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

Step-by-step solution

We have considered E^ _ [Sn(OH)₆]²⁻/HSnO₂^- = -0.9 V Pt | HSnO₂^-(aq), [Sn(OH)₆]²⁻(aq), OH^-(aq) | Bi₂O₃(s) | Bi(s) | 0.5M 0.05M E^ _ cell = +0.9 - 0.44 = 0.46 V Oxidation Half : HSnO₂^- + H₂O + 3OH^- [Sn(OH)₆]²⁻ + 2e^- Reduction Half : Bi₂O₃ + 3H₂O + 6e^- 2Bi + 6OH^- --------------------------------------- 3HSnO₂^-(aq) + Bi₂O₃(s) + 6H₂O + 3OH^-(aq) 3[Sn(OH)₆]²⁻(aq) + 2Bi(s) E_ cell = E^ _ cell - 0.059 6 (0.5)^3 (0.05)^3 [OH^-]^3 0.2353 = 0.46 - 0.059 6 3 [ 10 [OH^-] ] [ 10 OH^- ] = 2 0.2247 0.059 = 7.6 1 + pOH = 7