Using the Henderson-Hasselbalch equation, what is the pH of a buffer system that contains 0.87 M dimethylamine ((CH3)2NH) and 0.68 M dimethylammonium ion ((CH3)2NH2+)? The Kb of dimethylamine is 5.9 × 10⁻⁴.

• A. pH = 9.78
• B. pH = 10.22
• C. pH = 8.50
• D. pH = 11.00

Answer: (B)

Verified step by step guidance

1. Start by understanding the Henderson-Hasselbalch equation, which is used to calculate the pH of a buffer solution: $\mathrm{pH}=\mathrm{pK}a+\frac{\log }{\frac{\mathrm{[A}-]}{\mathrm{[HA]}}}$.
2. Since we are given the Kb of dimethylamine, we need to convert it to Ka using the relation: $Kw=Ka\times Kb$, where $Kw$ is the ion product of water, $1.0\times {10}^{-}$.
3. Calculate $Ka$ using the formula: $Ka=\frac{Kw}{Kb}$. Substitute the values: $\frac{1.0\times {10}^{-}}{5.9\times {10}^{-}}$.
4. Once you have $Ka$, calculate $\mathrm{pK}a$ using the formula: $\mathrm{pK}a=-\log \left(Ka\right)$.
5. Finally, substitute the values into the Henderson-Hasselbalch equation: $\mathrm{pH}=\mathrm{pK}a+\log \left(\frac{0.87}{0.68}\right)$ to find the pH of the buffer system.