The `P^(H)` of saturated aqueous solution of `Ba(OH)_(2)` is `10`. If the `K_(sp)` of `Ba(OH)_(2)` is `5 xx 10^(13)`, then the concentration of `Ba^(2+)` ions in the solution is

Options

A. `1xx10^(-5)`
B. `1xx10^(-3)`
C. `5xx10^(-5)`
D. `1xx10^(-2)`

Answer

C

Explanation

`Ba(OH)_(2)hArrBa^(+2)+3OH^(-)`
given `P^(H)=10` then
`P^(OH)=4`
`2S=[OH^(-)]=10^(-4)`
`[Ba^(+2)][2S]^(2)=K^(sp)`
`[Ba^(+2)]=(5xx10^(-13))/(10^(-8))=5xx10^(-5)`