Question

Equal volumes of 0.02 M `Na_(2)SO_(4)` solution and 0.02 M `BaCI_(2)` solution are mixed together . Predict whether a precipitate will get formed or not. `K_(sp)` value of `BaSO_(4)` is `1.5 xx 10^(-9)`

Answer 1

Correct Answer - Yes ionic product `=10^(-4)`
the precipitate of `BaSO_(4)` is to be formed.
The solubility equilibriuma may be represented as :
`BaSO_(4)(s) hArr Ba^(2+) (aq) +SO_(4)^(2-) (aq)`
Now , `Ba^(+)` ions are to be provided by `BaCI_(2)` solution while `SO_(4)^(2-)` ions by `Na_(2)SO_(4)` solution as a result of dissociation.
`BaCI_(2) (s) overset((aq))(to) Ba^(2+) (aq) +2CI^(-) (aq) , Na_(2)SO_(4) overset((aq))(to) 2Na^(+) (aq) +SO_(4)^(2-) (aq)`
Since equal volumes of the solution have been mixed the concentration of both `Ba^(2+)` ions and `SO_(4)^(2-)` ions after mixing will be reduced to halt.
Thus , `[Ba^(2+)]` after mixing `=(0.02)/(2) =0.01 M , [SO_(4)^(2-)` after mixing `=(0.02)/(2) =0.01 M`
The ionic product `=[Ba^(2+)][SO_(4)^(2-)] =0.01=10^(-4)`
As the ionic product is more the solubility product this means that `BaSO_(4)` will get precipitated.