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. `5xx10^(-9)M`
B. `2xx10^(-9)M`
C. `1.1xx10^(-9)M`
D. `1.0xx10^(M)`

Answer

C

Explanation

As 50 mL of 1 M `Na_(2)SO_(4)` solution are present in 500 mL of final solution, this means that 1 M `Na_(2)SO_(4)` has been diluted 10 times. Hence, Conc. Of `[SO_(4)^(2-)]` in final solution `= (1)/(10) M = 0.1 M`
`K_(sp) ` of `BaSO_(4) = 10^(-10)`
`:. [Ba^(2+)][SO_(4)^(2-)]=10^(-10)`
`[Ba^(2+)][0.1]=10^(-10)`
or `[Ba^(2+)]=10^(-9)M`
i.e., conc. of `Ba^(2+)` ions in final solution `=10^(-9)`M
Volume of final solution = 500 mL
Volume of original solution = 500 - 50 = 490 mL
Applying `underset("(Initial)") M_(1)V_(1)=underset("(Final)")M_(2)V_(2)`
`M_(1)xx900=(10^(-9))(500)`
or, `M_(1)=1.11xx10^(-9)M`