JEE Chemistry 2026 Question with Solution

Given: The required buffer has $\mathrm{pH} = 9.25$ and $\mathrm{p}K_b(\mathrm{NH_4OH}) = 4.75$.

Find: Which mixture gives this buffer solution.

Since the buffer consists of a weak base and its conjugate acid, we use the Henderson equation for basic buffer:

$$\text{pOH} = \text{p}K_b + \log\left(\frac{[\text{salt}]}{[\text{base}]}\right)$$

Step 1: Calculate pOH.

$$\text{pH} = 9.25 \Rightarrow \text{pOH} = 14 - 9.25 = 4.75$$

Step 2: Apply Henderson equation.

$$4.75 = 4.75 + \log\left(\frac{[\text{salt}]}{[\text{base}]}\right)$$

So,

$$\log\left(\frac{[\text{salt}]}{[\text{base}]}\right) = 0$$

Hence,

$$\frac{[\text{salt}]}{[\text{base}]} = 1$$

Thus, moles of $\mathrm{NH_4^+}$ formed must equal the remaining moles of $\mathrm{NH_4OH}$.

Step 3: Check option (A). Moles of $\mathrm{NH_4OH}$:

$$0.2 \times 0.5 = 0.1 \text{ mol}$$

Moles of $\mathrm{HCl}$:

$$0.1 \times 0.5 = 0.05 \text{ mol}$$

After neutralization:

$$\text{Remaining } \mathrm{NH_4OH} = 0.05 \text{ mol}, \quad \mathrm{NH_4^+} = 0.05 \text{ mol}$$

Therefore,

$$\frac{[\text{salt}]}{[\text{base}]} = 1$$

which gives the required pH.

Conclude: Option A forms a buffer of $\mathrm{pH} = 9.25$. The correct option is A.

Given: A basic buffer of $\mathrm{NH_4OH}$ and its salt is required to have $\mathrm{pH} = 9.25$.

Find: Which option gives equal conjugate acid and weak base.

For a basic buffer, when salt concentration equals base concentration,

$$\text{pOH} = \text{p}K_b$$

So,

$$\text{pH} = 14 - \text{p}K_b = 14 - 4.75 = 9.25$$

Therefore, we only need the option in which neutralization produces equal moles of $\mathrm{NH_4^+}$ and remaining $\mathrm{NH_4OH}$.

In option A:

• initial $\mathrm{NH_4OH} = 0.1$ mol
• $\mathrm{HCl} = 0.05$ mol
• after reaction, $0.05$ mol base remains and $0.05$ mol salt forms

So the ratio is $1:1$ and the correct option is A.