.
a. KCHO2 is the salt of the weak acid HCHO2, and since it is a salt of a weak acid, it will hydrolyze in water to produce a basic solution. The hydrolysis reaction is:
CHO2- (aq) + H2O (l) ⇌ HCHO2 (aq) + OH- (aq)
The equilibrium constant for this reaction is:
Kb = Kw / Ka = 1.0 × 10^-14 / 1.8 × 10^-4 = 5.56 × 10^-11
The initial concentration of CHO2- is 0.22 M, and since it hydrolyzes to produce OH-, the concentration of OH- at equilibrium is also equal to x. Let's define x as the concentration of OH- and set up an ICE table:
I: CHO2- H2O ⇌ HCHO2 OH-
C: 0.22 M -- -- --
E: 0.22-x -- x x
The equilibrium constant expression for the hydrolysis reaction is:
Kb = [HCHO2][OH-] / [CHO2-] = x^2 / (0.22 - x)
Solving for x, we get:
x = sqrt(Kb * (0.22 - x)) = sqrt(5.56 × 10^-11 * 0.22) = 1.32 × 10^-6 M
The pH of the solution can be calculated from the concentration of OH-:
pOH = -log[OH-] = -log(1.32 × 10^-6) = 5.88
pH = 14 - pOH = 8.12
Therefore, the pH of the 0.22 M KCHO2 solution is 8.12.
b. CH3NH3I is the salt of the weak base CH3NH2, and since it is a salt of a weak base, it will hydrolyze in water to produce an acidic solution. The hydrolysis reaction is:
CH3NH3+ (aq) + H2O (l) ⇌ CH3NH2 (aq) + H3O+ (aq)
The equilibrium constant for this reaction is:
Ka = Kw / Kb = 1.0 × 10^-14 / 4.4 × 10^-4 = 2.27 × 10^-11
The initial concentration of CH3NH3+ is 0.25 M, and since it hydrolyzes to produce H3O+, the concentration of H3O+ at equilibrium is also equal to x. Let's define x as the concentration of H3O+ and set up an ICE table:
I: CH3NH3+ H2O ⇌ CH3NH2 H3O+
C: 0.25 M -- -- --
E: 0.25-x -- x x
The equilibrium constant expression for the hydrolysis reaction is:
Ka = [CH3NH2][H3O+] / [CH3NH3+] = x^2 / (0.25 - x)
Solving for x, we get:
x = sqrt(Ka * (0.25 - x)) = sqrt(2.27 × 10^-11 * 0.25) = 7.52 × 10^-6 M
The pH of the solution can be calculated from the concentration of H3O+:
pH = -log[H3O+] = -log(7.52 × 10^-6) =