1. Ionic Equilibria in Aqueous Systems
CHEM 1B: GENERAL CHEMISTRY
Chapter 19:
Ionic Equilibria in Aqueous Systems
Instructor: Dr. Orlando E. Raola
Santa Rosa Junior College

2. Chapter 19
Ionic Equilibria in Aqueous Systems

3. 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

4. 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, or a weak base and its conjugate acid.

5. The effect of adding acid or base to an unbuffered solution.
Figure 19.1
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.

6. The effect of adding acid or base to a buffered solution.
Figure 19.2
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 mol/L CH3COOH ( a weak acid) with 1 mol/L CH3COONa (which provides the conjugate base, CH3COO-).

7. 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.

8. Table 19.1 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

9. 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.

10. Figure 19.3 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

11. 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+] decreases.
[HA]
[A-]
If the ratio decreases, [H3O+] increases.
[HA]
[A-]

12. Sample Problem 19.1
Calculating the Effect of Added H3O+ or OH- on Buffer pH
PROBLEM:
(a) Of a buffer solution consisting of 0.50 mol/L CH3COOH and mol/L CH3COONa
(b) After adding mol of solid NaOH to 1.0 L of the buffer solution
(c) After adding mol of HCl to 1.0 L of the buffer solution in (a).
Ka of CH3COOH = 1.8 x (Assume the additions cause a negligible change in volume.)
Calculate the pH:
PLAN:
We can calculate [CH3COOH]init and [CH3COO-]init from the given information. From this we can find the starting pH. For (b) and (c) we assume that the added OH- or H3O+ reacts completely with the buffer components. We write a balanced equation in each case, set up a reaction table, and calculate the new [H3O+].

13. Sample Problem 19.1
SOLUTION: (a)
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
Concentration (mol/L)
Initial
0.50 -
Change - −x +x
Equilibrium - x x x
Since Ka is small, x is small, so we assume
[CH3COOH] = 0.50 – x ≈ 0.50 mol/L and [CH3COO-] = x ≈ 0.50 mol/L
x = [H3O+] =
Ka x
[CH3COOH]
[CH3COO-]
≈ 1.8x10-5 x
0.50
= 1.8x10-5 mol/L
pH = -log(1.8x10-5) = 4.74
Checking the assumption:
1.8x10-5 mol/L
0.50 mol/L
x 100 = 3.6x10-3% (< 5%; assumption is justified.)

14. Sample Problem 19.1
(b)
[OH−]added =
0.020 mol
1.0 L soln
= mol/L OH−
Setting up a reaction table for the stoichiometry:
CH3COOH(aq) + OH-(aq) → CH3COO-(aq) + H2O(l)
Concentration (mol/L)
Initial
0.50
0.020 -
Change
-0.020
+0.020 -
Equilibrium
0.48 -
0.52
Setting up a reaction table for the acid dissociation, using new initial [ ]:
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
Concentration (mol/L)
Initial
0.48 -
0.52
Change - −x +x
Equilibrium - x x x

15. Sample Problem 19.1
Since Ka is small, x is small, so we assume
[CH3COOH] = 0.48 – x ≈ 0.48 mol/L and [CH3COO-] = x ≈ 0.52 mol/L
x = [H3O+] =
Ka x
[CH3COOH]
[CH3COO-]
≈ 1.8x10-5 x
0.48
0.52
= 1.7x10-5 mol/L
pH = -log(1.7x10-5) = 4.77

16. Sample Problem 19.1
(c)
[H3O+]added =
0.020 mol
1.0 L soln
= mol/L H3O+
Setting up a reaction table for the stoichiometry:
CH3COO-(aq) + H3O+(aq) → CH3COOH(aq) + H2O(l)
Concentration (mol/L)
Initial
0.50
0.020 -
Change
-0.020
+0.020 -
Equilibrium
0.48 -
0.52
Setting up a reaction table for the acid dissociation, using new initial [ ]:
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
Concentration (mol/L)
Initial
0.52 -
0.48
Change - −x +x
Equilibrium - x x x

17. Sample Problem 19.1
Since Ka is small, x is small, so we assume
[CH3COOH] = 0.52 – x ≈ 0.52 mol/L and [CH3COO-] = x ≈ 0.48 mol/L
x = [H3O+] =
Ka x
[CH3COOH]
[CH3COO-]
≈ 1.8x10-5 x
0.52
0.48
= 2.0x10-5 mol/L
pH = -log(2.0x10-5) = 4.70

18. 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]

19. 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.

20. The relation between buffer capacity and pH change.
Figure 19.4
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.

21. 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.

22. Sample Problem 19.2
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)?

23. Sample Problem 19.2
PLAN: Since the volumes of the solutions are equal, the scenes represent molarities as well as numbers. We count the particles of each species present in each scene and calculate the ratio of the buffer components.
SOLUTION:
[A-]/[HA] ratios: sample 1, 3/3 = 1; sample 2, 2/4 = 0.5; sample 3, 4/4 = 1; and sample 4, 4/2 = 2.

24. Sample Problem 19.2
(a) As the pH rises, more HA will be converted to A-. The scene with the highest [A-]/[HA] ratio is at the highest pH. Sample 4 has the highest pH because it has the highest ratio.
(b) The buffer with the greatest capacity is the one with the [A-]/[HA] closest to 1. Sample 3 has the greatest buffer capacity.
(c) Sample 2 has a lower [A-]/[HA] ratio than sample 1, so we need to increase the [A-] and decrease the [HA]. This is achieved by adding strong acid to sample 1.

25. 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]

26. Sample Problem 19.3
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 mol/L NaHCO3 to make the buffer? Ka of HCO3- is 4.7x10-11.
PLAN:
The conjugate pair is HCO3- (acid) and CO32- (base), and we know both the buffer volume and the concentration of HCO3-. We can calculate the ratio of components that gives a pH of 10.00, and hence the mass of Na2CO3 that must be added to make 1.5 L of solution.
SOLUTION:
[H3O+] = 10-pH = = 1.0x10-10 mol/L
Ka =
[CO32-][H3O+]
[HCO3-]
HCO3-(aq) + H2O(l) H3O+(aq) + CO32-(aq)

27. Sample Problem 19.3
Preparing a Buffer
[CO32-] =
Ka[HCO3-]
[H3O+] =
(4.7x10-11)(0.20)
1.0x10-10
= mol/L
0.094 mol CO32-
1 L soln
Amount (mol) of CO32- needed = 1.5 L soln x
= 0.14 mol CO32-
g Na2CO3
1 mol Na2CO3
0.14 mol Na2CO3 x
= 15 g Na2CO3
The chemist should dissolve 15 g Na2CO3 in about 1.3 L of 0.20 mol/L NaHCO3 and add more 0.20 mol/L NaHCO3 to make 1.5 L. Using a pH meter, she can then adjust the pH to by dropwise addition of concentrated strong acid or base.

28. 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.

29. Colors and approximate pH range of some common acid-base indicators.
Figure 19.5
Colors and approximate pH range of some common acid-base indicators. pH

30. The color change of the indicator bromthymol blue.
Figure 19.6
The color change of the indicator bromthymol blue.
pH < 6.0
pH =
pH > 7.5

31. 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.

32. Curve for a strong acid–strong base titration.
Figure 19.7
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.

33. 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

34. Calculating the pH during a strong acid–strong base titration
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

35. Example:
40.00 mL of mol/L HCl is titrated with mol/L NaOH.
The initial pH is simply the pH of the HCl solution:
[H3O+] = [HCl]init = mol/L and pH = -log(0.1000) = 1.00
To calculate the pH after mL of NaOH solution has been added:
Initial mol of H3O+ = L HCl x
mol
1 L
= 4.000x10-3 mol H3O+
OH- added = L NaOH x
mol
1 L
= 2.000x10-3 mol OH-
The OH- ions react with an equal amount of H3O+ ions, so
H3O+ remaining = 4.000x10-3 – 2.000x10-3 = 2.000x10-3 mol H3O+

36. [H3O+] =
2.000x10-3 mol
L L
= mol/L
pH = -log( ) = 1.48
The equivalence point occurs when mol of OH- added = initial mol of HCl, so when mL of NaOH has been added.
To calculate the pH after mL of NaOH solution has been added:
OH- added = L NaOH x
mol
1 L
= 5.000x10-3 mol OH-
OH- in excess = 5.000x10-3 – 4.000x10-3 = 1.000x10-3 mol OH-
[OH-] =
1.000x10-3 mol
L L
= mol/L
pOH = -log( ) = 1.95
pH = – 1.95 = 12.05

37. Curve for a weak acid–strong base titration.
Figure 19.8
Curve for a weak acid–strong base titration.
The pH increases slowly beyond the equivalence point.
The pH at the equivalent point is > 7.00 due to the reaction of the conjugate base with H2O.
The curve rises gradually in the buffer region. The weak acid and its conjugate base are both present in solution.
The initial pH is higher than for the strong acid solution.

38. Calculating the pH during a weak acid–strong base titration
Initial pH
[H3O+] =
pH = -log[H3O+]
[H3O+][A-]
[HA]
Ka =
pH before equivalence point
[H3O+] = Ka x or
pH = pKa + log
[base]
[acid]
[HA]
[A-]

39. Calculating the pH during a weak acid–strong base titration
pH at the equivalence point
[OH- ] =
where [A-] = and Kw Ka
Kb =
mol HAinit
Vacid + Vbase
[H3O+] ≈ and pH = -log[H3O+]
A-(aq) + H2O(l) HA(aq) + OH-(aq)
pH beyond the equivalence point
[OH-] =
pH = -log[H3O+]
mol OH-excess
Vacid + Vbase
[H3O+] = Kw
[OH-]

40. Sample Problem 19.4
Finding the pH During a Weak Acid–Strong Base Titration
PROBLEM:
Calculate the pH during the titration of mL of mol/L propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of mol/L NaOH:
(a) 0.00 mL (b) mL (c) mL (d) mL
PLAN:
The initial pH must be calculated using the Ka value for the weak acid. We then calculate the number of moles of HPr present initially and the number of moles of OH- added. Once we know the volume of base required to reach the equivalence point we can calculate the pH based on the species present in solution.
SOLUTION:
(a) [H3O+] =
= 1.1x10-3 mol/L
pH = -log(1.1x10-3) = 2.96

41. Sample Problem 19.4
(b) mL of mol/L NaOH has been added.
Initial amount of HPr = L x mol/L = 4.000x10-3 mol HPr
Amount of NaOH added = L x mol/L = 3.000x10-3 mol OH-
Each mol of OH- reacts to form 1 mol of Pr-, so
HPr(aq) OH-(aq) → Pr-(aq) + H2O(l)
Concentration (mol/L)
Initial -
Change − -
Equilibrium -
[H3O+] = Ka x
[HPr]
[Pr-]
= (1.3x10-5) x
= 4.3x10-6 mol/L
pH = -log(4.3x10-6) = 5.37

42. Sample Problem 19.4
(c) mL of mol/L NaOH has been added.
This is the equivalence point because mol of OH- added = = mol of HAinit.
All the OH- added has reacted with HA to form mol of Pr-.
mol
L L
[Pr-] =
= mol/L
Pr- is a weak base, so we calculate Kb = Kw Ka =
1.0x10-14
1.3x10-5
= 7.7x10-10
[H3O+] ≈ Kw =
1.0x10-14
= 1.6x10-9 mol/L
pH = -log(1.6x10-9) = 8.80

43. (d) 50.00 mL of 0.1000 mol/L NaOH has been added.
Sample Problem 19.4
(d) mL of mol/L NaOH has been added.
Amount of OH- added = L x mol/L = mol
Excess OH- = OH-added – HAinit = – = mol
[OH-] =
mol OH-excess
total volume =
mol L
= mol/L
[H3O+] = Kw
[OH-] =
1x10-14
= 9.0x10-13 mol/L
pH = -log(9.0x10-13) = 12.05

44. Curve for a weak base–strong acid titration.
Figure 19.9
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.

45. Titration of 40.00 mL of 0.1000 mol/L H2SO3 with 0.1000 mol/L NaOH
Figure 19.10
Curve for the titration of a weak polyprotic acid.
Titration of mL of mol/L H2SO3 with mol/L 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.
Figure 19.11
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.
We assume that the 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)
Qc =
[Pb2+][F-]2
[PbF2]
Qsp = Qc[PbF2] = [Pb2+][F-]2

49. Qsp and Ksp
Qsp is called the ion-product expression for a slightly soluble ionic compound.
For any slightly soluble compound MpXq, which consists of ions Mn+ and Xz-,
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
Metal sulfides behave differently from most other slightly soluble ionic compounds, since the S2- ion is strongly basic.
We can think of 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. Sample Problem 19.5
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
PLAN:
We write an equation for a saturated solution of each compound, and then write the ion-product expression at equilibrium, Ksp. Note the sulfide in part (d).
SOLUTION:
(a) MgCO3(s) Mg2+(aq) + CO32-(aq)
Ksp = [Mg2+][CO32-]
(b) Fe(OH)2(s) Fe2+(aq) + 2OH-(aq)
Ksp = [Fe2+][OH-]2
(c) Ca3(PO4)2(s) Ca2+(aq) + 2PO43-(aq)
Ksp = [Ca2+]3[PO43-]2

52. Sample Problem 19.5
(d) Ag2S(s) Ag+(aq) + S2-(aq)
S2-(aq) + H2O(l) → HS-(aq) + OH-(aq)
Ag2S(s) + H2O(l) Ag+(aq) + HS-(aq) + OH-(aq)
Ksp = [Ag+]2[HS-][OH-]

53. Table 19.2 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

54. Sample Problem 19.6
Determining Ksp from Solubility
PROBLEM:
(a) 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?
(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.
PLAN:
We write the dissolution equation and the ion-product expression for each compound. This tells us the number of moles of each ion formed. We use the molar mass to convert the solubility of the compound to molar solubility (molarity), then use it to find the molarity of each ion, which we can substitute into the Ksp expression.

55. Sample Problem 19.6
SOLUTION:
(a) PbSO4(s) Pb2+(aq) + SO42-(aq)
Ksp = [Pb2+][SO42-]
Converting from g/mL to mol/L:
4.25x10-3g PbSO4
100 mL soln x
1000 mL
1 L x
1 mol PbSO4
303.3 g PbSO4
= 1.40x10-4 mol/L PbSO4
Each mol of PbSO4 produces 1 mol of Pb2+ and 1 mol of SO42-, so [Pb2+] = [SO42-] = 1.40x10-4 mol/L
Ksp = [Pb2+][SO42-] = (1.40x10-4)2
= 1.96x10-8

56. Sample Problem 19.6
(b) PbF2(s) Pb2+(aq) + F-(aq)
Ksp = [Pb2+][F-]2
Converting from g/L to mol/L:
0.64 g PbF2
1 L soln x
1 mol PbF2
245.2 g PbF2
= 2.6x10-3 mol/L PbF2
Each mol of PbF2 produces 1 mol of Pb2+ and 2 mol of F-, so
[Pb2+] = 2.6x10-3 mol/L and [F-] = 2(2.6x10-3) = 5.2x10-3 mol/L
Ksp = [Pb2+][F-]2 = (2.6x10-3)(5.2x10-3)2
= 7.0x10-8

57. Sample Problem 19.7
Determining Solubility from Ksp
PROBLEM:
Calcium hydroxide (slaked lime) is a major component of mortar, plaster, and cement, and solutions of Ca(OH)2 are used in industry as a strong, inexpensive base. Calculate the molar solubility of Ca(OH)2 in water if the Ksp is 6.5x10-6.
PLAN:
We write the dissolution equation and the expression for Ksp. We know the value of Ksp, so we set up a reaction table that expresses [Ca2+] and [OH-] in terms of S, the molar solubility. We then substitute these expressions into the Ksp expression and solve for S.
SOLUTION:
Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)
Ksp = [Ca2+][OH-]2 = 6.5x10-6

58. Sample Problem 19.7
Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)
Concentration (mol/L) -
Initial
Change - +S
+ 2S
Equilibrium - S 2S
Ksp = [Ca2+][OH-]2 = (S)(2S)2 = 4S3 = 6.5x10-6
S = =
= 1.2x10-2 mol/L

59. Table 19.3 Relationship Between Ksp and Solubility at 25°C
No. of Ions
Formula
Cation/Anion
Ksp
Solubility (mol/L) 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.

60. The effect of a common ion on solubility.
Figure 19.12
The effect of a common ion on solubility.
PbCrO4(s) Pb2+(aq) + CrO42-(aq)
If Na2CrO4 solution is added to a saturated solution of PbCrO4, it provides the common ion CrO42-, causing the equilibrium to shift to the left. Solubility decreases and solid PbCrO4 precipitates.

61. Sample Problem 19.8
Calculating the Effect of a Common Ion on Solubility
PROBLEM:
In Sample Problem 19.7, we calculated the solubility of Ca(OH)2 in water. What is its solubility in 0.10 mol/L Ca(NO3)2? Ksp of Ca(OH)2 is 6.5x10-6.
PLAN:
The addition of Ca2+, an ion common to both solutions, should lower the solubility of Ca(OH)2. We write the equation and Ksp expression for the dissolution and set up a reaction table in terms of S, the molar solubility of Ca(OH)2. We make the assumption that S is small relative to [Ca2+]init because Ksp is low. We can then solve for S and check the assumption.
SOLUTION:
Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)
Ksp = [Ca2+][OH-]2

62. Sample Problem 19.8
[Ca2+]init = 0.10 mol/L because Ca(NO3)2 is a soluble salt, and dissociates completely in solution.
Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)
Concentration (mol/L) -
Initial
0.10
Change - +S
+ 2S
Equilibrium - S 2S
Ksp = [Ca2+][OH-]2 = 6.5x10-6 ≈ (0.10)(2S)2 = (0.10)(4S2)
so S ≈
4S2 ≈
6.5x10-6
0.10
= 4.0x10-3 mol/L
Checking the assumption:
4.0x10-3 mol/L
0.10 mol/L
x 100
= 4.0% < 5%

63. 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.

64. Figure 19.13
Test for the presence of a carbonate.
When a carbonate mineral is treated with HCl, bubbles of CO2 form.

65. Sample Problem 19.9
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
PLAN:
We write the balanced dissolution equation for each compound and note the anion. The anion of a weak acid reacts with H3O+, causing an increase in solubility.
SOLUTION:
(a) PbBr2(s) Pb2+(aq) + 2Br-(aq)
Br- is the anion of HBr, a strong acid, so it does not react with H3O+. The addition of strong acid has no effect on its solubility.

66. Sample Problem 19.9
(b) Cu(OH)2(s) Cu2+(aq) + 2OH-(aq)
OH- is the anion of H2O, a very weak acid, and is in fact a strong base. It will react with H3O+:
OH-(aq) + H3O+(aq) → 2H2O(l)
The addition of strong acid will cause an increase in solubility.
(c) FeS(s) Fe2+(aq) + S2-(aq)
S2- is the anion of HS-, a weak acid, and is a strong base. It will react completely with water to form HS- and OH-. Both these ions will react with added H3O+:
HS-(aq) + H3O+(aq) → H2S(aq) + H2O(l)
OH-(aq) + H3O+(aq) → 2H2O(l)
The addition of strong acid will cause an increase in solubility.

67. Limestone cave in Nerja, Málaga, Spain.
Figure 19.14
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 P 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
CO2(g) CO2(aq)
CO2(aq) + 2H2O(l) H3O+(aq) + HCO3-(aq)
CaCO3(s) + CO2(aq) + H2O(l) Ca2+(aq) + 2HCO3-(aq)

68. Predicting the Formation of a Precipitate
For a saturated solution of a slightly soluble ionic salt, Qsp = Ksp.
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 until the remaining solution is saturated.
If Qsp =< Ksp,
no precipitate will form because the solution is unsaturated.

69. Sample Problem 19.10
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 mol/L Ca(NO3)2 is mixed with L of mol/L NaF?
PLAN:
First we need to decide which slightly soluble salt could form, look up its Ksp value in Appendix C, and write the dissolution equation and Ksp expression. We find the initial ion concentrations from the given volumes and molarities of the two solutions, calculate the value for Qsp and compare it to Ksp.
SOLUTION:
The ions present are Ca2+, NO3-, Na+, and F-. All Na+ and NO3- salts are soluble, so the only possible precipitate is CaF2 (Ksp = 3.2x10-11).
CaF2(s) Ca2+(aq) + 2F-(aq) Ksp = [Ca2+][F-]2

70. Sample Problem 19.10
Ca(NO3)2 and NaF are soluble, and dissociate completely in solution.
We need to calculate [Ca2+] and [F-] in the final solution.
Amount (mol) of Ca2+ = mol/L Ca2+ x L = mol Ca2+.
[Ca2+]init =
0.030 mol Ca2+
0.100 L L
= 0.10 mol/L Ca2+
Amount (mol) of F- = mol/L F- x L = mol F-.
[F-]init =
0.012 mol F-
0.100 L L
= mol/L F-
Qsp = [Ca2+]init[F-]2init = (0.10)(0.040)2 = 1.6x10-4
Since Qsp > Ksp, CaF2 will precipitate until Qsp = 3.2x10-11.

71. Sample Problem 19.11
Using Molecular Scenes to Predict Whether a Precipitate Will Form
PROBLEM:
These four scenes represent solutions of silver (gray) and carbonate (black and red) ions above solid silver carbonate. (The solid, other ions, and water are not shown.)
(a) Which scene best represents the solution in equilibrium with the solid?
(b) In which, if any, other scene(s) will additional solid silver carbonate form?
(c) Explain how, if at all, addition of a small volume of concentrated strong acid affects the [Ag+] in scene 4 and the mass of solid present.

72. Sample Problem 19.11
PLAN: We need to determine the ratio of the different types of ion in each solution. A saturated solution of Ag2CO3 should have 2Ag+ ions for every 1 CO32- ion. For (b) we need to compare Qsp to Ksp. For (c) we recall that CO32- reacts with H3O+.
SOLUTION:
First we determine the Ag+/CO32- ratios for each scene.
Scene 1: 2/4 or 1/2 Scene 2: 3/3 or 1/1
Scene 3: 4/2 or 2/1 Scene 4: 3/4
(a) Scene 3 is the only one that has an Ag+/CO32- ratio of 2/1, so this scene represents the solution in equilibrium with the solid.

73. Sample Problem 19.11
(b) We use the ion count for each solution to determine Qsp for each one. Since Scene 3 is at equilibrium, its Qsp value = Ksp.
Scene 1: Qsp = (2)2(4) = 16
Scene 2: Qsp = (3)2(3) = 27
Scene 3: Qsp = (4)2(2) = 32
Scene 4: Qsp = (3)2(4) = 36
Scene 4 is the only one that has Qsp > Ksp, so a precipitate forms in this solution.
(c) Ag2CO3(s) Ag+(aq) + CO32-(aq)
CO32-(aq) + 2H3O+(aq) → [H2CO3(aq)] + 2H2O(l) → 3H2O(l) + CO2(g)
The CO2 leaves as a gas, so adding H3O+ decreases the [CO32-] in solution, causing more Ag2CO3 to dissolve.
[Ag+] increases and the mass of Ag2CO3 decreases.

74. Selective Precipitation
Selective precipitation is used to separate a solution containing a mixture of ions.
A precipitating ion is added to the solution until the Qsp of the more soluble compound is almost equal to its Ksp.
The less soluble compound will precipitate in as large a quantity as possible, leaving behind the ion of the more soluble compound.

75. Sample Problem 19.12
Separating Ions by Selective Precipitation
PROBLEM:
A solution consists of 0.20 mol/L MgCl2 and 0.10 mol/L 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.
PLAN:
Both compounds have 1/2 ratios of cation/anion, so we can compare their solubilities by comparing their Ksp values. Mg(OH)2 is 1010 times more soluble than Cu(OH)2, so Cu(OH)2 will precipitate first. We write the dissolution equations and Ksp expressions. Using the given cation concentrations, we solve for the [OH-] that gives a saturated solution of Mg(OH)2. Then we calculate the [Cu2+] remaining to see if the separation was successful.

76. Sample Problem 19.12
SOLUTION:
Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) Ksp = [Mg2+][OH-]2 = 6.3x10-10
Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) Ksp = [Cu2+][OH-]2 = 2.2x10-20
[OH-] = =
= 5.6x10-5 mol/L
This is the maximum [OH-] that will not precipitate Mg2+ ion.
Calculating the [Cu2+] remaining in solution with this [OH-]
Ksp
[OH-]2
[Cu2+] = =
2.2x10-20
(5.6x10-5)2
= 7.0x10-12 mol/L
Since the initial [Cu2+] is 0.10 mol/L, virtually all the Cu2+ ion is precipitated.

77. Formation of acidic precipitation.
Chemical Connections
Figure B19.1
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.

78. Figure 19.15
Cr(NH3)63+, a typical complex ion.
A complex ion consists of a central metal ion covalently bonded to two or more anions or molecules, called ligands.

79. Figure 19.16
The stepwise exchange of NH3 for H2O in M(H2O)42+.
The overall formation constant is given by
Kf =
[M(NH3)42+]
[M(H2O)42+][NH3]4

80. Table 19.4 Formation Constants (Kf) of Some Complex Ions at 25°C

81. Sample Problem 19.13
Calculating the Concentration of a Complex Ion
PROBLEM:
An industrial chemist converts Zn(H2O)42+ to the more stable Zn(NH3)42+ by mixing 50.0 L of mol/L Zn(H2O)42+ and 25.0 L of 0.15 mol/L NH3. What is the final [Zn(H2O)42+] at equilibrium? Kf of Zn(NH3)42+ is 7.8x108.
PLAN:
We write the reaction equation and the Kf expression, and use a reaction table to calculate equilibrium concentrations. To set up the table, we must first find [Zn(H2O)42+]init and [NH3]init using the given volumes and molarities. With a large excess of NH3 and a high Kf, we assume that almost all the Zn(H2O)42+ is converted to Zn(NH3)42+.
SOLUTION:
Zn(H2O)42+(aq) + 4NH3(aq) Zn(NH3)42+(aq) + 4H2O(l)
Kf =
[Zn(NH3)42+]
[Zn(H2O)42+][NH3]4

82. 4 mol of NH3 is needed per mol of Zn(H2O4)2+, so
Sample Problem 19.13
[Zn(H2O)42+]initial =
= 1.3x10-3 mol/L
50.0 L x mol/L
50.0 L L
[NH3]initial =
= 5.0x10-2 mol/L
25.0 L x 0.15 mol/L
50.0 L L
4 mol of NH3 is needed per mol of Zn(H2O4)2+, so
[NH3]reacted = 4(1.3x10-3 mol/L) = 5.2x10-3 mol/L and
[Zn(NH3)42+] ≈ 1.3x10-3 mol/L
Zn(H2O)42+(aq) + 4NH3(aq) Zn(NH3)42+(aq) + 4H2O(l)
Concentration (mol/L)
Initial
1.3x10-3
5.0x10-2 -
Change
~(-1.3x10-3)
~(-5.2x10-3)
~(+1.3x10-3) -
Equilibrium x
4.5x10-2
1.3x10-3 -

83. Sample Problem 19.13
Kf =
[Zn(NH3)42+]
[Zn(H2O)42+][NH3]4
= 7.8x108 =
(1.3x10-3)
x(4.5x10-2)4
x = [Zn(H2O)42+ = 4.1x10-7 mol/L

84. Sample Problem 19.14
Calculating the Effect of Complex-Ion Formation on Solubility
PROBLEM:
In black-and-white film developing, excess AgBr is removed from the film negative by “hypo”, an aqueous solution of sodium thiosulfate (Na2S2O3), which forms the complex ion Ag(S2O3)23-. Calculate the solubility of AgBr in (a) H2O; (b) 1.0 mol/L hypo.
Kf of Ag(S2O3)23- is 4.7x1013 and Ksp AgBr is 5.0x10-13.
PLAN:
After writing the equation and the Ksp expression, we use the given Ksp value to solve for S, the molar solubility of AgBr. For (b) we note that AgBr forms a complex ion with S2O32-, which shifts the equilibrium and dissolves more AgBr. We write an overall equation for the process and set up a reaction table to solve for S.

85. Sample Problem 19.14
SOLUTION:
(a) AgBr(s) Ag+(aq) + Br-(aq) Ksp = [Ag+][Br-] = 5.0x10-13
S = [AgBr]dissolved = [Ag+] = [Br-]
Ksp = [Ag+][Br-] = S2 = 5.0x10-13
S = 7.1x10-7 mol/L
(b) Write the overall equation:
AgBr(s) Ag+(aq) + Br-(aq)
Ag+(aq) + 2S2O32-(aq) Ag(S2O3)23-(aq)
AgBr(s) + 2S2O32-(aq) Br-(aq) + Ag(S2O3)23-(aq)
[Br-][Ag(S2O3)23-]
[S2O32-]2
Koverall = Ksp x Kf =
= (5.0x10-13)(4.7x1013)
= 24

86. Sample Problem 19.14 Initial - 1.0 Change - -2S +S Equilibrium -
AgBr(s) + 2S2O32-(aq) Br-(aq) + Ag(S2O3)23-(aq)
Concentration (mol/L)
Initial -
1.0
Change -
-2S +S
Equilibrium - S S
Koverall = S2
( S)2
= 24 S S
= = 4.9 so
S = 4.9 mol/L – 0.9S and 10.9S = 4.9 mol/L
S = [Ag(S2O3)23-] = 0.45 mol/L

87. The amphoteric behavior of aluminum hydroxide.
Figure 19.17
The amphoteric behavior of aluminum hydroxide.
When solid Al(OH)3 is treated with H3O+ (left) or with OH- (right), it dissolves as a result of the formation of soluble complex ions.