50 mL of 0.01 M solution of `Ca(NO_(3))_(2)` is added to 150 mL of 0.08 M solution of `(NH_(4))_(2)SO_(4)`. Predict Whether `CaSO_(4)` will be precipitated or not. `K_(sp) " of " CaSO_(4) =4 xx 10^(-5)`

Answer

Yes , ionic product `=1.5 xx 10^(-3)`

Explanation

the precipitate of `CaSO_(4)` is to be formed . The solubility equilibrium may be represented as :
`CaSO_(4) (s) hArr Ca^(2+) (aq) +SO_(4)^(2-) (aq)`
Now `Ca^(2+)` ions are to be provided by `Ca(NO_(3))_(2)` solution while `SO_(4)^(2-)` ions by `(NH_(4))_(2)SO_(4)` solution as a result of dissociation.
`Ca(NO_(3))_(2) overset((aq))(to) Ca^(2+)(aq) + 2NO_(3)^(-)(aq) , (NH_(4))_(2)SO_(4) (s) overset((aq))(to) 2NH_(4)^(+) (aq) + SO_(4)^(2-) (aq)`
The total volume of the solution after mixing =(50 +150) = 200 mL
In this solution , the concentration of `Ca^(2+)` ions after mixing will be `1//4 (50//200)` while that `SO_(4)^(2-)` ions will be `3//4(150//200)`
Thus `[Ca^(2+)]` before mixing =0.01 M`
`[Ca^(2+)]` after mixing `=0.01 xx 1/4 =0.025 M`
Similarly , `[SO_(4)^(2-)]` before mixing =0.08 M
`[SO_(4)^(2-)]` after mixing `=(0.08 xx 150)/(200) =0.06 M`
The ionic product `=[Ca^(2)][SO_(4^(2-)]=(0.025)xx (0.06) =1.5 xx 10^(-3)`
`K_(sp)` value of `CaSO_(4) =4 xx 10^(-5)` (Given)
Since the ionic product is more than the solubility product this means that `CaSO_(4)` will get precipitaed.