Question

A $25mL$ solution of $0.2MN{H}_{4}OH$ is titrated against $0.2MHCl$. Find the $pH$ of the solution when $25mL$ of $HCl$ is added.
$p{K}_b\left(N{H}_{4}OH\right)=4.74$

• A. 8.37
• B. 5.13
• C. 4.74
• D. 3.55

Answer 1

The correct option is B 5.13

mmol of $N{H}_{4}OH=25\times 0.2=5$
mmol of $HCl=25\times 0.2=5$
$N{H}_{4}OH\left(aq\right)+HCl\left(aq\right)\to N{H}_{4}Cl\left(aq\right)+H_{2}O\left(l\right)Initially:5500Final:0055$
Since only salt of weak base and strong acid i.e. $N{H}_{4}Cl$ is left, it will undergo hydrolysis.
$\text{Concentration}=\frac{\text{Moles}}{\text{Total Volume}}$

$\left[N{H}_{4}Cl\right]=\left[N{H}_4^{+}\right]=\frac{5}{50}=0.1M$

$N{H}_4^{+}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons N{H}_{4}OH\left(aq\right)+H^{+}\left(aq\right)$
$pH=7-\frac{1}{2}(p{K}_b+logC)pH=7-\frac{1}{2}(4.74+log(0.1\left)\right)pH=7-\frac{1}{2}(4.74-1)pH=7-1.87pH=5.13$
Theory:
Generally, Methyl Orange is used for weak base and strong acid titrations.