JEE MAIN - Chemistry (2025 - 7th April Evening Shift - No. 24)
One 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 $\log _{10} 3=0.477$
Explanation
$$\begin{aligned} &\begin{aligned} & \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{\left[\mathrm{NH}_4^{+}\right]}{\left[\mathrm{NH}_3\right]} \\ & \mathbf{p O H}=4.745 \end{aligned}\\ &\text { on adding } 0.05 \text { mole } \mathrm{HCl} \end{aligned}$$
$$\begin{array}{llll} \mathrm{NH}_3+ & \mathrm{H}^{\oplus} & \rightarrow & \mathrm{NH}_4^{\oplus} \\ 0.1 & 0.05 & & 0.1 \\ 0.05 & 0 & & 0.15 \end{array}$$
$$\begin{aligned} & \mathrm{pOH}^{\prime}=4.745+\log 3 \\ & \mathrm{pOH}^{\prime}-\mathrm{pOH}=0.477 \\ & 14-\mathrm{pH}^{\prime}-14+\mathrm{pH}=0.477 \\ & \Delta \mathrm{pH}=0.477 \\ & =47.7 \times 10^{-2} \approx 48 \times 10^{-2} \end{aligned}$$
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