What is the pH of a 0.145 M solution of (CH3)3N (trimethylamine), given that Kb for (CH3)3N is 6.4 × 10^{-5}?

• A. 11.87
• B. 8.13
• C. 5.00
• D. 2.13

Answer: (A)

Verified step by step guidance

1. Identify the species involved: (CH3)3N (trimethylamine) is a weak base. We are given its concentration (0.145 M) and its base dissociation constant, \(K_b = 6.4 \times 10^{-5}\).
2. Write the base dissociation equilibrium for trimethylamine in water: \(\mathrm{(CH_3)_3N + H_2O \rightleftharpoons (CH_3)_3NH^+ + OH^-}\).
3. Set up an expression for the base dissociation constant \(K_b\) in terms of the hydroxide ion concentration \([OH^-]\): \(K_b = \frac{[OH^-]^2}{[\text{initial base}] - [OH^-]}\). Since \(K_b\) is small, assume \([OH^-]\) is much smaller than the initial concentration, so \([\text{initial base}] - [OH^-] \approx [\text{initial base}]\).
4. Solve for \([OH^-]\) using the simplified expression: \([OH^-] = \sqrt{K_b \times [\text{initial base}]}\).
5. Calculate the pOH from \([OH^-]\) using \(pOH = -\log [OH^-]\), then find the pH using the relation \(pH = 14 - pOH\).