Question

0.1 mole of $C{H}_{3}N{H}_2(K_b=5\times {10}^{-4})$ is mixed with 0.08 mole of HCl and diluted to one litre. The $\left[H^{+}\right]$ in solution is:

• A. $8\times {10}^{-2}M$
• B. $8\times {10}^{-11}M$
• C. $1.6\times {10}^{-11}M$
• D. $8\times {10}^{-5}M$

Answer 1

Answer: (B)

$8\times {10}^{-11}M$

The reaction occuring is :

$\underset{0.02}{\underset{0.1}{C{H}_{3}N{H}_2}}+\underset{0}{\underset{0.08}{HCl}}\to \underset{0.08}{\underset{0}{C{H}_{3}N{H}_3^{+}Cl}}$

Thus, the reaction mixture contains unreacted methylamine and its hydrochloride salt.

Thus, it is a basic buffer solution.

The expression for the hydroxide ion concentration is $\left[O{H}^{-}\right]=K_b\times \frac{\left[Base\right]}{\left[\text{Conjugate acid}\right]}$

Substitute values in the above expression.

$\left[O{H}^{-}\right]=5\times {10}^{-4}\times \frac{0.02}{0.08}=1.25\times {10}^{-4}$

But $\left[H^{+}\right]\left[O{H}^{-}\right]=k_w={10}^{-14}$

$\therefore \left[H^{+}\right]=\frac{{10}^{-14}}{\left[O{H}^{-}\right]}=\frac{{10}^{-14}}{1.25\times {10}^{-4}}=8\times {10}^{-11}M$