One litre of `0.5 M KC1` is electrolysed by passing `9650` coulombs of electricity. The `pH of resulting solution is

Options

A. `1.0`
B. `2.0`
C. `7.0`
D. `13.0`

Answer

D

Explanation

`NaC1 (aq):`
Reduction potention of `H_(2)Ogt ` reduction potential of `Na^(o+)ion`.
Hence water undergoes reduciton to given `H_(2)(g)` and `overset(Theta)OH` ions at calhode.
Cathode `rArr H_(2)O +e^(-) rarr overset(Theta)OH +(1)/(2)H_(2)`
So solution becomes basic
Anode `rArr C1^(Theta) rarr (1)/(2) C1_(2) + e^(-)`
`1F = 1 "Eq of" overset(Theta)OH`
Number of Faraday's `= (9650)/(96500) = 0.1`
`overset(Theta)OH = 0.1 Eq or 0.1 mol`
`V_(sol) = 1L`
`[overset(Theta)OH] = (0.1Eq)/(1L) = 0.1N or M`
`pOH =- log (10^(-1)) = 1`
`pH = 14 -1 = 13`