InterviewSolution
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InterviewSolution
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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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Answer» Correct Answer - Yes , ionic product `=1.5 xx 10^(-3)` 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. |
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