Question

$30mL$ of $0.02M$ ammonium hydroxide is mixed with $15mL$ of $0.02M$ $HCl$. 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)

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

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