Question

0.1 mol of $C{H}_{3}N{H}_2(K_b=6\times {10}^{-4})$ is added to 0.08 mol of $HCl$ and the solution is diluted to $1L$. The hydrogen ion concentration in the solution is:

• A. $5\times {10}^{-5}M$
• B. $8\times {10}^{-2}M$
• C. $1.6\times {10}^{-11}M$
• D. $6.7\times {10}^{-11}M$

Answer 1

The correct option is D $6.7\times {10}^{-11}M$

According to the question, 0.1 mol of $C{H}_{3}N{H}_2$ is added to 0.08 mol of $HCl$.
So,
$C{H}_{3}N{H}_2+HCl\to C{H}_3\stackrel{+}{N}{H}_3+C{l}^{-}$
$Initially0.1mol0.08mol00$
$Finally0.02mol00.08mol0.08mol$

The expression of $K_b$ can be written as,

$C{H}_{3}N{H}_2+H_{2}O\rightleftharpoons C{H}_{3}N{H}_3^{+}+O{H}^{-}$

Hence, $K_b=\frac{\left[C{H}_{3}N{H}_3^{+}\right]\left[O{H}^{-}\right]}{\left[C{H}_{3}N{H}_2\right]}$
$\Rightarrow 6\times {10}^{-4}=\frac{\left(0.08\right)\left[O{H}^{-}\right]}{\left(0.02\right)}$
$\Rightarrow \left[O{H}^{-}\right]=\frac{K_b}{4}=\frac{6\times {10}^{-4}}{4}=1.5\times {10}^{-4}M$
We know, $pH+pOH=14$
$\Rightarrow \left[H^{+}\right]=\frac{{10}^{-14}}{1.5\times {10}^{-4}}\equiv 6.7\times {10}^{-11}M$