# Question #26dfc

Apr 27, 2017

$C .$

#### Explanation:

We use the buffer equation where,

$p H = p {K}_{a} + {\log}_{10} \left\{\frac{\left[{H}_{3} C N {H}_{2}\right]}{\left[{H}_{3} C N {H}_{3}^{+}\right]}\right\}$

But we were quoted $p {K}_{b}$ for $\text{methylamine}$ NOT $p {K}_{a}$ for $\text{methyl ammonium chloride}$.

Since in water, $p {K}_{a} + p {K}_{b} = 14$, clearly, however, $p {K}_{a} = 14 - 3.32 = 10.68$. Thus the concentration of $\text{methylamine}$ is greater than that of $\text{methylamine hydrochloride}$.

Apr 27, 2017

Excellent job. I just want to add that the same conclusion can be drawn using the pOH version of the Henderson - Hasselbalch Equation. pOH = pKb + $\log \left(\frac{C o n j A c i d}{W k B a s e}\right)$. It's just another approach in the mix.

#### Explanation:

Using:
pOH = pKb + ${\log}_{10} \left(\frac{C o n j A c i d}{W k B a s e}\right)$
Conj Acid => $C {H}_{3} N {H}_{3}^{+}$
Wk Base => $C {H}_{3} N {H}_{2}$

2.8 = 3.36 + ${\log}_{10} \left(\frac{C {H}_{3} N {H}_{3}^{+}}{C {H}_{3} N {H}_{2}}\right)$
${\log}_{10} \left(\frac{C {H}_{3} N {H}_{3}^{+}}{C {H}_{3} N {H}_{2}}\right)$ = 2.8 - 3.36 = -0.56
$\left(\frac{C {H}_{3} N {H}_{3}^{+}}{C {H}_{3} N {H}_{2}}\right)$ = ${10}^{- 0.56}$ = 0.275
=> $\left[C {H}_{3} N {H}_{3}^{+}\right] : \left[C {H}_{3} N {H}_{2}\right] =$ 0.275 : 1.00

=> $\left[C {H}_{3} N {H}_{3}^{+}\right] < \left[C {H}_{3} N {H}_{2}\right]$