Question

`0.1` millie moles of `CdSO_(4)` are present in `10ml` acid solution of `0.08 N HCI`. Now `H_(2)S` si passed to precipitate all the `Cd^(2+)` ions. What would be the `pH` of solution after filtering off percipitate, boilling of `H_(2)S` and making the solution `100ml` by adding `H_(2)S`?

Answer 1

`{:(,CdSO_(4)+,HC1+H_(2)Srarr,CdS+H_(2)SO_(4),),("mmol addedr",rArr0.1,10xx0.08,,),("mmol after reaction",rArr0,=0.8,0.1,0.1):}`
mmoles of `H^(o+)` = [From HC1] +[From `H_(2)SO_(4)`]
`= 0.8 +0.1xx2 = 1.0`
Total volume `=100mL`
`:. [H^(o+)] = 1//100 = 10^(-2) M :. pH = 2`