What is the pH of a 0.280 M solution of (CH_3)_2NH (dimethylamine)? The K_b of (CH_3)_2NH is 5.4 \times 10^{-4}.

• A. 12.10
• B. 2.90
• C. 11.23
• D. 9.85

Answer: (A)

Verified step by step guidance

1. Identify the species involved: (CH_3)_2NH (dimethylamine) is a weak base. We are given its concentration (0.280 M) and its base dissociation constant, K_b = 5.4 \times 10^{-4}.
2. Write the base dissociation equilibrium for dimethylamine in water: \(\mathrm{(CH_3)_2NH + H_2O \rightleftharpoons (CH_3)_2NH_2^+ + OH^-}\).
3. Set up an ICE table (Initial, Change, Equilibrium) to express the concentrations of species at equilibrium. Let \(x\) be the concentration of OH\(^-\) produced:
4. \[\begin{array}{c|ccc} & \mathrm{(CH_3)_2NH} & \mathrm{OH^-} & \mathrm{(CH_3)_2NH_2^+} \\ \hline \text{Initial (M)} & 0.280 & 0 & 0 \\ \text{Change (M)} & -x & +x & +x \\ \text{Equilibrium (M)} & 0.280 - x & x & x \end{array}\]
5. Write the expression for the base dissociation constant \(K_b\) in terms of \(x\): \[K_b = \frac{[\mathrm{OH^-}][(CH_3)_2NH_2^+]}{[(CH_3)_2NH]} = \frac{x \times x}{0.280 - x} \approx \frac{x^2}{0.280}\] (assuming \(x\) is small compared to 0.280).
6. Solve for \(x\) (the OH\(^-\) concentration) by rearranging the equation: \[x = \sqrt{K_b \times 0.280}.\] Then calculate the pOH using \[\mathrm{pOH} = -\log[OH^-] = -\log x,\] and finally find the pH using \[\mathrm{pH} = 14 - \mathrm{pOH}.\]