A volume of x\, mL of 5\, M \, NaHCO_3 solution was mixed with 10\, mL of 2\, M \, H_2CO_3 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)

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A volume of x\, mL of 5\, M \, NaHCO_3 solution was mixed with 10\, mL of 2\, M \, H_2CO_3 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)_6^ 2- (0.5\, M ) \; | \; HSnO_2^- (0.05\, M ) \; | \; OH^- \; | \; Bi_2O_3(s) \; | \; Bi(s) Consider up to one place of decimal for intermediate calculations. array ll Given: & E^ _ [Sn(OH)_6]^ 2- /HSnO_2^- = -0.90\, V \[4pt]& E^ _ Bi_2O_3/Bi = -0.44\, V \[4pt]& p K_a( H_2CO_3 ) = 6.11 \[4pt]& 2.303\,RT F = 0.059\, V \[6pt]& Antilog (1.29) = 19.5 array

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Correct answer: A. A

We have considered E^ _ [Sn(OH)_6]^ 2- /HSnO_2^- = -0.9 V Pt | HSnO_2^-(aq), [Sn(OH)_6]^ 2- (aq), OH^-(aq) | Bi_2O_3(s) | Bi(s) | 0.5M 0.05M E^ _ cell = +0.9 - 0.44 = 0.46 V Oxidation Half : HSnO_2^- + H_2O + 3OH^- [Sn(OH)_6]^ 2- + 2e^- Reduction Half : Bi_2O_3 + 3H_2O + 6e^- 2Bi + 6OH^- --------------------------------------- 3HSnO_2^-(aq) + Bi_2O_3(s) + 6H_2O + 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 = 7.6 pOH = 6.6 pH = 14 - 6.6 = 7.4 pH = pK_ a_1 + [HCO_3^-] [H_2CO_3] 7.4 = 6.11 + 5x 20 1.29 = x 4 x 4 = 19.5 x = 78

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