1.

The voltage of the cell Pb(s)|PbSO_(4)(s)|NaHSO_(4)(0.600M)||Pb^(2+)(2.50 xx 10^(-5)M)|Pb(s) is E = +0.061V. Calculate K_(2) = [H^(+)] [SO_(4)^(2-)]//[HSO_(4)^(-)] the dissociation constant for HSO_(4)^(-). Given Pb(s) +SO_(4)^(2-) rarr PbSO_(4) +2e^(-) (E^(@) = 0.356V),E^(@) (Pb^(2+)//Pb) =- 0.126V.

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ANSWER :`1.06 XX 10^(-2)`


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