Question

A solution of 0.15M NH3 is only 1.1% ionized in solution. Calculate the pH. The Kb of ammonia is 1.8*10^-5

Answer 1

The reaction is:-

$NH_{{3}_{(aq)}}+H_2O_{(l)}\rightarrow NH^+_{4_{(aq})}+OH^-$

Given concentration of $NH_3=0.15M$

As It ionize 1.1% only So Concentration of $NH_4^+$ will be-

$[NH_4^+]=\dfrac{1.1\times 0.15}{100}=1.65\times 10^{-3}M$

Also, $k_b=\dfrac{[NH_4^+][OH^-]}{[NH_3]}$

$1.8\times 10^{-5}=\dfrac{1.65\times 10^{-3}[OH^-]}{.15}$

$\Rightarrow [OH^-]=\dfrac{0.27\times 10^{-5}}{1.65\times 10^{-3}}=1.63\times 10^{-3}M$

$P_{OH}=-log([OH^-])=-log(1.63\times 10^{-3})$

$=2.79$

As we know, $P_H+P_{OH}=14\\$

$P_H=14-P_{OH}=14-2.79=11.21$

Hence $P_H$ is 11.21.