Given that solubility product of `BaSO_(4)` is `1 xx 10^(-10)`, will precipiate form when
a. Equal volumes of `2 xx 10^(-3)M BaC1_(2)` solution and `2 xx 10^(-4)M Na_(2)SO_(4)` solution, are mixed?
b. Equal volumes of `2 xx 10^(-8) M BaC1_(2)` solution and `2 xx 10^(-3)M Na_(2)SO_(4)` solution, are mixed?
c. `100mL` of `10^(-3)M BaC1_(2)`and `400mL` of `10^(-6)M Na_(2)SO_(4)` are mixed?

Explanation

(i) `BaCl_(2)` ionizes completely in the solution as :
`BaCl_(2) rarr Ba^(2) + 2 Cl^(-) :. [Ba^(2+)]=[BaCl_(2)]=2xx10^(-3)M` (Given)
`Na_(2)SO_(4)` ionizes completely in the solution as :
`Na_(2)SO_(4) rarr 2Na^(+)+SO_(4)^(2-) :. [SO^(4)^(2-) :. [SO_(4)^(2-)] = [ Na_(2)SO_(4)]=2xx10^(-4)M` (Given)
Since equal volumes of the two solutions are mixed together, therefore, the concentrations of `Ba^(2+)` ions and `SO_(4)^(2-)` ions after mixing will be
`[Ba^(2+)]=(2xx10^(-3))/(2) = 10^(-3) M and [SO_(4)^(2-)]=(2xx10^(-4))/(2) = 10^(-4)M`
`:. ` Ionic product of `BaSO_(4) = [ Ba^(2+)][SO_(4)^(2-)]=10^(-3)xx10^(-4)=10^(-7)`
which is greater than the solubility product `(1xx10^(-10)) ` of `BaSO_(4)` . Hence, a precipitate of `BaSO_(4)` will be formed.
(ii) Here, the concentrations before mixing are :
`[Ba^(2+)]=[BaCl_(2)]=2xx10^(-8) M ` (Given), `[SO_(4)^(2-)]=[Na_(2)SO_(4)]=2xx10^(-3)M ` (Given)
`:.` Concentrations after mixing equal volumes will be :
`[Ba^(2+)]= (2xx10^(-8))/(2) = 10^(-8)M and [SO_(4)^(2-)] = (2xx10^(-3))/(2) = 10^(-3) M`
`:.` Ionic product of `BaSO_(4) = [ Ba^(2+)][SO_(4)^(2-)]=10^(-8) xx 10^(-3)=10^(-11)`
which is less than the solubility product `(1xx10^(-10))`. Hence, no ppt. will be formed in this case.