Question

A $30mL$ solution of $0.2MN{H}_{4}OH$ is titrated with $0.2MHCl$. Find the $pH$ of the solution when $10mL$ of $HCl$ has been added.
Given $p{K}_b\left(N{H}_{4}OH\right)=4.74$

• A. 4.44
• B. 9.56
• C. 5.44
• D. 8.56

Answer 1

The correct option is B 9.56

mmol of $N{H}_{4}OH=30\times 0.2=6$
mmol of $HCl=10\times 0.2=2$

$N{H}_{4}OH\left(aq\right)+HCl\left(aq\right)\to N{H}_{4}Cl\left(aq\right)+H_{2}O\left(l\right)Initial:6200Final:4022$

After titration, $N{H}_{4}Cl$ and $N{H}_{4}OH$ is there which constitutes a basic buffer.
$\text{Final concentration}=\frac{\text{Moles}}{\text{Total Volume}}$
Total Volume $=30+10=40mL$
$[N{H}_{4}OH{]}_{final}=\frac{4}{40}M$
$[N{H}_{4}Cl{]}_{final}=[N{H}_4^{+}{]}_{final}=\frac{2}{40}M$
For a basic buffer,
$pOH=p{K}_b+log\left(\frac{\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}\right)pOH=4.74+log\left(\frac{\frac{2}{40}}{\frac{4}{40}}\right)pOH=4.74+log\left(\frac{1}{2}\right)=4.74-0.3pOH=4.44pH=14-4.44=9.56$