`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`?

`0.1` millie moles of `CdSO_(4)` are present in `10ml` acid solution o

1.	`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» `{:(,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`

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`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`?

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