A current strength of `1.0 A` is passed for `96.5s` through `200mL` of a solution of `0.05 M KCl` . Find
`a.` The amoudn of gases produced
`b.` The concentration of final solution `w.r.t. overset(c-)(O)H` ions
`c. pH` of the solution.

Explanation

`a." "Cl^(c-)rarr (1)/(2)Cl_(2)+e^(-)" "(`Anode`)`
`H_(2)O+e^(-)rarr2H_(2)+overset(c-)(O)H." ".(`Cathode`)`
Electricity passed `=(96.5xx1)/(96500)=(10^(-3))/(2)mol `of `e^(-)`
`:.10^(-3) mol ` of `e^(-)` produces `(10^(-3))/(2) mol` of `Cl_(2)`
`=(10^(-3))/(2)xx71=35.5xx10^(-3)g`
Also, `10^(-3)mol `of `e^(-)` produces `=(10^(-3))/(2) mol` of `H_(2)`
`=(10_(3))/(2)xx2=10^(-3)g`
Total weight of gases `=10^(-3)xx35.5+10^(-3)=0.0365g`
`b.` `[overset(x-)(O)H]=(10^(-3)mol)/(200mL(volume))=(1mmol)/(200mL)=0.005M`
`c. pOH=-log[0.005]=2.3impliespH=14-2.3=11.7`