JEE Chemistry 2026 Question with Solution

Given:

• Buffer is prepared by mixing NaHCO$_3$ and H$_2$CO$_3$.
• Recorded cell potential is $0.2535 \, \text{V}$.
• $E^\circ(\ce{HSnO2^- / Sn(OH)2}) = -0.90 \, \text{V}$
• $E^\circ(\ce{Bi2O3 / Bi}) = -0.44 \, \text{V}$
• $pK_a(\ce{H2CO3}) = 6.11$

Find: The value of $x$ in mL.

Step 1: Cell potential relation Cell potential is given by:

$$E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} + \frac{0.059}{n} \log \left(\frac{\text{oxidized}}{\text{reduced}}\right)$$

Standard cell potential:

$$E^\circ_{\text{cell}} = (-0.44) - (-0.90) = 0.46 \, \text{V}$$

Observed cell potential:

$$E_{\text{cell}} = 0.2535 \, \text{V}$$

Step 2: Relation between cell potential and pH The decrease in cell potential from standard value is due to pH effect:

$$E_{\text{cell}} = E^\circ_{\text{cell}} - 0.059 \times \text{pH}$$

Substituting values:

$$0.2535 = 0.46 - 0.059 \times \text{pH}$$ $$0.059 \times \text{pH} = 0.2065$$ $$\text{pH} = \frac{0.2065}{0.059} = 3.5$$

Step 3: Use Henderson–Hasselbalch equation

$$\text{pH} = pK_a + \log \left(\frac{\text{NaHCO}_3}{\text{H}_2\text{CO}_3}\right)$$

So,

$$3.5 = 6.11 + \log \left(\frac{\text{NaHCO}_3}{\text{H}_2\text{CO}_3}\right)$$ $$\log \left(\frac{\text{NaHCO}_3}{\text{H}_2\text{CO}_3}\right) = -2.61$$ $$\frac{\text{NaHCO}_3}{\text{H}_2\text{CO}_3} = 10^{-2.61} \approx 0.00245$$

Step 4: Calculate moles Moles of H$_2$CO$_3$:

$$2 \times 0.01 = 0.02 \, \text{mol}$$

Moles of NaHCO$_3$:

$$5 \times \frac{x}{1000} = 0.005x$$

Step 5: Apply the ratio

$$\frac{0.005x}{0.02} = 0.00245$$ $$x = \frac{0.02 \times 0.00245}{0.005} = 0.0098 \, \text{L}$$ $$x = 9.8 \, \text{mL}$$

Therefore, the value of $x$ is $10$ mL to the nearest integer.

Given: The cell potential depends on the buffer pH, and the buffer pH depends on the ratio of NaHCO$_3$ to H$_2$CO$_3$.

Find: First determine pH from the electrochemical data, then determine $x$ from the buffer ratio.

The standard cell potential is:

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (-0.44) - (-0.90) = 0.46 \, \text{V}$$

The measured value is smaller, so the solution pH contributes a subtraction term:

$$E_{\text{cell}} = E^\circ_{\text{cell}} - 0.059 \times \text{pH}$$

Hence,

$$0.2535 = 0.46 - 0.059 \times \text{pH}$$

which gives

$$\text{pH} = 3.5$$

Now apply the Henderson–Hasselbalch equation:

$$\text{pH} = pK_a + \log \left(\frac{\text{salt}}{\text{acid}}\right)$$ $$3.5 = 6.11 + \log \left(\frac{\text{NaHCO}_3}{\text{H}_2\text{CO}_3}\right)$$

Thus,

$$\frac{\text{NaHCO}_3}{\text{H}_2\text{CO}_3} = 10^{-2.61} \approx 0.00245$$

The acid moles are:

$$10 \, \text{mL of } 2 \, \text{M} = 0.01 \, \text{L} \times 2 = 0.02 \, \text{mol}$$

The salt moles are:

$$\frac{x}{1000} \times 5 = 0.005x$$

Now use the mole ratio:

$$\frac{0.005x}{0.02} = 0.00245$$

Solving,

$$x = 9.8 \, \text{mL}$$

Therefore, the nearest integer value is $10$.