A current of `1.0A` is passed for `96.5s` trhough a `200mL` solution of `0.05 M LiCl` solution. Find
`a.` The volume of gases produced at STP
`b.` The `pH` of solution at the end of electrolysis

Explanation

Number of faradays passed
`=(It)/(96500)=(1.0xx96.5)/(96500)=10^(-3)F`
Cathode`: 2H^(o+)+2e^(-) rarr H_(2) `
`(Li^(o+)` will remain in solution `)`
`2F-=1 mol `of `H_(2)`
or `1xx10^(-3)F-=0.5xx10^(-3)` mole of `H_(2)`
`-=0.5xx10^(-3)xx22400mL Cl_(2)` at `STP`
`=11.2mL Cl_(2)` at STP
In solution, `Li^(o+)` and `overset(c-)(O)H` are left.
To calculate the `pH` of solution, first calculate the millimoles of `H^(o+)` ions electrolyzed.
`implies` mmoles `H^(o+)` ions electrolyzed `=mEq` of `H^(o+)` ions electrolyzed `=` Number of faradays passed `=10^(-3)F`
Since `H_(2)O` produces equal number of `H^(o+)` and `overset(c-)(O)H` ions,mmoles `overset(c-)(O)H` ions left in excess `=10^(-3)`
`implies [overset(c-)(O)H]~~(10^(-3))/(200//1000)=5xx10^(-3)M`
`[` Neglect `overset(c-)(O)H` from dissociation of `H_(2)O]`
`implies pOH=-log (5xx10^(3))=3-log 5=2.3`
`impliespH=14-pOH=14-2.3=11.7`