1. Ionic Equilibria in Aqueous Systems

2. Ionic Equilibria in Aqueous Systems
19.1 Equilibria of Acid-Base Buffers
19.2 Acid-Base Titration Curves
19.3 Equilibria of Slightly Soluble Ionic Compounds
19.4 Equilibria Involving Complex Ions

3. Acid-Base Buffers
An acid-base buffer is a solution that lessens the impact of pH from the addition of acid or base.
An acid-base buffer usually consists of a conjugate acid-base pair where both species are present in appreciable quantities in solution.
An acid-base buffer is therefore a solution of a weak acid and its conjugate base (e.g., HOAC/OAc-), or a weak base and its conjugate acid (e.g., NH3/NH4+).

4. The effect of adding acid or base to an unbuffered solution.
A 100-mL sample of dilute HCl is adjusted to pH 5.00.
The addition of 1 mL of strong acid (left) or strong base (right) changes the pH by several units.

5. The effect of adding acid or base to a buffered solution.
A 100-mL sample of an acetate buffer is adjusted to pH 5.00.
The addition of 1 mL of strong acid (left) or strong base (right) changes the pH very little.
The acetate buffer is made by mixing 1 M CH3COOH ( a weak acid) with 1 M CH3COONa (which provides the conjugate base, CH3COO-).

6. Buffers and the Common-ion Effect
A buffer works through the common-ion effect.
Acetic acid in water dissociates slightly to produce some acetate ion:
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
acetic acid
acetate ion
If NaCH3COO is added, it provides a source of CH3COO- ion, and the equilibrium shifts to the left. CH3COO- is common to both solutions.
The addition of CH3COO- reduces the % dissociation of the acid.

7. The Effect of Added Acetate Ion on the Dissociation of Acetic Acid
[CH3COOH]init
[CH3COO-]added
% Dissociation*
[H3O+] pH
0.10
0.00
1.3
1.3x10-3
2.89
0.050
0.036
3.6x10-5
4.44
0.018
1.8x10-5
4.74
0.15
0.012
1.2x1015
4.92
* % Dissociation =
[CH3COOH]dissoc
[CH3COOH]init
x 100

8. How a Buffer Works
The buffer components (HA and A-) are able to consume small amounts of added OH- or H3O+ by a shift in equilibrium position.
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
Added OH- reacts with CH3COOH, causing a shift to the right.
Added H3O+ reacts with CH3COO-, causing a shift to the left.
The shift in equilibrium position absorbs the change in [H3O+] or [OH-], and the pH changes only slightly.

9. How a buffer works. H3O+ Buffer has more HA after addition of H3O+.
H2O + CH3COOH ← H3O+ + CH3COO-
Buffer has equal concentrations of A- and HA.
OH-
Buffer has more A- after addition of OH-.
CH3COOH + OH- → CH3COO- + H2O

10. Relative Concentrations of Buffer Components
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
Ka =
[CH3COO-][H3O+]
[CH3COOH]
[H3O+] = Ka x
[CH3COOH]
[CH3COO-]
Since Ka is constant, the [H3O+] of the solution depends on the ratio of buffer component concentrations.
If the ratio increases, [H3O+] increases.
[HA]
[A-]
If the ratio decreases, [H3O+] decreases.
[HA]
[A-]

11. The Henderson-Hasselbalch Equation
HA(aq) + H2O(l) A-(aq) + H3O+(aq)
Ka =
[H3O+][A-]
[HA]
[H3O+] = Ka x
[HA]
[A-]
-log[H3O+] = -logKa – log
[HA]
[A-]
pH = pKa + log
[base]
[acid]

12. Calculating the pH of a Buffer solution (Assume the additions cause a negligible change in volume.)
PROBLEM:
(a) Of a buffer solution consisting of 0.50 M CH3COOH (HOAc) and M CH3COONa (NaOAc)
Calculate the pH:
pH = 4.74

13. Calculating the Effect of Added OH- on Buffer pH (pKa = 4.74)
PROBLEM:
(b) After adding mol of solid NaOH to 1.0 L of the buffer solution
Consider chemical reaction first. Which, NaOAc or HOAc, is the
Reagent? How much excess reagent remains after reaction?
THE ICE table:
pH = 4.77

14. Calculating the Effect of Added H3O+ on Buffer pH
PROBLEM:
(c) After adding mol of HCl to 1.0 L of the buffer solution in (a). Ka of CH3COOH = 1.8 x 10-5.
Consider which, NaOAc or HOAc, is the Reagent?
How much excess reagent remains after reaction?
The ICE table again (group practice)
pH = 4.70

15. Buffer Capacity
The buffer capacity is a measure of the “strength” of the buffer, its ability to maintain the pH following addition of strong acid or base.
The greater the concentrations of the buffer components, the greater its capacity to resist pH changes.
The closer the component concentrations are to each other, the greater the buffer capacity.

16. The relation between buffer capacity and pH change.
When strong base is added, the pH increases least for the most concentrated buffer.
This graph shows the final pH values for four different buffer solutions after the addition of strong base.

17. Buffer Range
The buffer range is the pH range over which the buffer is effective.
Buffer range is related to the ratio of buffer component concentrations.
[HA]
[A-]
The closer is to 1, the more effective the buffer.
If one component is more than 10 times the other, buffering action is poor. Since log10 = 1, buffers have a usable range within ± 1 pH unit of the pKa of the acid component.

18. Using Molecular Scenes to Examine Buffers
PROBLEM:
The molecular scenes below represent samples of four HA/A- buffers. (HA is blue and green, A- is green, and other ions and water are not shown.)
(a) Which buffer has the highest pH?
(b) Which buffer has the greatest capacity?
(c) Should we add a small amount of concentrated strong acid or strong base to convert sample 1 to sample 2 (assuming no volume changes)?

19. Preparing a Buffer Choose the conjugate acid-base pair.
The pKa of the weak acid component should be close to the desired pH.
Calculate the ratio of buffer component concentrations.
Determine the buffer concentration, and calculate the required volume of stock solutions and/or masses of components.
Mix the solution and correct the pH.
pH = pKa + log
[base]
[acid]

20. Preparing a Buffer
PROBLEM:
An environmental chemist needs a carbonate buffer of pH to study the effects of the acid rain on limsetone-rich soils. How many grams of Na2CO3 must she add to 1.5 L of freshly prepared 0.20 M NaHCO3 to make the buffer? Ka of HCO3- is 4.7x10-11.
15 g Na2CO3

21. Acid-Base Indicators
An acid-base indicator is a weak organic acid (HIn) whose color differs from that of its conjugate base (In-).
The ratio [HIn]/[In-] is governed by the [H3O+] of the solution. Indicators can therefore be used to monitor the pH change during an acid-base reaction.
The color of an indicator changes over a specific, narrow pH range, a range of about 2 pH units.

22. Colors and approximate pH range of some common acid-base indicators.

23. The color change of the indicator bromthymol blue.
pH < 6.0
pH =
pH > 7.5

24. Acid-Base Titrations
In an acid-base titration, the concentration of an acid (or a base) is determined by neutralizing the acid (or base) with a solution of base (or acid) of known concentration.
The equivalence point of the reaction occurs when the number of moles of OH- added equals the number of moles of H3O+ originally present, or vice versa.
The end point occurs when the indicator changes color.
- The indicator should be selected so that its color change occurs at a pH close to that of the equivalence point.

25. How to calculate pH during an Acid-Base Titrations: HA + OH-
Setting up an ICE table for the neutralization reaction:
H+ + OH-  H2O or HA + OH-  A- + H2O
When an acid and a base is mixed, the neutralization reaction proceeds quickly to reach completion.
The limiting reagent is consumed instantaneously (zero molarity in the ICE table).
Molarity of each species in the solution is affected by the increased volume (VHA + VOH-).
After reaching the equivalence point, pH of the titration mixture depends on the molarity of excess reagent.

26. Curve for a strong acid–strong base titration.
The pH increases gradually when excess base has been added.
The pH rises very rapidly at the equivalence point, which occurs at pH = 7.00.
The initial pH is low.

27. Calculating the pH during a strong acid–strong base titration
Initial pH
[H3O+] = [HA]init
pH = -log[H3O+]
pH before equivalence point
initial mol H3O+ = Vacid x Macid
mol OH- added = Vbase x Mbase
mol H3O+remaining = (mol H3O+init) – (mol OH-added)
[H3O+] = pH = -log[H3O+]
mol H3O+remaining
Vacid + Vbase

28. Calculating the pH during a strong acid–strong base titration (contd.)
pH at the equivalence point
pH = 7.00 for a strong acid-strong base titration.
pH beyond the equivalence point
initial mol H3O+ = Vacid x Macid
mol OH- added = Vbase x Mbase
mol OH-excess = (mol OH-added) – (mol H3O+init)
[OH-] =
pOH = -log[OH-] and pH = pOH
mol OH-excess
Vacid + Vbase

29. Example:
40.00 mL of M HCl is titrated with M NaOH.
Calculate
The initial pH
[H+] = [HCl] = M, pH = 1.00
The pH after mL of NaOH solution has been added:
First, consider the combined volume after addition of NaOH.
ICE table:
x = M, [H+] = M, pH = 1.48

30. Example:
40.00 mL of M HCl is titrated with M NaOH.
Calculate the pH after mL of NaOH solution has been added.
First, consider the combined volume after addition of NaOH.
ICE table:
x = M, [OH-] = M, pH = 12.05

31. Titration Curve for a weak acid–strong base titration
The pH increases slowly beyond the equivalence point.
The curve rises gradually in the buffer region. The weak acid and its conjugate base are both present in solution.
The pH at the equivalent point is > 7.00 due to the reaction of the conjugate base with H2O.
The initial pH is higher than for the strong acid solution.

32. Before titration: Weak acid only
Calculating the pH during a weak acid–strong base titration: HA + OH-  H2O + A-
Before titration: Weak acid only
Valid approximation? pH = -log[H3O+]
pH before equivalence point (Vbase < Veq): conjugate base A- (from rxn of HA and added OH- ) and excess HA yields a Buffer

33. Calculating the pH during a weak acid–strong base titration (contd.)
pH at the equivalence point: Ionization of the conjugate base A-
Where
Finally
A-(aq) + H2O(l) HA(aq) + OH-(aq)
pH beyond the equivalence point depends on excess OH-

34. Finding the pH During a Weak Acid–Strong Base Titration
Example:
Calculate the pH during the titration of mL of M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of M NaOH: (a) 0.00 mL;
Solution: Note that the neutralization rxn of HA with NaOH is instantaneous. Find Veq first.
Before titration begins, only weak acid HPr exists
[H+]  = 1.1 10-3 M,
Is the approximation valid?
[H+]/[HA]0 = < 5%, so yes
pH = -log[H+] = 2.96

35. Finding the pH During a Weak Acid–Strong Base Titration (contd.)
Example:
Calculate the pH during the titration of mL of M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of M NaOH: (b) mL
Solution: (b) Note that the outcome of the reaction depends on the limiting reagent. Use of ICE table (solution table), I + C = E
Find [HA] and [OH-] after mixing HA with NaOH (before reaction):
Volume after mixing = mL mL = mL
[HA] = mL HA x M/70.00mL = M
[OH-] = mL x M/70.00 mL = M
HA +
OH- → A-
+ H2O
Initial
Change -x +x
End

36. Finding the pH During a Weak Acid–Strong Base Titration (contd.)
Example:
Calculate the pH during the titration of mL of M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of M NaOH: (b) mL
Solution: (b) Note that reactant OH- is lower than HA, OH- is the limiting reagent and becomes zero after reaction is complete.
Solve for x =
HA +
OH- → A-
+ H2O
Initial
Change -x +x
End x
Find
[A-] = M
[HA] = M
pH = pKa + log([A-]/[HA]) = 5.37

37. Time to Discuss and Practice : Step by step before eqiv. point
Calculate the pH when the following volume of NaOH is added to mL M HPr solution: a) mL; b) mL; c) mL. Ka = 1.3x10-5
HA +
OH- → A-
+H2O I
[HA]
[OH-] C ? E
L.R.?
Step by step before eqiv. point
Find total volume after adding base
Find [HA] and [OH-] before rxn: Dilution
ICE table and find limiting reagent (L.R.)
Solve ICE table: find [HA] and [A-] after rxn. I + C = E
Apply Henderson-Hasselbalch equation
Answer keys: a) 4.41; b) 4.89; c) 5.73

38. Finding the pH During a Weak Acid–Strong Base Titration (contd
Finding the pH During a Weak Acid–Strong Base Titration (contd., almost!)
PROBLEM:
Calculate the pH during the titration of mL of M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of M NaOH: (c) mL,
(c) Note this is the equivalence point! 
All HA is now consumed and converted into the conjugate base
Total volume = mL mL = mL
Mol A- = mol HA (stoichiometry)
[A-] = mL HA x mol/L /80.00 mL = M
[OH-] = 6.21x10-6 M, pOH = 5.21, pH = 8.79

39. Finding the pH During a Weak Acid–Strong Base Titration (contd
Finding the pH During a Weak Acid–Strong Base Titration (contd., finally!)
PROBLEM:
Calculate the pH during the titration of mL of M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of M NaOH: (d) mL
(d) Note Vbase > Veq, so it is beyond equivalence point
All HA has been consumed and there is excess NaOH in the solution.
The unreacted NaOH determine the final pH:
Mol NaOH (reacted) = mol HA x 1 mol NaOH/1 mol HA = mol
Mol NaOH excess = mol NaOH – mol NaOH(reacted) = mol
Total volume of the reaction = mL mL = mL = L
[OH-] = mol/ L = M
pOH = 1.95, pH = 12.05

40. Curve for a weak base–strong acid titration.
The pH decreases gradually in the buffer region. The weak base and its conjugate acid are both present in solution.
The pH at the equivalence point is < 7.00 due to the reaction of the conjugate acid with H2O.

41. pH before equivalence point: Use ICE table
Calculating the pH during a weak base-strong acid titration: Common Ion Effect
Initial pH
[OH-] =
pH = 14 - pOH
[OH-][HB+]
[B]
Kb =
pH before equivalence point: Use ICE table
Ka =
pH = pKa + log
[B]
[HB+] Kw Kb

42. pH at the equivalence point
Calculating the pH during a weak base-strong acid titration (continued)
pH at the equivalence point
[H+] = pH = -log[H3O+]
HB+(aq) + H2O(l) B(aq) + H3O+(aq)
pH beyond the equivalence point
[H+] =
pH = -log[H3O+]
mol H+excess
Vacid + Vbase

43. Finding the pH During a Weak base-strong acid Titration
PROBLEM:
Calculate the pH during the titration of mL of M methylamine (MeNH2; Kb = 4.4x10-4) after adding the following volumes of M HCl: (a) 0.00 mL; (b) mL
(a) [OH-] = 6.6x10-3M, pH = 11.82
(b) B H+ →
HB+
Initial
40.00 mL x M/70.00 mL
30.00 mL x M/70.00 mL
Change -x +x
End x
0 (L. R.)
x = , pH = pKa + log([B]/[HB+]) = 10.16

44. Finding the pH during a Weak base-strong acid Titration
PROBLEM:
Calculate the pH during the titration of mL of M methylamine (Kb = 4.4 x10-4) after adding the following volumes of M HCl: (c) mL, (d) mL
(c) [HB+] = M, [H+] = = 1.1E-6 M, pH = 5.97
(d) B H+ →
HB+
Initial
40.00 mL x M/90.00 mL
50.00 mL x M/90.00 mL
Change -x +x
End
0 (L. R.) x
[H+] = M, pH = 1.95

45. Titration of 40.00 mL of 0.1000 M H2SO3 with 0.1000 M NaOH
Curve for the titration of a weak polyprotic acid.
Titration of mL of M H2SO3 with M NaOH
pKa2 = 7.19
pKa1 = 1.85

46. Amino Acids as Polyprotic Acids
An amino acid contains a weak base (-NH2) and a weak acid (-COOH) in the same molecule.
Both groups are protonated at low pH and the amino acid behaves like a polyprotic acid.

47. Abnormal shape of red blood cells in sickle cell anemia.
Several amino acids have charged R groups in addition to the NH2 and COOH group. These are essential to the normal structure of many proteins.
In sickle cell anemia, the hemoglobin has two amino acids with neutral R groups instead of charged groups. The abnormal hemoglobin causes the red blood cells to have a sickle shape, as seen here.

48. Equilibria of Slightly Soluble Ionic Compounds
Any “insoluble” ionic compound is actually slightly soluble in aqueous solution: very small amount of such a compound that dissolves will dissociate completely.
For a slightly soluble ionic compound in water, equilibrium exists between solid solute and aqueous ions.
PbF2(s) Pb2+(aq) + 2F-(aq)
Qsp = Qc[PbF2] = [Pb2+][F-]2
Qc =
[Pb2+][F-]2
[PbF2]

49. Qsp and Ksp
Qsp is called the ion-product expression for a slightly soluble ionic compound.
For any slightly soluble compound MpXq:
MpXq(s) pMn+(aq) + qXz-(aq)
Qsp = [Mn+]p[Xz-]q
When the solution is saturated, the system is at equilibrium, and Qsp = Ksp, the solubility product constant.
The Ksp value of a salt indicates how far the dissolution proceeds at equilibrium (saturation).

50. Metal Sulfides: Sulfide Ion Interacts with Water
Metal sulfides behave differently from most other slightly soluble ionic compounds, since the S2- ion is strongly basic.
The dissolution of a metal sulfide as a two-step process:
MnS(s) Mn2+(aq) + S2-(aq)
S2-(aq) + H2O(l) → HS-(aq) + OH-(aq)
MnS(s) + H2O(l) Mn2+(aq) + HS-(aq) + OH-(aq)
Ksp = [Mn2+][HS-][OH-]

51. Writing Ion-Product Expressions
PROBLEM:
Write the ion-product expression at equilibrium for each compound:
(a) magnesium carbonate (b) iron(II) hydroxide
(c) calcium phosphate (d) silver sulfide
Ksp = [Mg2+][CO32-]
Ksp = [Fe2+][OH-]2
Ksp = [Ca2+]3[PO43-]2
Ksp = [Ag+]2[HS-][OH-]

52. Solubility-Product Constants (Ksp) of Selected Ionic Compounds at 25°C
Name, Formula
Ksp
Aluminum hydroxide, Al(OH)3
Cobalt(II) carbonate, CoCO3
Iron(II) hydroxide, Fe(OH)2
Lead(II) fluoride, PbF2
Lead(II) sulfate, PbSO4
Silver sulfide, Ag2S
Zinc iodate, Zn(IO3)2
3x10-34
1.0x10-10
4.1x10-15
3.6x10-8
1.6x10-8
4.7x10-29
8x10-48
Mercury(I) iodide, Hg2I2
3.9x10-6

53. PROBLEM: Determining Ksp from Solubility
Lead(II) sulfate (PbSO4) is a key component in lead-acid car batteries. Its solubility in water at 25°C is 4.25x10-3 g/100 mL solution. What is the Ksp of PbSO4?
[PbSO4(aq)] = 1.40x10-4 M [Pb2+] = [SO42-] = 1.40x10-4 M Ksp = 1.96x10-8

54. Determining Ksp from Solubility
PROBLEM:
(b) When lead(II) fluoride (PbF2) is shaken with pure water at 25°C, the solubility is found to be 0.64 g/L. Calculate the Ksp of PbF2.
[PbF2(aq)] = 2.6 x10-3 M [Pb2+] = 2.6 x10-3 M [F-] = 5.2 x10-3 M Ksp = 7.0 x10-8

55. Determining Solubility from Ksp
PROBLEM:
Calculate the molar solubility of Ca(OH)2 in water if the Ksp is 6.5x10-6.
S = =
= 1.2x10-2 M

56. Relationship Between Ksp and Solubility at 25°C
No. of Ions
Formula
Cation/Anion
Ksp
Solubility (M) 2
MgCO3
1/1
3.5x10-8
1.9x10-4
PbSO4
1.6x10-8
1.3x10-4
BaCrO4
2.1x10-10
1.4x10-5 3
Ca(OH)2
1/2
6.5x10-6
1.2x10-2
BaF2
1.5x10-6
7.2x10-3
CaF2
3.2x10-11
2.0x10-4
Ag2CrO4
2/1
2.6x10-12
8.7x10-5
The higher the Ksp value, the greater the solubility, as long as we compare compounds that have the same total number of ions in their formulas.

57. Predicting the Formation of a Precipitate
For a saturated solution of a slightly soluble ionic salt, Qsp = Ksp.
For equilibrium MpXq(s) pMn+(aq) + qXz-(aq)
Shift to the right = Dissolving occurs
Shift to the left = Precipitation occurs
When two solutions containing the ions of slightly soluble salts are mixed,
If Qsp = Ksp, the solution is saturated and no change will occur.
If Qsp > Ksp, a precipitate will form
If Qsp =< Ksp, no precipitate will form

58. How to Predict Whether a Precipitate Will Form?
Find [ions] after two solutions are mixed. Consider the volume change if applicable.
Calculate solubility quotient Qsp from the formula of possible products
If calculated Qsp is larger than Ksp, precipitation will form

59. Predicting Whether a Precipitate Will Form
PROBLEM:
A common laboratory method for preparing a precipitate is to mix solutions containing the component ions. Does a precipitate form when L of 0.30 M Ca(NO3)2 is mixed with L of M NaF? Ksp(CaF2) = 3.2x10-11.
[Ca2+] = 0.10 M, [F-] = M
Q(CaF2) = 1.6x10-5

60. Selective Precipitation
Selective precipitation is used to separate a solution containing a mixture of ions.
After adding a precipitating ion that form precipitate with both ions, Qsp increases from 0.
Once Qsp > Ksp(less soluble compd), precipitate of that compound forms.
As long as Qsp < Ksp(more soluble compd), the less soluble compound will precipitate in as large a quantity as possible, leaving behind the ion of the more soluble compound.

61. Selective Precipitation
To completely separate two metal ions, the concentration of the precipitating ion should results in
Ksp(more soluble precipitate)  Qsp > Ksp(less soluble precipitate)
The maximal concentration of the precipitating ion can be determined from the Ksp of the more soluble precipitate.

62. Separating Ions by Selective Precipitation
PROBLEM:
A solution consists of 0.20 M MgCl2 and 0.10 M CuCl2. Calculate the [OH-] that would separate the metal ions as their hydroxides. Ksp of Mg(OH)2= is 6.3x10-10; Ksp of Cu(OH)2 is 2.2x10-20.
[OH-] = 5.6x10-5 M

63. The effect of a common ion on solubility
Add Na2CrO4(aq)
saturated PbCrO4(aq)
More PbCrO4(s)
PbCrO4(s) Pb2+(aq) + CrO42-(aq)
Why? The added common ion CrO42- causes the equilibrium to shift to the left. Solubility decreases and solid PbCrO4 precipitates.

64. Calculating the Effect of a Common Ion on Solubility
PROBLEM:
In the previous practice, we calculated the solubility of Ca(OH)2 in water as M. What is its solubility in 0.10 M Ca(NO3)2? Ksp of Ca(OH)2 is 6.5x10-6.
Using approximation, s = 4.0x10-3 M
Compare with solubility in pure water (0.012 M), why this value is much smaller?

65. Effect of pH on Solubility
Changes in pH affects the solubility of many slightly soluble ionic compounds.
The addition of H3O+ will increase the solubility of a salt that contains the anion of a weak acid.
CaCO3(s) Ca2+(aq) + CO32-(aq)
CO32-(aq) + H3O+(aq) → HCO3-(aq) + H2O(l)
HCO3-(aq) + H3O+(aq) → [H2CO3(aq)] + H2O(l) → CO2(g) + 2H2O(l)
The net effect of adding H3O+ to CaCO3 is the removal of CO32- ions, which causes an equilibrium shift to the right. More CaCO3 will dissolve.

66. Test for the presence of a carbonate.
When a carbonate mineral is treated with HCl, bubbles of CO2 form.

67. Predicting the Effect on Solubility of Adding Strong Acid
PROBLEM:
Write balanced equations to explain whether addition of H3O+ from a strong acid affects the solubility of each ionic compound:
(a) lead(II) bromide (b) copper(II) hydroxide (c) iron(II) sulfide
(a) No effect
(b) [OH-] decreases, causing increase in solubility.
(c) [OH-] and [HS-] decreases, causing increase in solubility.

68. Limestone cave in Nerja, Málaga, Spain.
Limestone is mostly CaCO3 (Ksp = 3.3x10-9).
Ground water rich in CO2 trickles over CaCO3, causing it to dissolve. This gradually carves out a cave.
Water containing HCO3- and Ca2+ ions drips from the cave ceiling. The air has a lower PCO2 than the soil, causing CO2 to come out of solution. A shift in equilibrium results in the precipitation of CaCO3 to form stalagmites and stalactites.
CO2(g) CO2(aq)
CO2(aq) + 2H2O(l) H3O+(aq) + HCO3-(aq)
CaCO3(s) + CO2(aq) + H2O(l) Ca2+(aq) + 2HCO3-(aq)

69. Formation of acidic precipitation.
Chemical Connections
Formation of acidic precipitation.
Since pH affects the solubility of many slightly soluble ionic compounds, acid rain has far-reaching effects on many aspects of our environment.

70. Formation of Complex ion affects Solubility equilibrium
Recall in Chapter 21 metal ion forms complex ions with ligands: Mx+ + n L→ MLnx+
Starting at equilibrium where Qsp = Ksp, formation of complex ion from the metal ion will affect solubility equilibrium by reducing [Mx+].
After adding a reagent that form complex ion with metal cation Mx+, [Mx+] , so Qsp .
Since Qsp < Ksp, the equilibrium shifts towards right, causing more dissolving.

71. Examples of Formation of Complex ion affecting Solubility equilibrium
Addition of ammonia helps insoluble silver chloride to dissolve: Ag+ + 2 NH3 → Ag(NH3)2+
Addition of excess hydroxide ion helps insoluble aluminum hydroxide to dissolve:
Al(OH)3 + 3 OH- → Al(OH)63-

72. HOAc + OH- → OAc- + H2O Initial 0.50 M Change -x x Equilibrium
Calculating the Effect of Added OH- on Buffer pH (pKa = 4.74)
PROBLEM:
(b) After adding mol of solid NaOH to 1.0 L of the buffer solution
Consider chemical reaction first. Which, NaOAc or HOAc, is the
Reagent? How much excess reagent remains after reaction?
HOAc
+ OH- →
OAc
H2O
Initial
0.50 M
0.020 mol/1.0 L
Change -x x
Equilibrium x
0 (L. R.) x
pH = 4.77

73. OAc- + H+ → HOAc Initial 0.50 0.020 mol/1.0 L Change -x x Equilibrium
Calculating the Effect of Added H3O+ on Buffer pH
PROBLEM:
(c) After adding mol of HCl to 1.0 L of the buffer solution in (a). Ka of CH3COOH = 1.8 x 10-5.
Consider which, NaOAc or HOAc, is the Reagent?
How much excess reagent remains after reaction?
OAc- + H+ →
HOAc
Initial
0.50
0.020 mol/1.0 L
Change -x x
Equilibrium x
0 (L. R.) x
pH = 4.70

74. H+ + OH- → H2O Initial Change -x End 0.06667 - x 0 (L. R.) Example:
40.00 mL of M HCl is titrated with M NaOH.
Calculate
The initial pH
[H+] = [HCl] = M, pH = 1.00
The pH after mL of NaOH solution has been added:
First, consider the combined volume after addition of NaOH. H+
OH- →
H2O
Initial
40.00 mL x M/60.00 mL
20.00 mL x M/60.00 mL
Change -x
End x
0 (L. R.)
x = M, [H+] = M, pH = 1.48

75. H+ + OH- → H2O Initial Change -x End 0 (L. R.) 0.05556 - x Example:
40.00 mL of M HCl is titrated with M NaOH.
Calculate the pH after mL of NaOH solution has been added.
First, consider the combined volume after addition of NaOH. H+
OH- →
H2O
Initial
40.00 mL x M/90.00 mL
50.00 mL x M/90.00 mL
Change -x
End
0 (L. R.) x
x = M, [OH-] = M, pH = 12.05

76. Finding the pH During a Weak Acid–Strong Base Titration
Example:
Calculate the pH during the titration of mL of M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of M NaOH: (a) 0.00 mL;
Solution: Note that the neutralization rxn of HA with NaOH is instantaneous.
Before titration begins, only weak acid HPr exists
[H+]  = 1.1 10-3 M,
Is the approximation valid?
[H+]/[HA]0 = < 5%, so yes
pH = -log[H+] = 2.96

77. Finding the pH During a Weak Acid–Strong Base Titration (contd.)
Example:
Calculate the pH during the titration of mL of M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of M NaOH: (b) mL
Solution: (b) Note that the outcome of the reaction depends on the limiting reagent. Use of ICE table (solution table), I + C = E
Find [HA] and [OH-] after mixing HA with NaOH (before reaction):
Volume after mixing = mL mL = mL
[HA] = mL HA x M/70.00mL = M
[OH-] = mL x M/70.00 mL = M
HA +
OH- → A-
+ H2O
Initial
Change -x +x
End

78. Finding the pH During a Weak Acid–Strong Base Titration (contd.)
Example:
Calculate the pH during the titration of mL of M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of M NaOH: (b) mL
Solution: (b) Note that reactant OH- is lower than HA, OH- is the limiting reagent and becomes zero after reaction is complete.
Solve for x =
HA +
OH- → A-
+ H2O
Initial
Change -x +x
End x
Find
[A-] = M
[HA] = M
pH = pKa + log([A-]/[HA]) = 5.37

79. Time to Discuss and Practice : Step by step before eqiv. point
Calculate the pH when the following volume of NaOH is added to mL M HPr solution: a) mL; b) mL; c) mL. Ka = 1.3x10-5
HA +
OH- → A-
+H2O I
[HA]
[OH-] C ? E
L.R.?
Step by step before eqiv. point
Find total volume after adding base
Find [HA] and [OH-] before rxn: Dilution
ICE table and find limiting reagent (L.R.)
Solve ICE table: find [HA] and [A-] after rxn. I + C = E
Apply Henderson-Hasselbalch equation
Answer keys: a) 4.41; b) 4.89; c) 5.73

80. Finding the pH During a Weak base-strong acid Titration
PROBLEM:
Calculate the pH during the titration of mL of M methylamine (MeNH2; Kb = 4.4x10-4) after adding the following volumes of M HCl: (a) 0.00 mL; (b) mL
(a) [OH-] = 6.6x10-3M, pH = 11.82
(b) B H+ →
HB+
Initial
40.00 mL x M/70.00 mL
30.00 mL x M/70.00 mL
Change -x +x
End x
0 (L. R.)
x = , pH = pKa + log([B]/[HB+]) = 10.16

81. Finding the pH during a Weak base-strong acid Titration
PROBLEM:
Calculate the pH during the titration of mL of M methylamine (Kb = 4.4 x10-4) after adding the following volumes of M HCl: (c) mL, (d) mL
(c) [HB+] = M, [H+] = = 1.1E-6 M, pH = 5.97
(d) B H+ →
HB+
Initial
40.00 mL x M/90.00 mL
50.00 mL x M/90.00 mL
Change -x +x
End
0 (L. R.) x
[H+] = M, pH = 1.95