1. Aim # 9: How are acids and bases formed from salts?
H.W. # 9
Study pp (sec. 14.8) Ans. ques. p. 706 # 113,119,123,129
# 126 (Hint: find Kb, then Ka, etc.)

2. I The relationship between Ka and Kb
For any weak acid,
(1) HA(aq) H+(aq) + A-(aq) Ka
For the salt of its conjugate base,
(2) NaA(aq) → Na+(aq) + A-(aq)
and (3) A-(aq) + H2O(ℓ) HA(aq) + OH-(aq) Kb

3. Adding reactions (1) and (3)
HA(aq) H+(aq) + A-(aq)
A-(aq) + H2O(ℓ) HA(aq) + OH-(aq)
H2O(ℓ) H+(aq) + OH-(aq) Kw
Ka = [H+][A-] Kb = [HA][OH-] [HA] [A-]
Ka x Kb = [H+][A-][HA][OH-] [HA][A-]
Ka x Kb = [H+][OH-] = Kw

4. For any acid (base) and its conjugate base (acid)
Ka x Kb = Kw = 1.0 x 10-14
Problem: Calculate the value of Kb for acetate ion, C2H3O2-.
Ans: For acetic acid, HC2H3O2, Ka = 1.8 x 10-5
KaKb = Kw 1.0 x 10-14
Kb = Kw = 1.0 x = 5.6 x Ka x 10-5
II Some salts in aqueous solution react with water to produce an acidic solution; Some salts react with water to produce a basic solution. The process by which either forms is called hydrolysis.

5. Other salts do not react, and produce a neutral solution.
III Salts that produce basic solutions
sodium acetate, NaC2H3O2, is the salt of a strong base (NaOH) and a weak acid (HC2H3O2).
1. NaC2H3O2(s) + H2O(ℓ) → Na+(aq) + C2H3O2-(aq)
hydrolysis 2. C2H3O2-(aq) + H2O(ℓ) HC2H3O2(aq) + OH-(aq)
Salts formed from a strong base and a weak acid produce basic solutions (pH >7).

6. Problem: Calculate the pH of a 0
Problem: Calculate the pH of a 0.20 M sodium acetate (NaC2H3O2) solution.
Ans: NaC2H3O2(s) + H2O(ℓ) → Na+(aq) + C2H3O2-(aq)
C2H3O2-(aq) + H2O(ℓ) → HC2H3O2(aq) + OH-(aq)
Kb = [HC2H3O2][OH-] = (x)(x) ͘ [C2H3O2] – x
Ka for acetic acid, HC2H3O2, = 1.8 x 10-5
Kb = Kw = 1.0 x = 5.55 x Ka x 10-5
5.55 x = x ≈ x2 ͘ – x

7. x2 = 1.11 x 10-10
[OH-] = x = 1.05 x 10-5 M
pOH = -log(1.05 x 10-5) = 4.95
pH = – 4.98 = 9.02
check: x 10-5 x 100 = .005%
IV Salts that produce acidic solutions
pyridine hydrochloride is the salt of weak base (pyridine, C5H5N) and a strong acid (HCl).
1. C5H5NHCl(s) + H2O(ℓ) → C5H5NH+(aq) + Cl-(aq)
hydrolysis 2. C5H5NH+(aq) C5H5N(aq) + H+(aq)
Salts formed from a weak base and a strong acid
produce acidic solutions (pH < 7)

8. Problem: Calculate the pH of a 0
Problem: Calculate the pH of a 0.30 M pyridine hydrochloride (C5H5NHCl) solution.
Ans: C5H5NHCl(s) + H2O(ℓ) → C5H5NH+(aq) + Cl-(aq)
C5H5NH+(aq) C5H5N(aq) + H+(aq)
Ka = [C5H5N][H+] = (x)(x) ͘ [C5H5NH+] – x
Kb for pyridine, C5H5N, = 1.7 x 10-9
Ka = Kw = 1.0 x = 5.88 x Kb x 10-9
5.88 x 10-6 = (x)2 ≈ x2͘ ͘ – x
x2 = 1.76 x 10-6
[H+] = x = 1.32 x 10-3

9. Check: 1.3 x 10-3 x 100 = .43% .30 pH = -log(1.32 x 10-3) = 2.88
V Salts that produce neutral solutions
sodium chloride is the salt of a strong base (NaOH) and a strong acid (HCl)
1. NaCl(s) + HCl(ℓ) → Na+(aq) + Cl-(aq)
2. Cl-(aq) + H2O(aq) x
Hydrolysis does not occur because HCl is a strong acid that dissociates completely.
The solution is neutral (pH = 7)

10. VI Salts formed from a weak base and a weak acid
ammonium cyanide is the salt of the weak base ammonia (NH3) and the weak acid hydrocyanic acid (HCN)
1. NH3(aq) + HCN(g) → NH4CH(s) 2. NH4CN(s) + H2O(ℓ) → NH4+(aq) + CN-(aq) acidic ion basic ion
If Ka of the acidic ion > Kb of the basic ion, the solution will be acidic.
If Kb of the basic ion > Ka of the acidic ion,
the solution will be basic.

11. For NH4CN,
Ka NH4+ = Kw = 1.0 x = 5.6 x KbNH x 10-5
KbCN- = Kw = 1.0 x = 2.0 x KaHCN x 10-10
KbCN- > KaNH4+ therefore the solution will be basic.

12. Kb pyridine = 1.7 x 10-9 Ka nitrous acid = 4.5 x 10-4
Problem: Will a salt solution of the salt pyridinium nitrite be acidic or basic?
Kb pyridine = 1.7 x Ka nitrous acid = 4.5 x 10-4
Ans: Ka = Kw = 1.0 x = 5.9 x Kb x 10-9
Kb = Kw = 1.0 x = 2.2 x Ka x 10-4
Ka>Kb therefore, the salt solution will be acidic.

13. Problem: Will a salt solution of NaH2PO4 be acidic or basic?
Ans: NaH2PO4(s) + H2O(ℓ) → Na+(aq) + H2PO4-(aq)
H2PO4-(aq) ↔ HPO42-(aq) + H+(aq)
Ka2 = 6.2 x 10-8
H2PO4-(aq) + H2O(ℓ) ↔ H3PO4(aq) + OH-(aq) Kb = Kw = 1.0 x = 1.3 x Ka x 10-3
Because Ka>Kb , the solution will be acidic.

14. Practice Problems
Zumdahl (8th ed.) p. 693 # 126,118,120