17.1 The Common-Ion Effect
• “Whenever a weak electrolyte and a strong electrolyte
containing a common ion are together in solution, the
weak electrolyte ionizes less than it would if it were alone
in solution.”
• The equilibrium constant does not change.
• The relative concentrations or reactant and product in the
equilibrium expression change.
• Changes follow the Le Chatelier Principle.
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Initial
Concentration (M)
Change in
Concentration (M)
Equilibrium
Concentration (M)
CH3COOH
(M)
0.30
H+ (M)
CH3COO– (M)
0
0.30
–x
+x
+x
0.30 – x
x
0.30 + x
4) Ka = [H+][CH3COO–] / [CH3COO–] = 1.8 × 10–5
1.8 × 10–5 = (x)(0.3+x)/(0.30-x)
Assume that adding and substracting x from 0.30 will not change
0.30 enough to matter and the equation becomes
1.8 × 10–5 = (x)(0.30)/(0.30) which results in x = [H+] = 1.8 × 10–5
So pH = -log[H+] = 4.74
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Aqueous
Equilibria
17.2 Buffers
• Solutions of a weak conjugate
acid–base pair that resist
drastic changes in pH are
called buffers.
• These solutions contain
relatively high concentrations
(10 −3 M or more)
of both the acid and base.
Their concentrations are
approximately equal.
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Ways to Make a Buffer
1) Mix a weak acid and a salt of its conjugate base or
a weak base and a salt of its conjugate acid.
(CH3COOH with NaCH3COO or NH3 with NH4Cl )
2) Add strong acid and partially neutralize a weak
base or add strong base and partially neutralize a
weak acid.
CH3COOH(aq) + OH- ⇌ H2O(l) + CH3COO–(aq)
For laboratory work, prepackaged buffers at specific pH
may be purchased.
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Aqueous
Equilibria
Effect of Acetate on the Acetic Acid
Equilibrium
• Acetic acid is a weak acid:
➢ CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)
• Sodium acetate is a strong electrolyte:
➢ NaCH3COO(aq) → Na+(aq) + CH3COO–(aq)
• The presence of acetate from sodium acetate in an
acetic acid solution will shift the equilibrium,
according to LeChâtelier’s Principle:
➢ CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)
Aqueous
Equilibria
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Sample Exercise 17.1 Calculating the pH When a
Common Ion Is Involved
What is the pH of a solution made by adding 0.30 mol of
acetic acid and 0.30 mol of sodium acetate to enough
water to make 1.0 L of solution?
NaCH3COO(aq) → Na+(aq) + CH3COO–(aq)
0.3 M
0.3 M
0.3 M
CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)
Ka = [H3O+][CH3COO–] / [CH3COO–] = 1.8 × 10–5
Aqueous
Equilibria
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How a Buffer Works
• Adding a small amount of acid or base only slightly
neutralizes one component of the buffer, so the pH
doesn’t change much.
Aqueous
Equilibria
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Calculating the pH of a Buffer
• For a weak acid: Ka = [H+][A–]/[HA]
• Take –log of both sides:
–log Ka = –log[H+] + –log([A–]/[HA])
• Rearrange:
–log[H+] = –log Ka +log([A–]/[HA])
• Which is:
pH = pKa + log([A–]/[HA])
• This equation is known as the Henderson–Hasselbalch
equation. This applies only to buffers.
Aqueous
Equilibria
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Sample Exercise 17.3 Calculating the pH of a Buffer
What is the pH of a buffer that is 0.12 M in lactic acid
[CH3CH(OH)COOH, or HC3H5O3] and 0.10 M in sodium
lactate [CH3CH(OH)COONa or NaC3H5O3]? For lactic acid,
Ka = 1.4 × 10–4. Lactic acid: HA
Sodium lactate: MA; MA → M+ + A- (fully dissociates)
pH = pKa + log([A–]/[HA]) (Handerson-hasselbach equation assumes x is small
compared to 0.12 M and 0.10 M)
pH = -log(1.4 × 10–4 )+ log((0.10)/(0.12)) = 3.85 +(-0.08) = 3.77
Initial
Concentration (M)
Change in
Concentration (M)
Equilibrium
Concentration (M)
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HA (M)
0.12
H+ (M)
0
A– (M)
0.10
–x
+x
+x
0.12 – x
x
Aqueous
0.10 + x Equilibria
Sample Exercise 17.4 Calculating pH When the
Henderson–Hasselbalch Equation May Not Be Accurate
Calculate the pH of a buffer that initially contains
1.00 × 10–3 M CH3COOH and 1.00 × 10–4 M
CH3COONa in the following two ways: (i) using the
Henderson–Hasselbalch equation and (ii) making
no assumptions about quantities (which means
you will need to use the quadratic equation). The
Ka of CH3COOH is 1.80 × 10–5.
Aqueous
Equilibria
© 2015 Pearson Education, Inc.
Calculate the pH of a buffer that initially contains 1.00 × 10–3 M CH3COOH and
1.00 × 10–4 M CH3COONa in the following two ways: (i) using the Henderson–
Hasselbalch equation and (ii) making no assumptions about quantities (which
means you will need to use the quadratic equation). The Ka of CH3COOH is 1.80
× 10–5.
pH = pKa + log([A–]/[HA])
pKa = –log Ka = –log (1.80 × 10–5 ) = 4.742
Aqueous
Equilibria
© 2015 Pearson Education, Inc.
pH = pKa + log([A–]/[HA])
pKa = –log Ka = –log (1.80 × 10–5 ) = 4.742
Aqueous
Equilibria
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Titrations of Polyprotic Acids
• Using the Henderson–
Hasselbalch equation,
we can see that half way
to each equivalence point
gives us the pKa for
that step.
the concentrations of the each acid and base conjugate pairsAqueous
Equilibria
were identical for each equilibrium, log (1) = 0 and so pH = pKa.
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Solubility Equilibria
• Because ionic compounds are strong
electrolytes, they dissociate completely to the
extent that they dissolve.
• When an equilibrium equation is written, the
solid is the reactant and the ions in solution
are the products.
• The equilibrium constant expression is called
the solubility-product constant. It is
represented as Ksp.
Aqueous
• Constant for a solute at a given temperature.
Equilibria
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Solubility Product
➢For example:
BaSO4(s) ⇌ Ba2+(aq) + SO42–(aq)
➢The equilibrium constant expression is
Ksp = [Ba2+][SO42−]
➢Another example:
Ba3(PO4)2(s) ⇌ 3 Ba2+(aq) + 2 PO43–(aq)
➢The equilibrium constant expression is
Ksp = [Ba2+]3[PO43−]2
Aqueous
Equilibria
© 2015 Pearson Education, Inc.
Sample Exercise 17.10 Writing Solubility-Product (Ksp)
Expressions
Write the expression for the solubility-product constant for
CaF2, and look up the corresponding Ksp value in Appendix D.
Solution
Analyze We are asked to write an equilibrium-constant expression for the process by which CaF2
dissolves in water.
Plan We apply the general rules for writing an equilibrium-constant expression, excluding the solid
reactant from the expression. We assume that the compound dissociates completely into its component
ions:
CaF2(s)
Ca2+(aq) + 2 F–(aq)
Solve The expression for Ksp is
Ksp = [Ca2+][F–]2
Appendix D gives 3.9 × 10–11 for this Ksp.
Practice Exercise 1
Which of these expressions correctly expresses the solubility-product constant for Ag3PO4 in water?
(a) [Ag][PO4], (b) [Ag+][PO43–], (c) [Ag+]3[PO43–], (d) [Ag+][PO43–]3, (e) [Ag+]3[PO43–]3.
Practice Exercise 2
Aqueous
Equilibria
© 2015
Pearson
Education, Inc.
Give
the
solubility-product-constant
expressions and Ksp values (from Appendix D) for (a) barium carbonate and
Solubility vs. Solubility Product
• Ksp is not the same as solubility.
• Solubility is the quantity of a substance that
dissolves to form a saturated solution
• Common units for solubility:
➢ Grams per liter (g/L)
➢ Moles per liter (mol/L)
Aqueous
Equilibria
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Sample Exercise 17.11 Calculating Ksp from Solubility
Solid silver chromate is added to pure water at 25 °C, and
some of the solid remains undissolved. The mixture is
stirred for several days to ensure that equilibrium is
achieved between the undissolved Ag2CrO4(s) and the
solution. Analysis of the equilibrated solution shows that its
silver ion concentration is 1.3 × 10–4 M. Assuming that the
Ag2CrO4 solution is saturated and that there are no other
important equilibria involving Ag+ or CrO42– ions in the
solution, calculate Ksp for this compound.
Ag2CrO4(s)
↔ 2Ag+(aq) + CrO42–(aq)
Ksp = [Ag+]2[CrO42–]
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Aqueous
Equilibria
Solid silver chromate is added to pure water at 25 °C, and some of the solid
remains undissolved. The mixture is stirred for several days to ensure that
equilibrium is achieved between the undissolved Ag2CrO4(s) and the solution.
Analysis of the equilibrated solution shows that its silver ion concentration is 1.3
× 10–4 M. Assuming that the Ag2CrO4 solution is saturated and that there are
no other important equilibria involving Ag+ or CrO42– ions in the solution,
calculate Ksp for this compound.
Ag2CrO4(s) ↔ 2Ag+(aq) + CrO42–(aq)
Ksp = [Ag+]2[CrO42–]
Solve From the chemical formula of silver chromate, we
know that there must be two Ag+ ions in solution for each
CrO42– ion in solution. Consequently, the concentration of
CrO42– is half the concentration of Ag+:
Ksp = [Ag+]2[CrO42–] = (1.3 × 10–4)2(6.5 × 10–5) = 1.1 × 10–12
Aqueous
Equilibria
© 2015 Pearson Education, Inc.
Sample Exercise 17.12 Calculating Solubility from Ksp
The Ksp for CaF2 is 3.9 × 10–11 at 25 °C. Assuming
equilibrium is established between solid and dissolved
CaF2, and that there are no other important equilibria
affecting its solubility, calculate the solubility of CaF2 in
grams per liter.
Plan To go from Ksp to solubility, we follow the steps indicated by the red arrows in
Figure 17.16. We first write the chemical equation for the dissolution and set up a
table of initial and equilibrium concentrations. We then use the equilibriumconstant expression. In this case we know Ksp, and so we solve for the
concentrations of the ions in solution. Once we know these concentrations, we
use the formula weight to determine solubility in g/L.
Aqueous
Equilibria
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Sample Exercise 17.12 Calculating Solubility from Ksp
The Ksp for CaF2 is 3.9 × 10–11 at 25 °C. Assuming
equilibrium is established between solid and dissolved
CaF2, and that there are no other important equilibria
affecting its solubility, calculate the solubility of CaF2 in
grams per liter.
Plan To go from Ksp to solubility, we follow the steps indicated by the red arrows in
Figure 17.16. We first write the chemical equation for the dissolution and set up a
table of initial and equilibrium concentrations. We then use the equilibriumconstant expression. In this case we know Ksp, and so we solve for the
concentrations of the ions in solution. Once we know these concentrations, we
use the formula weight to determine solubility in g/L.
Aqueous
Equilibria
© 2015 Pearson Education, Inc.
Sample Exercise 17.12 Calculating Solubility from Ksp
The Ksp for CaF2 is 3.9 × 10–11 at 25 °C. Assuming
equilibrium is established between solid and dissolved
CaF2, and that there are no other important equilibria
affecting its solubility, calculate the solubility of CaF2 in
grams per liter.
Aqueous
Equilibria
© 2015 Pearson Education, Inc.
Factors Affecting Solubility
• Three factors that affect the
solubility of ionic compounds:
(1)Presence of common ions,
(2) solution pH, and
(3)presence of complexing
agents.
Aqueous
Equilibria
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Factors Affecting Solubility
• The Common-Ion Effect
– If one of the ions in a solution
equilibrium is already dissolved in
the solution, the solubility of the
salt will decrease.
– If either calcium ions or fluoride
ions are present, then calcium
fluoride will be less soluble.
Aqueous
Equilibria
© 2015 Pearson Education, Inc.
Sample Exercise 17.13 Calculating the Effect of a
Common Ion on Solubility
Calculate the molar solubility of CaF2 at 25 °C in a solution
that is (a) 0.010 M in Ca(NO3)2 and (b) 0.010 M in NaF.
Aqueous
Equilibria
© 2015 Pearson Education, Inc.
Sample Exercise 17.13 Calculating the Effect of a
Common Ion on Solubility
Calculate the molar solubility of CaF2 at 25 °C in a solution
that is (a) 0.010 M in Ca(NO3)2 and (b) 0.010 M in NaF.
b) The common ion is F–, and at
equilibrium we have [Ca2+] = x and [F–] = 0.010 + 2x
Assuming that 2x is much smaller than
0.010 M (that is, 0.010 + 2x ≃ 0.010),
we have
Thus, 3.9 × 10–7 mol of solid CaF2 should dissolve per liter of
0.010 M NaF solution.
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Aqueous
Equilibria
Sample Integrative Exercise Putting Concepts Together
A sample of 1.25 L of HCl gas at 21 °C and 0.950 atm is
bubbled through 0.500 L of 0.150 M NH3 solution. Calculate
the pH of the resulting solution assuming that all the HCl
dissolves and that the volume of the solution remains 0.500 L.
HCl(g) + NH3(aq) → NH4+(aq) + Cl–(aq)
Aqueous
Equilibria
© 2015 Pearson Education, Inc.
Sample Integrative Exercise Putting Concepts Together
A sample of 1.25 L of HCl gas at 21 °C and 0.950 atm is bubbled through 0.500 L
of 0.150 M NH3 solution. Calculate the pH of the resulting solution assuming that
all the HCl dissolves and that the volume of the solution remains 0.500 L.
Thus, the reaction produces a solution containing a mixture of NH3, NH4+,
and Cl–. The NH3 is a weak base (Kb = 1.8 × 10–5), NH4+ is its conjugate acid,
and Cl– is neither acidic nor basic. Consequently, the pH depends on [NH3] and
[NH4+],
Aqueous
Equilibria
© 2015 Pearson Education, Inc.
Sample Integrative Exercise Putting Concepts Together
A sample of 1.25 L of HCl gas at 21 °C and 0.950 atm is bubbled through 0.500 L
of 0.150 M NH3 solution. Calculate the pH of the resulting solution assuming that
all the HCl dissolves and that the volume of the solution remains 0.500 L.
We can calculate the pH using either Kb for NH3 or Ka for NH4+. Using the Kb
expression, we have
Hence, pOH = –log(9.4 × 10–6) = 5.03 and pH = 14.00 – pOH = 14.00 – 5.03 =
8.97.
Aqueous
Equilibria
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