When HCl gas is passed through a saturated aqueous solution of NaCl

Explanation

`CuSO_(4) rarr Cu^(2+) + SO_(4)^(2-)`
At cathode: `Cu^(2=) + 2e^(-) rarr Cu(s)`
At anode `2H_(2)O rarr O_(2) + 4H^(+) + 4e^(-)`
Number of Faradays `= (I xx t)/(96500C) = (5A xx 10 xx 60s)/(96500C mol^(-1))= 0.031F`
1 Faraday =1 equivalent of `H^(+)` ions
0.031F= 0.031 equivalent of `H^(+)= 0.031` mol of `H^(+)`
Therefore `[H^(+)]= (0.031)/(3.1)= 0.01 = 10^(-2)M`
`pH= -log [H^(+)] = log [10^(-2)]=2`