What is the pH of a 0.145 M solution of (CH3)3N, given that the Kb of (CH3)3N is 6.4 × 10⁻⁵?

• A. 12.45
• B. 7.00
• C. 11.12
• D. 2.88

Answer: (C)

Verified step by step guidance

1. Identify the base dissociation reaction for (CH3)3N in water: (CH3)3N + H2O ⇌ (CH3)3NH⁺ + OH⁻.
2. Write the expression for the base dissociation constant (Kb): Kb = [ (CH3)3NH⁺ ][ OH⁻ ] / [ (CH3)3N ].
3. Set up an ICE (Initial, Change, Equilibrium) table to determine the concentrations of species at equilibrium. Initially, [ (CH3)3N ] = 0.145 M, and [ (CH3)3NH⁺ ] = [ OH⁻ ] = 0 M.
4. Assume that x is the change in concentration for (CH3)3NH⁺ and OH⁻ at equilibrium. Therefore, [ (CH3)3NH⁺ ] = x, [ OH⁻ ] = x, and [ (CH3)3N ] = 0.145 - x.
5. Substitute the equilibrium concentrations into the Kb expression and solve for x: 6.4 × 10⁻⁵ = (x)(x) / (0.145 - x). Once x is found, calculate pOH = -log[OH⁻] and then find pH using pH = 14 - pOH.