To prepare a buffer of $\mathrm{pH}8.26$ amount of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$ to be added to $500\mathrm{mL}\mathrm{of}0.01\mathrm{M}{\mathrm{NH}}_4\mathrm{OH}$ solution is $\left[{\mathrm{pK}}_{\mathrm{a}}\right({\mathrm{NH}}_4^{+})=9.26]$:

The correct option is B

$0.025\mathrm{mole}$

Explanation for correct option

(b) $0.025\mathrm{mole}$

Buffer solution:

• Buffer solution or buffer is a solution which is a combination of either weak acid and it's salt (conjugate base) or weak base with it's salt (conjugate acid).

Step $1$: Finding ammonium ion concentration

• Consider ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$ (ammonium sulphate) is to be added to $500\mathrm{mL}$ solution= x mol
• Then, $\left[{\mathrm{NH}}_4^{+}\right]=4\mathrm{xM}$

Step $2$: Finding pOH

• Given $\mathrm{pH}=8.26$
• $\mathrm{pOH}=14-\mathrm{pH}$
• $\mathrm{pOH}=14-8.26$
• $\mathrm{pOH}=5.74$

Step $3$: Finding ${\mathrm{pK}}_{\mathrm{b}}$

• Given ${\mathrm{pK}}_a=9.26$
• ${\mathrm{pK}}_{\mathrm{b}}=14-{\mathrm{pK}}_{\mathrm{a}}$
• ${\mathrm{pK}}_{\mathrm{b}}=14-9.26$
• ${\mathrm{pK}}_{\mathrm{b}}=4.74$

Step $4$: Using formula, finding amount of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$ to be added

• $\mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[{\mathrm{NH}}_4^{+}\right]}{\left[{\mathrm{NH}}_4\mathrm{OH}\right]}$
• $5.74=4.74+\log \frac{4\mathrm{x}}{0.01}$
• $\log 4\mathrm{x}\times {10}^2=1$
• $4\mathrm{x}\times {10}^2=1$
• $\mathrm{x}=\frac{1}{40}\mathrm{mol}$
• $\mathrm{x}=0.025\mathrm{mol}$

Hence the amount of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$ to be added is $0.025\mathrm{mole}$.

Explanation for incorrect options

(a) $0.05\mathrm{mole}$

• After calculation, the amount of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$ to be added came out to be $0.025\mathrm{mole}$.
• So option (a) is incorrect.

(c) $0.10\mathrm{mole}$

• By the given data, the amount of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$ to be added is $\text{0.025 mole}$.
• Hence option (c) is incorrect.

(d) $0.005\mathrm{mole}$

• From the given details, the amount of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$ to be added is $0.025\mathrm{mole}$.
• Hence option (d) is incorrect.

Hence option (b) $0.025\mathrm{mole}$ is the correct option.