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50 mL of 0.01 M solution of Ca(NO(3))(2)...

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)`

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Calculation of `Ca^(2+)` concentration ,`M_(1)V_(1) = M_(2)V_(2)`
` 0.01 xx50 = M_(2)xx200`
` :. [Ca(NO_(3))_(2)]` after mixing ` = 2.5 xx10^(-3)` mol `L^(-1)`
Since ` Ca(NO_(3))_(2) ` is completely ionized , `[ Ca^(2+)] = [Ca(No_(3))_(2)] `
` = 2.5 xx10^(-3) ML^(-1)`
Calculation of `So_(4)^(2-)` in concentration
Applying `M_(1)V_(1) = M_(2)V_(2)`
` :. 0.08 xx 150 = M_(2) xx 200`
` :. M_(2) = (0.08 xx 150)/(200) = 6 xx 10^(-2)` M
` :. [ (NH_(4))_(2)SO_(4)] ` is completely ionized , ` [ SO_(4)^(2-)] = [(NH_(4))_(2) SO_(4)] = 6 xx 10^(-2) mol//L `
Ionic product ` = [Ca^(2+) ] [So_(4)^(2-)] = [2.5 xx10^(-3)] [ 6 xx 10^(-2)] = 1.5 xx10^(-4)`
Since Ionic product `(1.5 xx 10^(-4))` is greater than solubility product of `CaSo_(4)(4xx10^(-4)),` hence precipitate of `CaSO_(4)` will be formed .
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