Question

A $20mL$ sollution of $0.02MN{H}_{4}OH$ is titrated with $0.4MHCl$. Find the $pH$ of the solution when $80mL$ of $N{H}_{4}OH$ is added.
Given $p{K}_a\left(N{H}_{4}OH\right)=4.74$

• A. 0.5
• B. 2.5
• C. 7
• D. 4.4

Answer 1

The correct option is A 0.5

mmol of $N{H}_{4}OH=20\times 0.02=0.4$
mmol of $HCl=80\times 0.4=32$

$N{H}_{4}OH\left(aq\right)+HCl\left(aq\right)\to N{H}_{4}Cl\left(aq\right)+H_{2}O\left(l\right)Initially:0.43200Final:031.60.40.4$
$\text{Final Concentration}=\frac{\text{Moles}}{\text{Total Volume}}\left[HCl\right]=\frac{31.6}{100}=0.316M\left[N{H}_{4}Cl\right]=\frac{0.4}{100}=0.004M$
Since concentration of $HCl$ is fairly high in comparsion to $\left[N{H}_{4}Cl\right]$, we can neglect the concentration of $H^{+}$ that comes from hydrolysis of the salt $\left[N{H}_{4}Cl\right]$.
All the concentration of $H^{+}$ can be assumed from $HCl$ only.
$\left[HCl\right]=\left[H^{+}\right]=0.316MpH=-log\left[H^{+}\right]pH=-log\left(0.316\right)pH=0.5$