An aqueous solution contains an unknown concentration of `Ba^(2+)`. When 50 mL of a 1 M solution of `Na_(2)SO_(4)` is added, `BaSO_(4 )`just begins to precipitate. The final volume is 500 mL. The solubility product of `BaSO_(4)` is `1xx10^(–10)`. What is the original concentration of `Ba^(2+)`?

Options

A. `1.1xx10^(-9)M`
B. `1.1xx10^(-10)M`
C. `5.1xx10^(-9)M`
D. `2xx10^(-9)M`

Answer

A

Explanation

`BaSO_(4)(s) to Ba^(2+)(aq) K_(sp)=10^(-10)`
`Na_(2)SO_(4) to 2Na^(+) + SO_(4)^(2-)`
Conc. of `SO_(4)^(2-)` in final solution `=(50xx1)/500 =0.1M`
For final solution`implies [Ba^(2+)][SO_(4)^(-)]=10^(-10) implies[Ba^(2+)]=10^(-9)M`
`M_(i)V_(i)=M_(f)V_(f)`
`Cxx450=10^(-9)xx500implies C=1.1xx10^(-9)M`