Question

What is the $\left[{OH}^{-}\right]$ in a $0.01M$ solution of aniline hydrochloride? $K_b$ for aniline is $4.0\times {10}^{-10}$?

Answer 1

$C_6{H}_5{NH}_3^{+}+H_{2}O\rightleftharpoons C_6{H}_5{NH}_2+H_3{O}^{+}$

$C$

$C(1-h)$ $Ch$ $Ch$

$K_h=\frac{\left(Ch\right)\left(Ch\right)}{C(1-h)}$

Let '$h<<1$'

So $K_h=C{h}^2$

We know that for a salt of weak base and strong acid $K_h=\frac{K_w}{K_b}$

So $K_h=2.5\times {10}^{-5}$

$C{h}^2=2.5\times {10}^{-5}$

$h=0.05$

$\left[H^{+}\right]=Ch=5\times {10}^{-4}M$

We know that $\left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-14}$

So $\left[O{H}^{-}\right]=2\times {10}^{-11}M$