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 $\log _{10} 3=0.477$
Answer (integer)
48
Solution
<p>$$\begin{aligned}
&\begin{aligned}
& \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{\left[\mathrm{NH}_4^{+}\right]}{\left[\mathrm{NH}_3\right]} \\
& \mathbf{p O H}=4.745
\end{aligned}\\
&\text { on adding } 0.05 \text { mole } \mathrm{HCl}
\end{aligned}$$</p>
<p>$$\begin{array}{llll}
\mathrm{NH}_3+ & \mathrm{H}^{\oplus} & \rightarrow & \mathrm{NH}_4^{\oplus} \\
0.1 & 0.05 & & 0.1 \\
0.05 & 0 & & 0.15
\end{array}$$</p>
<p>$$\begin{aligned}
& \mathrm{pOH}^{\prime}=4.745+\log 3 \\
& \mathrm{pOH}^{\prime}-\mathrm{pOH}=0.477 \\
& 14-\mathrm{pH}^{\prime}-14+\mathrm{pH}=0.477 \\
& \Delta \mathrm{pH}=0.477 \\
& =47.7 \times 10^{-2} \approx 48 \times 10^{-2}
\end{aligned}$$</p>
About this question
Subject: Chemistry · Chapter: Equilibrium · Topic: Chemical Equilibrium and Kc, Kp
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