1. Chapter 13 Applications of Aqueous Equilibria
13.1 Solutions of Acids or Bases Containing a Common Ion
13.2 Buffered Solutions
13.3 Exact Treatment of Buffered Solutions (skip)
13.4 Buffer Capacity
13.5 Titrations and pH Curves
13.6 Acid-Base Indicators
13.7 Titration of Polyprotic Acids (skip)
13.8 Solubility Equilibria and the Solubility Product
13.9 Precipitation and Qualitative Analysis (skip)
13.10 Complex Ion Equilibria (skip)
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2. The Common Ion Effect (1)
This applies to weak acids, weak bases and solubility of salts when a common ion is added to the equilibrium reaction. You can change the pH, but adding salt.
HF (aq) H+ (aq) + F- (aq) Ka = 7.2 x 10-4
NaF (s) → Na+ (aq) + F– (aq)
When F- is added (from NaF), then [H+] must decrease (Le Chatelier’s principle). The pH increases. See example 13.1 in the text to see how to quantitatively determine this effect, with an ICE box)

3. The Common Ion Effect (2)
If a solution and a salt to be dissolved in it have an ion in common, then the solubility of the salt is depressed relative to pure.
General equation
AB (s) A+ (aq) + B- (aq)
If you add BH(aq), which dissociates into B- and H+, then the [A+] decreases, and AB is driven out of solution.

4. Human blood is a buffered solution
Applications of Aqueous Equilibrium
Buffered Solutions
The Common Ion Effect on Buffering
Buffers: resist change in pH when acid or base is added.
Buffer Solutions: contain a common ion and are important in biochemical and physiological processes
Organisms (and humans) have built-in buffers to protect them against changes in pH.
Blood: (pH 7.4) Death = 7.0 <pH > 7.8 = Death
Human blood is a buffered solution
Human blood is maintained by a combination of CO3-2, PO4-3 and protein buffers. 3

5. Unbuffered solution
A solution that is buffered by acetic acid/acetate

6. HA H+ + A – How Do Buffers Work?
HA = generic acid
Ka = [H+][A-]/[HA] => [H+] = Ka[HA]/[A-]
If Ka is small (weak acid) then [H+] does not change much when [HA] and [A-] change.
If [HA] and [A-] are large, and [HA]/[A-] ≈ 1,then small additions of acid ([H+]) or base ([OH-]) don’t change the ratio much.

7. Example: A solution of 0.5 M acetic acid plus 0.5 M acetate
Ka = 1.8x pKa = 4.7
HA H+ + A-
Ka = [H+][A-]/[HA] pH = pKa + log[A-]/[HA]
Use an ICE box to calculate the pH
[AH]ini = 0.5, [A-]ini = 0.5, [H+]ini = 0
=> => pH = i.e., pH=pKa

8. Example: A solution of 0.5 M acetic acid plus 0.5 M acetate
Ka = 1.8x pKa = 4.74
HA H+ + A-
Ka = [H+][A-]/[HA] pH = pKa + log[A-]/[HA]
Use an ICE box to calculate the pH
[AH]ini = 0.5, [A-]ini = 0.5, [H+]ini = 0
=> => pH = i.e., pH=pKa
Now add NaOH to 0.01 M
(in an unbuffered solution this would give a pOH of 2 and pH of 12)
Use HA + OH- → H2O + A-
Redo the ICE Box
[AH]ini = 0.49, [A-]ini = 0.51
=> => pH = 4.76

9. Buffer Calculation: Add Acid (#1) to a Buffered Solution
Acetic Acid: Ka = 1.8x pKa = 4.74
HA H+ + A-
Ka = [H+][A-]/[HA] => -pKa = -pH + log[A-]/[HA] =>
pH = pKa + log[A-]/[HA]
Case 1) [CH3COOH]tot = [CH3COOH] + [CH3COO-] = 1.0M
pH = pKa => [CH3COOH] = [CH3COO-] = 0.5
Now add 0.01 M HCl (strong acid)
[CH3COOH] = [CH3COO-] = 0.49
pH = pKa + log[A-]/[HA] = log(0.49/0.51)
= 4.74 – 0.02 = 4.72
ΔpH = -0.02

10. Buffer Calculation: Add Acid (#2) to a dilute Buffered Solution
Acetic Acid: Ka = 1.8x pKa = 4.74
HA H+ + A-
Ka = [H+][A-]/[HA] => -pKa = -pH + log[A-]/[HA] =>
pH = pKa + log[A-]/[HA]
Case 2) [CH3COOH]tot = [CH3COOH] + [CH3COO-] = 0.10 M
pH = pKa => [CH3COOH] = [CH3COO-] = 0.05
Now add 0.01 M HCl (strong acid)
[CH3COOH] = [CH3COO-] = 0.04
pH = pKa + log[A-]/[HA] = log(0.04/0.06)
= 4.74 – 1.5 = 3.54
ΔpH = -1.5

11. Adding Base to a Buffered Solution: OH- ions do not accumulate but are replaced by A- ions.
OH– + HA ⇌ A– + H2O

12. When the OH- is added, the concentrations of HA and A- change, but only by small amounts. Under these conditions the [HA]/[A-] ratio and thus the [H+] stay virtually constant.
OH– + HA ⇌ A– + H2O
When protons are added to a buffered solution, the conjugate base (A–) reacts
H+ + A– ⇌ HA

13. Characteristics of Buffered solutions
Contain relatively large concentrations of a weak acid and its conjugate base.
When acid is added, it reacts with the conjugate base
When base is added, it reacts with the acid.
pH is determined by the ratio of the base and acid.

14. Buffer Capacity
Buffer capacity is the amount of protons or hydroxide ions that can be absorbed without a significant change in pH.
pH is determined by the ratio of [A–]/[HA] and pKa
Capacity is determined by the magnitudes of [HA] and [A–].

15. Equivalence Point

16. [H+] or pH depends on Ka and the ratio of acid to salt or [A─].
Buffer Summary
Buffer Design: Add known amount of HA (weak acid) and salt of HA (its conjugate base, A─)
[H+] or pH depends on Ka and the ratio of acid to salt or [A─].
Thus if both conc. HA and A- are large then small additions of acid or base don’t change the ratio much
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18. How to Actually Make a Buffer (a buffered solution in a lab)
HA H+ + A –
Assume you want to make a 1L of of 1M buffer at pH Use acetic acid, because the pKa of acetic acid is 4.74.
To make the Buffer:
The easy way (this is how we actually do it in the lab): Weigh out the appropriate amount of Na+Acetate- (70 g). Add it to a volumetric flask. Add water to around 0.9 Liters. Add a stir bar and stir. Slowly add HCl, while monitoring the pH with a pH meter. When the pH reaches the pKa (4.74), stop. Take out the stir bar and add enough water to give 1L.
The hard way. Start with two solutions. Solution A is 1M acetic acid. Solution B is 1M Na+Acetate-. Add solution A to solution B. Then check the pH with a pH meter. Adjust the pH with HCl or NaOH as required, to get the pH to 4.74 But this will change the concentration of A- and HA.

19. A titration is complete when the second solute is fully consumed
Acid-Base Titrations
A controlled addition of measured volumes of a solution of known concentration (the Titrant) from a buret to a second solution of unknown concentration under conditions in which the solutes react cleanly (without side reactions), completely, and rapidly.
Titration:
A titration is complete when the second solute is fully consumed
Completion is signaled by a change in some physical property, such as the color of the reacting mixture or the color of an indicator that has been added to it.
[NaOH]
Indicator phenolphthalein
“X”
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20. Titrations and pH Curves
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21. OH– + HA ⇌ A– + H2O Strong acid
Titration of a weak acid with a strong base.
The base reacts with the acid. .
OH– + HA ⇌ A– + H2O
Strong acid
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22. number of moles of acid = number of moles of base
Equivalence point determined defined by the Stoichiometry, not by the pKa.
nbase = nacid
number of moles of acid = number of moles of base
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23. For a variety of weak acids:
The Equivalence point occurs at the same stoichiometric amount of base added
The weaker the acid, the greater the pH value for the equivalent point.

24. Similar for titration of weak bases with strong acids
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25. Acid-Base Indicators
The indicator phenolphthalein is pink in basic solution and colorless in acidic solution.

26. Indicators (Weak Acid Equilibria)
pH = -log10[H3O+]
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27. HIn(aq) + H2O(l) ↔ H3O+(aq) + In–(aq)
Indicators
A soluble compound, generally an organic dye, that changes its color noticeably over a fairly short range of pH.
Typically, Indicators are a weak organic acid that has a different color than its conjugate base.
HIn(aq) + H2O(l) ↔ H3O+(aq) + In–(aq)
Acid: HIn (aq)
Conjugate base: In– (aq)
Phenolphtalein
Indicator denoted by In
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28. HIn(aq) + H2O(l) ↔ H3O+(aq) + In-(aq)
Acid: HIn (aq)
Conjugate base: In- (aq)
HIn(aq) + H2O(l) ↔ H3O+(aq) + In-(aq)
[H3O+][OH-] = Kw
pH = -log10[H3O+]

29. HIn(aq) + H2O(l) ↔ H3O+(aq) + In-(aq)
Methyl Red
Bromothymol blue
Phenolphtalein
HIn(aq) + H2O(l) ↔ H3O+(aq) + In-(aq)
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30. The useful pH ranges for several common indicators
HIn(aq) + H2O(l) ↔ H3O+(aq) + In–(aq)
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31. Could use either indicator Methyl red changes color to early
Indicator Selection
Want indicator color change and titration equivalence point to be as close as possible
Easier with a large pH change at the equivalence point
Strong acid
Weak acid
Could use either indicator
Methyl red changes color to early
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32. Solubility Product Ksp
Describes a chemical equilibrium in which an excess solid salt is in equilibrium with a saturated aqueous solution of its separated ions.
General equation
AB (s) ↔ A+ (aq) + B- (aq)
The solubility expression controls the amount of solid that will dissolve
Ksp =
Ksp =
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33. Ksp Values at 25°C for Common Ionic Solids
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34. The Solubility of Ionic Solids AgCl(s) ↔Ag+(aq) + Cl-(aq)
The Solubility Product
AgCl(s) ↔Ag+(aq) + Cl-(aq)
excess
Ksp =
The solid AgCl, which is in excess, is understood to have a concentration of 1 mole per liter.
Ksp
= 1.6  at 25oC
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35. The Solubility of Ionic Solids
The Solubility Product
Ag2SO4(s) ↔2Ag+(aq) + SO42-(aq)
excess
Ksp =
Fe(OH)3(s) ↔Fe+3(aq) + 3OH-1(aq)
excess
Ksp =
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36. Let “y” = molar solubility in mol/L
Solubility and Ksp
Determine the mass of lead(II) iodate dissolved in 2.50 L of a saturated aqueous solution of Pb(IO3)2 at 25oC. The Ksp of Pb(IO3)2 is 2.6 
Pb(IO3)2(s) ↔ Pb2+(aq) IO3-(aq)
Let “y” = molar solubility in mol/L
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37. Pb(IO3)2(s) ↔ Pb2+(aq) + 2 IO3-(aq)
Determine the mass of lead(II) iodate dissolved in 2.50 L of a saturated aqueous solution of Pb(IO3)2 at 25oC. The Ksp of Pb(IO3)2 is 2.6 
Pb(IO3)2(s) ↔ Pb2+(aq) IO3-(aq)
[Pb2+][IO3-]2 = Ksp
“y” = molar solubility
[Pb2+][IO3-]2 =
y = 4.0  10-5
 [Pb(IO3)2] = [Pb2+] = y = 4.0  10-5 mol L-1
 [IO3-] = 2y = 8.0  10-5 mol L-1
Gram solubility of
Lead (II) iodate
= (4.0  10-5 mol L-1)  (557 g mol-1)
= g L-1
 2.50 L
Molar Mass of lead (II) iodate
Pb(IO3)2 = 557g per mole
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38. The Solubility of Salts
Solubility and Ksp
Exercise 9-3
Compute the Ksp of silver sulfate (Ag2SO4) at 25oC if its mass solubility is 8.3 g L-1.
1 Ag2SO4 (s) ↔ 2 Ag+(aq) SO42-(aq)
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39. Ksp = 1 Ag2SO4 (s) ↔ 2 Ag+(aq) + 1 SO42-(aq)
Compute the Ksp of silver sulfate (Ag2SO4) at 25oC if its mass solubility is 8.3 g L-1.
1 Ag2SO4 (s) ↔ 2 Ag+(aq) SO42-(aq)
[y] = (8.3 g Ag2SO4 L-1)
 (1 mol Ag2SO4/311.8 g)
[y] = 2.66  10-2 mol Ag2SO4 L-1
[Ag+]2[SO42-] = Ksp
Ksp =
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40. Dissolution (Solubility) General reaction
The Nature of Solubility Equilibria
Dissolution and precipitation are reverse of each other.
Dissolution (Solubility)
General reaction
X3Y2 (s) ↔ 3X+2 (aq) + 2Y-3 (aq)
[s]
Ksp =
s = molar solubility
expressed in moles per liter
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41. Relative Solubilities
A salt’s Ksp value gives us information about its solubility.
Salt ↔ Cation + Anion
If the salts being compared produce the same number of ions, eg., AgI, CuI, CaSO4
s = [cation]
s = [anion]
Ksp = [cation] [anion] = s2
Salt molar solubility = s = (Ksp)1/2
Salt Ksp Solubility
AgI 1.5 x x 10-8
CuI 5.0 x x 10-6
CaSO x x 10-3
Solubility CaSO4 > CuI > AgI
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42. [Zn2+][OH-]2 = Ksp The Effects of pH on Solubility [Zn2+][OH-]2 = Ksp
Solubility of Hydroxides
Many solids dissolve more readily in more acidic solutions
Zn(OH)2(s) ↔Zn2+(aq) + 2 OH-(aq)
[Zn2+][OH-]2 = Ksp = 4.5  10-17
If pH decreases (or made more acidic), the [OH-] decreases. In order to maintain Ksp the [Zn2+] must increase and consequently more solid Zn(OH)2 dissolves.
Make more acidic:
Zinc hydroxide is more soluble in acidic solution than in pure water.
[Zn2+][OH-]2 = Ksp
[Zn2+][OH-]2 = Ksp
[Zn2+][OH-]2 = Ksp
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43. The Effects of pH on Solubility
Estimate the molar solubility of Fe(OH)3 in a solution that is buffered to a pH of 2.9. Lookup Ksp = 1.1x10-36
Fe(OH)3(s) ↔Fe3+(aq) + 3 OH-(aq)
(pH = 2.9 and) pOH = 11.1
[OH-] = 7.9  mol L-1
[Fe3+][OH-]3 = Ksp
[Fe3+] = Ksp/[OH-]3 = 1.1  / (7.9  10-12)3
[Fe3+] = [Fe(OH)3] = 2.2  10-3 mol L-1 answer
In pure water:
[Fe3+] = y [OH-] = 3y
y(3y)3 = 27y4 = Ksp = 1.1  10-36
y = 4.5  mol L-1 = [Fe3+] = [Fe(OH)3]=
[OH-] = 3y = 1.3  10-9 mol L-1
pOH = 8.87
(and pH = 5.13)
In pure water, Fe(OH)3 is 5 x 10 6 less soluble than at pH = 2.9
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44. TlIO3(s) ↔ Tl+(aq) + IO3-(aq)
The Common Ion Effect
The Ksp of thallium(I) iodate (TlO3) is 3.1  10-6 at 25oC. Determine the molar solubility of TlIO3 in mol L-1 KIO3 at 25oC.
TlIO3(s) ↔ Tl+(aq) IO3-(aq)
[Tl+] (mol L-1) [IO3-] (mol L-1)
Initial concentration
Equilibrium concentration
Change in concentration
[Tl+][IO3-] = Ksp
s = [TlIO3] = 6.2 × 10-5 mol L-1 = molar solubility
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45. TlIO3(s) ↔ Tl+(aq) + IO3-(aq)
The Common Ion Effect
The Ksp of thallium(I) iodate (TlO3) is 3.1  10-6 at 25oC. Determine the molar solubility of TlIO3 in mol L-1 KIO3 at 25oC.
TlIO3(s) ↔ Tl+(aq) IO3-(aq)
[s]
With common ion (from previous calculation)
s = [TlIO3] = 6.2 × 10-5 mol L-1 = molar solubility
What if no common ion is added? i.e., dissolve thallium iodate in pure water
s= 1.76x10-3 mol L-1 = molar solubility
[Tl+][IO3-] = Ksp
With common ion, s = [TlIO3]= 6.2 × 10-5 mol L-1
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46. Chapter 8 Applications of Aqueous Equilibria
8.1 Solutions of Acids or Bases Containing a Common Ion
8.2 Buffered Solutions
8.3 Exact Treatment of Buffered Solutions (skip)
8.4 Buffer Capacity
8.5 Titrations and pH Curves
8.6 Acid-Base Indicators
8.7 Titration of Polyprotic Acids (skip)
8.8 Solubility Equilibria and the Solubility Product
8.9 Precipitation and Qualitative Analysis (skip)
8.10 Complex Ion Equilibria (skip)
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