Question

$C{H}_{3}N{H}_2$ ($0.1$ mole, $K_b=5\times {10}^{-4})$ is added to $0.08$ moles of $HCl$ and the solution is diluted to one litre. The resulting hydrogen ion concentration is :

• A. $1.6\times {10}^{-11}$
• B. 8 $\times {10}^{-11}$
• C. $5\times {10}^{-4}$
• D. $8\times 10^{-2}$

Answer 1

The correct option is B 8 $\times {10}^{-11}$

$C{H}_{3}N{H}_2+HCl\to C{H}_{3}N{H}_3{+Cl}^{-}$

Initially 0.1 0.08
After reaction 0.02 0 0.08

$K_b=\frac{\left[C{H}_{3}N{H}_3^{+}\right]\left[O{H}^{-}\right]}{\left[C{H}_{3}N{H}_2\right]}$

$\left[O{H}^{-}\right]=\frac{K_b\left[C{H}_{3}N{H}_2\right]}{C{H}_{3}N{H}_3^{+}}=\frac{5\times {10}^{-4}\times 0.02}{0.08}=1.25\times {10}^{-4}$

$\left[H^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}=\frac{1\times {10}^{-14}}{1.25\times {10}^{-4}}=8\times {10}^{-11}$

Therefore, the option is B.