Give reason why `BaSO_(4)` will precipitate out when equal volumes of `2xx10^(-3) M BaCl_(2)` solution and `2xx10^(-4)M Na_(2)SO_(4)` solution are mixed. Given that the solubility product of `BaSO_(4)` is `1xx10^(-10).`

Explanation

`BaCl_(2) and Na_(2)SO_(4)` are ionises completely in the solution
`BaCl_(2)to Ba^(+2)+2Cl^(-)`
`[Ba^(+2)]=[BaCl_(2)]=2xx10^(-3)M`
`Na_(2)SO_(4)to2Na^(+)+SO_(4)^(-2)`
`[Na_(2)SO_(4)]=[SO_(4)^(-2)]=2xx10^(-4)M`
Equal volumes of the two solutions are mixed together, therefore, the concentration 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`
`therefore"lonic product of"BaSO_(4)=[Ba^(+2)][SO_(4)^(-2)]=10^(-7)M`
Here, ionic proudct of `BaSO_(4)` is greater than solubility product value. hence, a precipitate of `BaSO_(4)` will be formed.