Q19 / 75 · JEE Main 2025 (7 Apr Shift 2) · Chemistry
Only litre buffer solution was prepared by adding 0.10 mol each of $\mathrm{NH}_3$ and $\mathrm{NH}_4 \mathrm{Cl}$ in deionised water. The change in pH on addition of 0.05 mol of HCl to the above solution is ________ $\times 10^{-2}$, (Nearest integer) (Given : $\mathrm{pK}_{\mathrm{b}}$ of $\mathrm{NH}_3=4.745$ and $\log _{10} 3=0.477$)
Correct answer: 48
Solution
$\begin{aligned}<br />& \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{\left[\mathrm{NH}_4^{+}\right]}{\left[\mathrm{NH}_3\right]} \\<br>& \mathbf{p O H}=4.745<br />\end{aligned}$
on adding 0.05 mole HCl
$\begin{aligned}<br />& \begin{array}{llll}<br />\mathrm{NH}_3+ & \mathrm{H}^{\oplus} & \rightarrow & \mathrm{NH}_4^{\oplus} \\<br>0.1 & 0.05 & & 0.1 \\<br>0.05 & 0 & & 0.15<br />\end{array} \\<br>& \mathrm{pOH}^{\prime}=4.745+\log 3 \\<br>& \mathrm{pOH}^{\prime}-\mathrm{pOH}=0.477 \\<br>& 14-\mathrm{pH}^{\prime}-14+\mathrm{pH}=0.477 \\<br>& \Delta \mathrm{pH}=0.477 \\<br>& =47.7 \times 10^{-2} \approx 48 \times 10^{-2}<br />\end{aligned}$
on adding 0.05 mole HCl
$\begin{aligned}<br />& \begin{array}{llll}<br />\mathrm{NH}_3+ & \mathrm{H}^{\oplus} & \rightarrow & \mathrm{NH}_4^{\oplus} \\<br>0.1 & 0.05 & & 0.1 \\<br>0.05 & 0 & & 0.15<br />\end{array} \\<br>& \mathrm{pOH}^{\prime}=4.745+\log 3 \\<br>& \mathrm{pOH}^{\prime}-\mathrm{pOH}=0.477 \\<br>& 14-\mathrm{pH}^{\prime}-14+\mathrm{pH}=0.477 \\<br>& \Delta \mathrm{pH}=0.477 \\<br>& =47.7 \times 10^{-2} \approx 48 \times 10^{-2}<br />\end{aligned}$
Hence, the correct answer is 48.