A volume of x mL of 5 M NaHCO3 solution was mixed with 10 mL of 2 M H2CO3 solution to make an electrolytic buffer

Q. A volume of x mL of 5 M NaHCO₃ solution was mixed with 10 mL of 2 M H₂CO₃ solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of 235.3 mV, then the value of x = ______ mL (nearest integer).

Sn(s) | Sn(OH)₆²⁻ (0.5 M) | HSnO₂⁻ (0.05 M) | OH⁻ | Bi₂O₃(s) | Bi(s)

Consider up to one place of decimal for intermediate calculations

Given :
E°_{HSnO₂⁻ / Sn(OH)₆²⁻} = −0.9 V
E°_{Bi₂O₃ / Bi} = −0.44 V
pK_a(H₂CO₃) = 6.11
2.303RT / F = 0.059 V
Antilog (1.29) = 19.5

Correct Answer: 78

Explanation

The given electrochemical cell consists of two half cells involving tin and bismuth. Using the given standard electrode potentials, the standard cell potential is calculated as:

E°_cell = (−0.44) − (−0.90) = +0.46 V

The observed cell potential is 0.2353 V, therefore using the Nernst equation:

E_cell = E°_cell − (0.059 / n) log Q

Substituting the given values and simplifying, the pH of the buffer solution is obtained as:

pH = 6.11 + log ( [HCO₃⁻] / [H₂CO₃] )

From the electrochemical data, log ( [HCO₃⁻] / [H₂CO₃] ) = 1.29

Thus,

[HCO₃⁻] / [H₂CO₃] = 19.5

Moles of H₂CO₃ = 2 × 0.010 = 0.020 mol

Required moles of HCO₃⁻ = 19.5 × 0.020 = 0.39 mol

Volume of 5 M NaHCO₃ needed:

x = 0.39 / 5 = 0.078 L = 78 mL

Hence, the required value of x is 78 mL.

Q. A volume of x mL of 5 M NaHCO3 solution was mixed with 10 mL of 2 M H2CO3 solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of 235.3 mV, then the value of x = ______ mL (nearest integer).

Explanation

Related JEE Main Questions

Q. A volume of x mL of 5 M NaHCO₃ solution was mixed with 10 mL of 2 M H₂CO₃ solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of 235.3 mV, then the value of x = ______ mL (nearest integer).