Question

A volume of x mL of 5 M NaHCO3 solution is mixed with 10 mL of 2 M H2CO3 solution to make an electrolytic buffer.
If the same buffer is used in the following electrochemical cell and the recorded cell potential is 253.5 mV, then the value of x (nearest integer) is ______ mL.

Electrochemical cell:
Sn(s) | Sn(OH)2(s) | HSnO2− (0.05 M) | OH− (0.05 M) || Bi2O3(s) | Bi(s)

Given:
Standard electrode potential of HSnO2− / Sn(OH)2 = −0.90 V
Standard electrode potential of Bi2O3 / Bi = −0.44 V

pKa of H2CO3 = 6.11
2.303 RT / F = 0.059 V
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.
Cell potential is given by:
Ecell = Ecathode − Eanode + (0.059 / n) × log (oxidized / reduced)

Standard cell potential:
Ecathode = −0.44 V
Eanode = −0.90 V
E°cell = (−0.44) − (−0.90) = 0.46 V

Observed cell potential:
Ecell = 0.2535 V

Step 2: Relation between cell potential and pH.
The decrease in cell potential from standard value is due to pH effect:
Ecell = E°cell − 0.059 × pH

0.2535 = 0.46 − 0.059 × pH
0.059 × pH = 0.2065
pH = 0.2065 / 0.059 = 3.5

Step 3: Henderson–Hasselbalch equation.
pH = pKa + log (NaHCO3 / H2CO3)

3.5 = 6.11 + log (NaHCO3 / H2CO3)
log (NaHCO3 / H2CO3) = −2.61
NaHCO3 / H2CO3 = 10^(−2.61) ≈ 0.00245

Step 4: Calculate moles.
Moles of H2CO3 = 2 × 0.01 = 0.02 mol
Moles of NaHCO3 = 5 × (x / 1000) = 0.005x

Step 5: Apply ratio.
0.005x / 0.02 = 0.00245
x = (0.02 × 0.00245) / 0.005 = 0.0098 L
x = 9.8 mL

Final Answer:
x ≈ 10 mL