JEE Chemistry 2025 Question with Solution

Given: A buffer is prepared with $0.10 \, \text{mol}$ each of $NH_3$ and $NH_4Cl$ in about $1 \, \text{L}$ solution. Then $0.05 \, \text{mol}$ of $HCl$ is added. Also, $pK_b(NH_3) = 4.745$ and $\log_{10}3 = 0.477$.

Find: The change in pH in the form _____ $\times 10^{-2}$.

For a basic buffer,

$$pOH = pK_b + \log_{10}\frac{[\text{salt}]}{[\text{base}]}$$

Initially, moles of salt and base are equal, so

$$pOH_{\text{initial}} = 4.745 + \log_{10}\frac{0.10}{0.10} = 4.745$$

Hence,

$$pH_{\text{initial}} = 14 - 4.745 = 9.255$$

Now the added acid reacts as

$$NH_3 + HCl \rightarrow NH_4^+ + Cl^-$$

After adding $0.05 \, \text{mol}$ of $HCl$:

• moles of $NH_3$ become $0.10 - 0.05 = 0.05$
• moles of $NH_4^+$ become $0.10 + 0.05 = 0.15$

So,

$$pOH_{\text{final}} = 4.745 + \log_{10}\frac{0.15}{0.05}$$ $$pOH_{\text{final}} = 4.745 + \log_{10}3 = 4.745 + 0.477 = 5.222$$

Therefore,

$$pH_{\text{final}} = 14 - 5.222 = 8.778$$

The magnitude of change in pH is

$$|\Delta pH| = |8.778 - 9.255| = 0.477$$

Now,

$$0.477 = 47.7 \times 10^{-2}$$

Nearest integer is $48$.

Therefore, the required numerical answer is 48.

Given: The buffer has equal initial moles of $NH_3$ and $NH_4Cl$, and $0.05 \, \text{mol}$ of $HCl$ is added.

Find: The change in pH.

Because the solution volume remains effectively constant, the Henderson-Hasselbalch relation can be applied directly using mole ratios.

Initially,

$$\frac{\text{salt}}{\text{base}} = \frac{0.10}{0.10} = 1$$

so,

$$pOH_{\text{initial}} = pK_b$$

After reaction with $HCl$,

$$NH_3: 0.10 \to 0.05, \qquad NH_4^+: 0.10 \to 0.15$$

Thus the new ratio is

$$\frac{\text{salt}}{\text{base}} = \frac{0.15}{0.05} = 3$$

Hence the increase in $pOH$ is exactly

$$\Delta pOH = \log_{10}3 = 0.477$$

Since $pH = 14 - pOH$, the magnitude of change in pH is also

$$|\Delta pH| = 0.477 = 47.7 \times 10^{-2}$$

So the nearest integer is 48.

Therefore, the required answer is 48.