Question

$10$ mL of $0.25M$ $H_2{SO}_4$ is completely neutralized by $0.125M$ solution of ${NH}_3$. If $K_b$ for ${NH}_3={10}^{-5}$, then the $pH$ of the solution at the equivalence point is:

Answer 1

Volume V of ammonia required for neutralization $=\frac{10ml\times 0.25M\times 2}{0.125M}=40$ mL

The concentration of ammonium ions in the solution is $\left[N{H}_4^{+}\right]=\frac{2\times 10\times 0.25}{10+40}=0.1M$

$N{H}_4^{+}+HOH\rightleftharpoons N{H}_{4}OH+H^{+}$
0.1 M 0 0
0.1-x x x

$K_a=\frac{K_w}{K_b}=\frac{{10}^{-14}}{{10}^{-5}}={10}^{-9}$

$\frac{\left[N{H}_{4}OH\right]\left[H^{+}\right]}{\left[N{H}_4^{+}\right]}=K_a$

$\frac{x\times x}{0.1-x}={10}^{-9}$

$0.1-x\simeq 0.1$

$\frac{x^2}{0.1}={10}^{-9}$

$\left[H^{+}\right]=x={10}^{-5}$

$pH=-log\left[H^{+}\right]=-log{10}^{-5}=5$