Question

Calculate $K_b$ for a base whose $0.1M$ solution has pH of $10.5$.

Answer 1

$BOH\rightleftharpoons B^{+}+O{H}^{-}$

$0.1$

$0.1(1-\alpha )$ $0.1\alpha$ $0.1\alpha$

$K_b=\frac{\left(0.1\alpha \right)\left(0.1\alpha \right)}{0.1(1-\alpha )}$

It is given that $pH=10.5$ so $-log\left(\right[H^{+}\left]\right)=10.5$

$\left[H^{+}\right]=3.16\times {10}^{-11}$

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

$\left[O{H}^{-}\right]=3.16\times {10}^{-4}M=0.1\alpha$

$\alpha =3.16\times {10}^{-3}$

$K_b=\frac{\left(0.1\alpha \right)\left(0.1\alpha \right)}{0.1(1-\alpha )}$

As $\alpha <<1$ so $1-\alpha =1$

$K_b=0.1{\alpha }^2={10}^{-6}$