How many moles of $\ce{HCl}$ must be added to $\pu{100 ml}$ of a $\pu{0.100 M}$ solution of methylamine ($\mathrm pK_\mathrm b = 3.36$) to give a buffer having a $\mathrm{pH}$ of $10.00$?

(The answer is supposed to be $\pu{8.1 mmol}$.)

My thought process:

First, I found the $\mathrm pK_\mathrm a$ of methylammonium ion:

$$\begin{align} \mathrm pK_\mathrm a(\ce{CH3NH3+}) &= 14 - \mathrm pK_\mathrm b(\ce{CH3NH2}) \\ &= 10.64 \end{align}$$

Substituting this into the Henderson–Hasselbalch equation

$$\begin{align} \mathrm{pH} &= \mathrm pK_\mathrm a + \log\left(\frac{[\ce{CH3NH2}]}{[\ce{CH3NH3+}]}\right) \\ 10.00 &= 10.64 + \log\left(\frac{\pu{10 mmol}}{x}\right) \end{align}$$

where $x$ is the amount of $\ce{HCl}$ that must be added. However, when I solve for $x$ I find $x = \pu{43.7 mmol}$, nowhere near the correct answer. What am I missing?


Since $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 14$, you get $\mathrm{p}K_\mathrm{a} = 10.64$ for the methylammonium cation.

$\ce{HCl}$ protonates methylamine. The amount of methylammonium increases by the same amount methylamine decreases. So

$$\mathrm{p}K_\mathrm{a} = \mathrm{pH} + \log\left(\frac{[\ce{HA}]}{[\ce{A-}]}\right) = \mathrm{pH} + \log\left(\frac{x}{10~\pu{mmol} - x}\right)$$

lets you calculate the value $x$ of $\ce{HCl}$ needed to obtain the desired $\mathrm{pH}$.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.