Question

Calculate the pH of a $0.01MN{H}_{4}OH$ solution containing 1 mol/litre of $N{H}_{4}Cl$ ? $(p{K}_b=4.76)$

• A. 6.8
• B. 7.24
• C. 8.4
• D. 7

Answer 1

The correct option is B 7.24

(i) $\begin{array}{ccccc}N{H}_{4}OH& \rightleftharpoons & N{H}_4^{+}& +& O{H}^{-}\\ (0.01-X)M& & XM& & XM\end{array}$

(ii) $\begin{array}{cccccc}& N{H}_{4}Cl& \rightleftharpoons & N{H}_4^{+}& +& C{l}^{-}\\ \text{Initial}& 1M& & 0M& & 0M\\ \text{Final}& 0.0M& & 1M& & 1M\end{array}$
So at equilibrium, $\left[O{H}^{-}\right]=XM$
$[N{H}_4^{+}{]}_{\text{total}}=[N{H}_4^{+}{]}_{\text{base}}+[N{H}_4^{+}{]}_{\text{salt}}$
$=XM+1M=1M$ (X+1 is close to 1 because $N{H}_{4}OH$ is a weak base )
$\left[N{H}_{4}OH\right]=0.01M=0.01M$
$pOH=p{K}_b+\text{log}\frac{\left[\text{Salt}\right]}{\left[\text{Base}\right]}$
$=4.76+\text{log}\frac{1.0}{0.01}$
$=4.76+\text{log}100=6.76$
$pH=14-pOH$
$=14-6.76=7.24$