`20 mL` of `1.5 xx 10^(-5)M` barium chloride solution is mixed with `40` mL of `0.9 xx 10^(-5)` sodium sulphate. Will a precipitate get formed ? `(K_(sp) " for " BaSO_(4) = 1xx 10^(-10))`

Explanation

the solubility equilibrium may be represented as :
`BaSO_(4)(s) overset(aq)(hArr) Ba^(2+) (aq) +SO_(4)^(2-) (aq)`
No `Ba^(2+)` ions are to be provided by `BaCl_(2)` solution while `SO_(4)^(2-)` ions by `Na_(2)SO_(4)` solutions as a result of dissociation.
`BaCl_(2)(s) overset(aq)(to) Ba^(2+) (aq) +2Cl^(-) (aq) , Na_(2)SO_(4) overset(aq)(to) 2Na^(+) (aq) +SO_(4)^(2-) (aq)`
The total volume of solution after mixing =(20 +40) =60 mL
In the solution the concentration of `Ba^(2+)` ions after mixing will be `1//3 (20//60)` while that of the `SO_(4)^(2-)` ions will be `2//3(40//60)`
`"Thus, "" " [Ba^(2+)]` before mixing `=1.5 xx 10^(-5) M`
`[Ba^(2+)]` after mixing `=1//3 xx 1.5 xx 10^(-5)M =0.5 xx 10^(-5) M`
Similarly , `[SO_(4)^(2-)]` before mixing `=0.9 xx 10^(-5) M`
`[SO_(4)^(2-)]` after mixing `=2//3 xx0.9 xx10^(-5) M = 0.6 xx 10^(-5)M`
The ionic product `=[Ba^(2+)][SO_(4)^(2-)] =(0.5 xx10^(-5)) xx(0.6 xx 10^(-5)) =3.0 xx 10^(-11)`
`K_(sp)` value of `BaSO_(4) =1xx10^(-10)`
Since the ionic product is less than teh solubility product, this means that `BaSO_(4)` will get precipitated.