐ 5.598x
6.598x = 5.598 x 10‐3
x = 8.5 x 10‐4 moles
4) Let us determine the volume of NaOH required:
8.5 x 10‐4 mol divided by 0.1 mol/L = 8.5 x 10‐3 L = 8.5 mL
Problem #2:
A buffer is prepared containing 1.00 molar acetic acid and 1.00 molar sodium acetate.
What is its pH?
Answer: pH = 4.752
Problem #3:
A buffer is prepared containing 0.800 molar acetic acid and 1.00 molar sodium acetate.
What is its pH?
Answer: 4.849
Problem #4:
A buffer is prepared containing 1.00 molar acetic acid and 0.800 molar sodium acetate.
What is its pH?
Answer: 4.655
Problem #5:
A buffer is prepared containing 1.00 molar anisic acid and 1.00 molar sodium anisate.
What is its pH?
Answer: pH = 4.471
Problem #6:
A buffer is prepared containing 0.600 molar anisic acid and 0.800 molar sodium anisate.
What is its pH?
Answer: pH = 4.596
Problem #7:
A buffer is prepared containing 0.700 molar anisic acid and 0.300 molar sodium anisate.
What is its pH?
The Ka for anisic acid is 3.38 x 10ˉ5
Answer: pH = 4.103
Problems using the Henderson‐Hasselbalch equation: a base and its salt
Problem #1:
A buffer is prepared containing 1.00 molar ammonia and 1.00 molar ammonium
chloride. What is its pH?
Solution
Solving problems with bases has (potentially) one more step that when working with
acids. This happens if you are given the Kb for the base.
You can't use the Kb for the base, you have to use the Ka for the conjugate acid (and
thence to the pKa.)
In this case, we know the Kb for ammonia; it is 1.77 x 10ˉ5.
We need to know the Ka for the ammonium ion (it is the conjugate acid). Form that we
get the pKa. We do that as follows:
Kw = KaKb
1.00 x 10ˉ14 = (x) (1.77 x 10ˉ5)
x = 5.65 x 10ˉ10
pKa = ‐ log Ka = ‐ log 5.65 x 10ˉ10 = 9.248
Now, we are ready to insert into the HH equation:
pH = 9.248 + log (1.00 / 1.00)
Since the log of 1 is zero, we have pH = 9.248
Before solving #2 and #3, make sure you know which the acid is and which the base is.
The acid is the substance with the proton. It goes in the denominator of the HH
equation. The base is the substance without the proton. It goes in the numerator of the
HH equation.
Problem #2:
A buffer is prepared containing 0.800 molar ammonia and 1.00 molar ammonium
chloride. What is its pH?
Answer: 9.151
Problem #3:
A buffer is prepared containing 1.00 molar ammonia and 0.800 molar ammonium
chloride. What is its pH?
Answer: 9.345
Problem #4:
A buffer is prepared containing 1.00 molar nicotine and 1.00 molar nicotine
hydrochloride. What is its pH?
Answer: pH = 8.021
Problem #5:
A buffer is prepared containing 0.600 molar nicotine and 0.800 molar nicotine
hydrochloride. What is its pH?
Answer: pH = 8.021 ‐ 0.125 = 7.896
Problem #6:
A buffer is prepared containing 0.700 molar nicotine and 0.300 molar nicotine
hydrochloride. What is its pH?
The Kb for nicotine is 1.05 x 10ˉ6
A brief word about nicotine hydrochloride: The formula for nicotine is C10H14N2. The salt
is nicotine hydrochloride and its formula looks like this: C10H15N2+Clˉ.
Compare this to ammonia (formula NH3) and ammonum chloride (formula NH4+Clˉ).
The Kb for nicotine is 1.05 x 10ˉ6; the Ka for nicotine hydrochloride is 9.524 x 10ˉ9.
Answer: 8.389
Other Henderson‐Hasselbalch Problems
A reminder of the Henderson‐Hasselbalch equation:
Problem #1: pKa for phenophthalein is 9.3 at room temp.
a) Calculate ratio of its anionic form to acid form at pH 8.2 and at pH 10.
b) Using these values, explain the colour change within this pH range.
Problem #2: Aspirin has a pKa of 3.4. What is the ratio of Aˉ to HA in:
(a) the blood (pH = 7.4)
(b) the stomach (pH = 1.4)
General comment about the solutions: You have to find the ratio between Aˉ and HA so
the concentrations are not needed
Problem #3: Calculate the pH of the solution that results from the addition of 0.040
moles of HNO3 to a buffer made by combining 0.500 L of 0.380 M HC3H5O2 ( Ka = 1.30 x
10ˉ5) and 0.500 L of 0.380 M NaC3H5O2
Assume addition of the nitric acid has no effect on volume.
Answer: pH = 4.700
Problem #4: You need to produce a buffer solution that has a pH of 5.27. You already
have a solution that contains 10.0 mmol (millimoles) of acetic acid. How many
millimoles of sodium acetate will you need to add to this solution? The pKa of acetic acid
is 4.75.
Answer: 33.1 millimoles of sodium acetate
Problem #5: What is the pH when 25.0 mL of 0.200 M of CH3COOH has been titrated
with 35.0 mL of 0.100 M NaOH?
Answer: pH = 5.120
Problem #6. If an acetate buffer solution was going to be prepared by neutralizing
HC2H3O2 with 0.10 M NaOH, what volume (in mL) of 0.10 M NaOH would need to be
added to 10.0 mL of 0.10 M HC2H3O2 to prepare a solution with pH = 5.50?
Answer: 8.5 mL
HINT: In doing the salt (sodium acetate) and the acid (acetic acid), use moles (n) rather
than molarity (M) for the H‐H equation. Since everything occurs in the same volume of
solution, the ratio of salt moles to acid moles (i.e. nA‐/nHA) is the same as the ratio of
molarities (i.e. [A‐]/[HA]. Besides, we don't know the final molarities since we are adding
an unknown volume of NaOH solution.
Problem #7: A beaker with 100.0 mL of an acetic acid buffer with a pH of 5.000 is
sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is
0.1000 M. A student adds 7.300 ml of a 0.3600 M HCl solution to the beaker. How much
will the pH change? The pKa of acetic acid is 4.752.
Answer: pH = 4.532
Problem #8. Calculate the pH of a buffer that is 0.10 M in KH2PO4 and 0.050 M in
Na2HPO4. What is the pH after adding 5.0 mL of 0.20 M HCl to 0.10 L of this buffer (Ka
for H2PO4– = 6.32 × 10–8)?
Note: The equilibrium constant for the reaction between H+ and HPO4- is 1.59 × 107.
Answer: See Notes Chapter 6.
Bonus Question
a) Calculate the pH of the following acid–base buffers.
i. 100 mL of 0.025 M formic acid and 0.015 M sodium formate
ii. 50.00 mL of 0.12 M NH3 and 5.30 mL of 1.0 M HCl
iii. 5.00 g of Na2CO3 and 5.00 g of NaHCO3 diluted to 100 mL
b) Calculate the pH of the buffers in a) after adding 5.0 mL of 0.10 M HCl. Be sure to
state and justify any assumptions you make in solving the problems.
c) Calculate the pH of the buffers in a) after adding 5.0 mL of 0.10 M NaOH. Be sure to
state and justify any assumptions you make in solving the problems.
Ionic Strength
Calculate the ionic strength of the following solutions
a. 0.050 M NaCl
b. 0.025 M CuCl2
c. 0.10 M Na2SO4