Question

For the preparation of a buffer of $pH=8.26$ , the amount of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$ required to be mixed with one litre of $0.1(\mathrm{M}){\mathrm{NH}}_3$ (aq), ${\mathrm{pK}}_{\mathrm{b}}=4.74$ is ?

• A. $1.0\text{ moles}$
• B. $10.0mole$ s
• C. $0.50mole$ s
• D. $5\text{ moles}$

Answer 1

Answer: (C)

$\text{pOH}={\text{pK}}_{\text{b}}+log\frac{\left[{\text{NH}}_4^{+}\right]}{\left[{\text{NH}}_3\right]}$

$5.74=4.74+log\frac{\left[{\text{NH}}_4^{+}\right]}{0.1}$

or $log\frac{\left[N{H}_4^{+}\right]}{0.1}=1$

or $\frac{\left[N{H}_4^{+}\right]}{0.1}=10$

or $\left[N{H}_4^{+}\right]=1\left(M\right)$

$\therefore {\left({\text{NH}}_4^{+}\right)}_2{\left(\text{SO}\right)}_4$ required $=0.5\text{ mole}$