Question

$30\, mL$ of $0.02\, M$ ammonium hydroxide is mixed with $15\, mL$ of $0.02\, M$ $HCl$. What will be the $pH$ of the solution $\left(p K_{b}=4.0\right)$ ?

• A. 4
• B. 8
• C. 6
• D. 10

Answer 1

Answer: (D)

$30\, mL$ of $0.02\, M\, NH _{4} OH +15\, mL$ of $0.02\, M\,HCl$ Number of millimoles of
$NH _{4} OH =30 \times 0.02=0.6\, mol$ Number of millimoles of
$HCl =15 \times 0.02=0.3\, mol In$ an acid base reaction, salt will always form but we
have to check what is left or consumed.
$HCl + NH _{4} OH \rightarrow NH _{4} Cl + H _{2} O$

Basic buffer will form as weak base
$NH _{4} OH$ is left. $\left[ NH _{4} Cl \right]=\frac{0.3}{\text { Total volume }}=\frac{0.3}{45}$
${\left[ NH _{4} OH \right]=\frac{0.3}{\text { Total volume }}=\frac{0.3}{45}}$
Applying $p OH =p K_{a}+\log \frac{[\text { Salt }]}{[\text { Base }]}$
$p O H=4+\log \frac{\frac{03}{45}}{\frac{03}{45}}$
$[\because \log 1=0] p O H=4+\log 1 p O H=4$
$\because p H=14-p O H$
$\therefore p H =14-4$
$p H=10$