Ionic Equilibria Notes

Salts

Salts are formed when acids react with bases in a neutralization reaction.
There are three different types of salts:
- Neutral salt (SA-SB): formed from strong acid and strong base.
- Acidic salt (SA-WB): formed from strong acid and weak base.
- Basic salt (SB-WA): formed from strong base and weak acid.

Hydrolysis is the reaction of a cation or an anion (or both) with water.
The reaction is reversible.
Example:
- X^+ (aq) + H_2O (l) \rightleftharpoons
- Y^- (aq) + H_2O (l) \rightleftharpoons

The relationship between the dissociation constants of acids (Ka) and their conjugate base (Kb) can be derived as follows:
- CH_3COO^- hydrolysis in water:
  - Acid: CH_3COOH
  - Conjugate Base: CH_3COO^-
  - Base: CH_3COO^-
  - Conjugate Acid: CH_3COOH
K_w = [H^+][OH^-]
Ka = \frac{[CH3COO^-][H^+]}{[CH_3COOH]}
Kb = \frac{[CH3COOH][OH^-]}{[CH_3COO^-]}
Ka \times Kb = \frac{[CH3COO^-][H^+]}{[CH3COOH]} \times \frac{[CH3COOH][OH^-]}{[CH3COO^-]} = [H^+][OH^-] = K_w
- Therefore, Ka \times Kb = K_w

Neutral salts are formed from the neutralization reaction of strong acids (SA) with strong bases (SB).
The pH of aqueous solutions of neutral salts is 7 (pH = 7).
Examples: NaCl, KBr, KI, RbBr, BaCl_2
HCl (aq) + NaOH (aq) \rightarrow NaCl (aq) + H_2O (l)
- Strong Acid + Strong Base -> Neutral Salt
In aqueous solution, NaCl dissociates completely to Na^+ and Cl^- ions.
Ions of neutral salt do not undergo hydrolysis
- Na^+ + H_2O \rightarrow No Reaction
- Cl^- + H_2O \rightarrow No Reaction

Acidic salts are formed when strong acids (SA) react with weak bases (WB).
The pH of aqueous solutions of acidic salts is less than 7 (pH < 7).
The cation of acidic salts will undergo hydrolysis.
Examples: NH4Cl, NH4NO_3
HCl (aq) + NH3 (aq) \rightarrow NH4Cl (aq)
- Strong Acid + Weak Base -> Acidic Salt
In aqueous solution, NH4Cl dissociates completely to NH4^+ and Cl^- ions.
- NH4Cl (aq) \rightarrow NH4^+ (aq) + Cl^- (aq)
The NH_4^+ ion of acidic salt undergoes hydrolysis.
In aqueous solution, the NH4^+ ion donates a proton to H2O to form NH3 and H3O^+ ions.
- NH4^+ (aq) + H2O (l) \rightleftharpoons NH3 (aq) + H3O^+ (aq)
Thus, the solution is acidic (acidic salt).

Given that Kb for NH3 is 1.8 \times 10^{-5}, calculate the pH of a 0.20 M NH_4Cl solution.
NH4Cl (aq) \rightarrow NH4^+ (aq) + Cl^- (aq)
- Initial: 0.20 M 0 M 0 M
Hydrolysis of NH_4^+:
- NH4^+ (aq) + H2O (l) \rightleftharpoons NH3 (aq) + H3O^+ (aq)
- Initial: 0.20 M 0 M 0 M
- Change: -x +x +x
- Final: (0.20 – x) M x M x M
Ka = \frac{Kw}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}
Ka = \frac{[H3O^+][NH3]}{[NH4^+]} = \frac{(x)(x)}{(0.20 - x)} \approx \frac{x^2}{0.20}
x = \sqrt{K_a \times 0.20} = \sqrt{5.56 \times 10^{-10} \times 0.20} = 1.05 \times 10^{-5}
- Assume that x is small (0.2- x) 0.2
[H_3O^+] = 1.05 \times 10^{-5} M
pH = -log[H_3O^+] = -log(1.05 \times 10^{-5}) = 4.98

Basic salts are formed from the neutralisation reaction of weak acids (WA) react with strong bases (SB).
The pH of aqueous solutions of basic salts is greater than 7 (pH > 7).
The anion of basic salts will undergo hydrolysis.
Examples: CH3COONa, KNO2
CH3COOH (aq) + NaOH (aq) \rightarrow CH3COONa (aq) + H_2O (l)
- Weak Acid + Strong Base -> Basic Salt
In aqueous solution, CH3COONa dissociates completely to Na^+ and CH3COO^- ions.
- CH3COONa (aq) \rightarrow Na^+ (aq) + CH3COO^- (aq)
The CH_3COO^- ion of basic salt undergoes hydrolysis.
In aqueous solution, the CH3COO^- ion accepts a proton from H2O to form CH_3COOH and OH^- ions.
- CH3COO^- (aq) + H2O (l) \rightleftharpoons CH_3COOH (aq) + OH^- (aq)
Thus, the solution is basic (basic salt).

0. 05 mol sodium benzoate (C6H5COONa) was dissolved in water in a 100 cm3 volumetric flask. What is the pH of the aqueous solution formed?
[Ka for C6H_5COOH is 6.5 \times 10^{-5}]
Concentration of C6H5COONa = \frac{0.05 mol}{100 cm^3} \times \frac{1000 cm^3}{1 dm^3} = 0.5 mol/dm^3 = 0.5 M
C6H5COONa (aq) \rightarrow C6H5COO^- (aq) + Na^+ (aq)
- Initial: 0.5 M 0 M 0 M
Hydrolysis of C6H5COO^-:
- C6H5COO^- (aq) + H2O (l) \rightleftharpoons C6H_5COOH (aq) + OH^- (aq)
- Initial: 0.5 M 0 M 0 M
- Change: -x +x +x
- Final: (0.5 – x) M x M x M
Kb = \frac{Kw}{K_a} = \frac{1.0 \times 10^{-14}}{6.5 \times 10^{-5}} = 1.54 \times 10^{-10}
Kb = \frac{[C6H5COOH][OH^-]}{[C6H_5COO^-]} = \frac{(x)(x)}{(0.5 - x)} \approx \frac{x^2}{0.5}
x = \sqrt{K_b \times 0.5} = \sqrt{1.54 \times 10^{-10} \times 0.5} = 8.77 \times 10^{-6}
- Assume that x is small (0.5- x) 0.5
[OH^-] = 8.77 \times 10^{-6} M
pOH = -log[OH^-] = -log(8.77 \times 10^{-6}) = 5.06
pH = 14 - pOH = 14 - 5.06 = 8.94

The neutralization reactions of weak acids (WA) with weak bases (WB) may produce neutral, basic, or acidic salt.
- Neutral salt – (Ka = Kb) – Ka approximately equal to Kb.
- Acidic salt – (Ka > Kb) – Cation hydrolysis will be more extensive than anion hydrolysis.
- Basic salt – (Ka < Kb) – Anion hydrolysis will be more extensive than cation hydrolysis.

C6H5COONH4 (ammonium benzoate) is an acidic salt because Ka for C6H5COOH (6.5 \times 10^{-5}) is greater than Kb for NH3 (1.8 \times 10^{-5}).

Calculate the pH of 0.10 M N2HCl solution. (K1 for N2H4 = 1.7 \times 10^{-7})
Calculate the pH of 0.42 M NH4NO3 solution. (Kb for NH3 = 1.8 \times 10^{-5})
Calculate the pH of 0.25 M CH3NH3NO3 solution. (Kb for CH3NH2 = 4.4 \times 10^{-4})
Calculate the pH of a 0.15 M sodium acetate solution (CH3COONa). (Ka for CH_3COOH is 1.8 \times 10^{-5})
Calculate the pH of a 0.24 M potassium formate solution (HCOOK). (K_a for HCOOH is 1.7 \times 10^{-4})
Calculate the pH of a 0.75 M potassium hypochlorite solution (KClO). (K_a for HClO is 3.0 \times 10^{-8})
Identify type of salt for the ammonium ethanoate solutions. (a: Ka = 1.8 \times 10^{-5}, b: Kb = 1.8 \times 10^{-5})
Identify type of salt for the ammonium nitrite solutions. (a: Ka = 4.2 \times 10^{-4}, b: Kb = 1.8 \times 10^{-5})
Predict the pH of the following solutions will be acidic, basic, or neutral:
- (a) NH_4I
- (b) CaCl_2
- (c) KCN
- (d) NaClO_4
- (e) CH3NH3Br
- (f) HCOOK