Question

A volume of \(x\) mL of 5 M NaHCO\(_3\) solution was mixed with 10 mL of 2 M H\(_2\)CO\(_3\) solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of 253.5 mV, then the value of \(x =\) ______ mL (nearest integer). \[ \ce{Sn(s) | Sn(OH)2(s) | HSnO2^- (0.05 M) | OH^- (0.05 M) || Bi2O3(s) | Bi(s)} \] Given: \[ E^\circ(\ce{HSnO2^- / Sn(OH)2}) = -0.90 \,\text{V}, \quad E^\circ(\ce{Bi2O3 / Bi}) = -0.44 \,\text{V} \] \[ pK_a(\ce{H2CO3}) = 6.11, \quad \frac{2.303RT}{F} = 0.059 \,\text{V}, \quad \text{Antilog}(1.29) = 19.5 \]

Hint:

For buffer problems in electrochemistry: 1. Use Henderson–Hasselbalch equation to relate pH with buffer ratio. 2. Connect pH to cell potential via Nernst equation. 3. Always check units carefully when converting volumes to moles.

Answer 1

Correct Answer: 10

Step 1: Cell potential relation. \[ E_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} + \frac{0.059}{n}\log\frac{[\text{oxidized species}]}{[\text{reduced species}]} \] Here, \[ E^\circ_{\text{cell}} = (-0.44) - (-0.90) = 0.46 \,\text{V} \] Observed cell potential: \[ E_{\text{cell}} = 0.2535 \,\text{V} \]
Step 2: Henderson–Hasselbalch equation for buffer. \[ pH = pK_a + \log\frac{[\ce{NaHCO3}]}{[\ce{H2CO3}]} \]
Step 3: Relating cell potential to pH. The difference between observed and standard potential is due to pH effect: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - 0.059 \times \text{pH} \] \[ 0.2535 = 0.46 - 0.059 \times \text{pH} \] \[ 0.059 \times \text{pH} = 0.2065 \quad \Rightarrow \quad \text{pH} = 3.5 \]
Step 4: Apply Henderson–Hasselbalch. \[ 3.5 = 6.11 + \log\frac{[\ce{NaHCO3}]}{[\ce{H2CO3}]} \] \[ \log\frac{[\ce{NaHCO3}]}{[\ce{H2CO3}]} = -2.61 \] \[ \frac{[\ce{NaHCO3}]}{[\ce{H2CO3}]} = 10^{-2.61} \approx 0.00245 \]
Step 5: Concentration ratio. Moles of \(\ce{H2CO3}\): \[ n = 2 \times 0.01 = 0.02 \,\text{mol} \] Moles of \(\ce{NaHCO3}\): \[ n = 5 \times \frac{x}{1000} = 0.005x \] Ratio: \[ \frac{0.005x}{0.02} = 0.00245 \quad \Rightarrow \quad x = \frac{0.02 \times 0.00245}{0.005} = 0.0098 \,\text{L} = 9.8 \,\text{mL} \] Final Answer: \[ x \approx 10 \,\text{mL} \]