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. `1 xx 10^(-5)`
B. `1 xx 10^(-3)`
C. `5 xx 10^(-5)`
D. `1 xx 10^(-2)`

Answer

C

Explanation

`Ba(OH)_(2) hArr Ba^(+2) + 2OH^(-)`
given `P^(H) = 10` then
`P^(OH) = 4`
`2S = [OH^(-)] = 10^(-4)`
`[Ba^(+2)] [2S^(2)] = K_(sp)`
`[Ba^(+2)] = (5 xx 10^(-13))/(10^(-5)) = 5 xx 10^(-5)`