1. [H3O+] =
2.000x10–3 mol
L L
= M
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
= M
pOH = –log( ) = 1.95
pH = – 1.95 = 12.05

2. Curve for a weak acid–strong base titration.
Figure 19.9
Curve for a weak acid–strong base titration.
The pH increases slowly beyond the equivalence
point.
The pH at the equivalence 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.

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

4. 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)
√Kb x [A–]
√Kb x [A–]
pH beyond the equivalence point
[OH–] =
pH = –log[H3O+]
mol OH–excess
Vacid + Vbase
[H3O+] = Kw
[OH–]

5. Sample Problem 19.4
Finding the pH During a Weak Acid–Strong Base Titration
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:
(a) 0.00 mL (b) mL (c) mL (d) mL

8. Curve for a weak base–strong acid titration.
Figure 19.10
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.

9. Figure 19.11
Curve for the titration of a weak polyprotic acid.

10. Start of Ksp

11. 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

12. 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).

13. 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–]

14. 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

15. 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

16. 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.

17. 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.

18. Table 19.3 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.

19. The effect of a common ion on solubility.
Figure 19.13
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.

20. 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 M Ca(NO3)2? Ksp of Ca(OH)2 is 6.5x10–6.

21. 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.

22. 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

23. 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.

24. 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 M Ca(NO3)2 is mixed with L of M NaF?

25. 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.

26. 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.

27. Sample Problem 19.12
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.

28. Cr(NH3)63+, a typical complex ion.
Figure 19.16
Cr(NH3)63+, a typical complex ion.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
A complex ion consists of a central metal ion covalently bonded to two or more anions or molecules, called ligands.

29. Figure 19.17
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

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

31. 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 M Zn(H2O)42+ and 25.0 L of 0.15 M NH3. What is the final [Zn(H2O)42+] at equilibrium? Kf of Zn(NH3)42+ is 7.8x108.

32. 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 M hypo.
Kf of Ag(S2O3)23– is 4.7x1013 and Ksp AgBr is 5.0x10–13.

33. The amphoteric behavior of aluminum hydroxide.
Figure 19.18
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.