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. `1.0xx10^(-10)` M
C. `5xx10^(-9)` M
D. `2xx10^(-9)` M

Answer

A

Explanation

`Ba^(+2)+SO_(4)^(2-)hArrunderset(ppt)(BaSO_(4(s))`
Final conc. Of `[SO_(4)^(2-)]=(MV_(1))/(V_(1)+V_(2))=(1xx50)/(500)=0.1M`
Final conc. Of `[Ba^(+2)]` when `BaSO_(4)` start precipitating
`K_(SP)=Q_(SP)=[Ba^(+2)][SO_(4)^(2-)]`
`10^(-10)=[Ba^(+2)](0.1M)`
`[Ba^(+2)]=10^(-9)M`
Initial conc. `[Ba^(+2)]` ,
Initial volume was = `500-50=450` mL
`M_(1)V_(1)=M_(2)V_(2)`
`M_(1)=(M_(2)V_(2))/(V_(1))=(10^(-9)xx500)/(450)`
`M_(1)=1.1xx10^(-9)` M