The solubility of `Sr(OH)_(2)` at `298 K` is `19.23 g L^(-1)` of solution. Calculate the concentrations cf strontium and hydroxyl ions and the `pH` of the solution.

Answer

`[Sr^(2+)] = 0.1581M, [OH^(-)] = 0.3162M, pH = 13.50`

Explanation

Solubility of `Sr(OH)_(2) = 19.23 g//L`
Then concentration of `Sr(OH)_(2)`
`= (19.23)/(121.63) M`
`Sr(OH)_(2(aq)) rarrSr_((aq))^(2+) + 2(OH^(-))_((aq))`
`:. [Sr^(2+)] = 0.1581 M`
`[OH^(-)] = 2 xx 0.1581 M = 0.3126 M`
Now, `K_(w) = [OH^(-)] [H^(+)]`
`(10^(-14))/(0.3126) = [H^(+)]`
`rArr [H^(+)] = 3.2 xx 10^(-14)`
`:. pH = 13.495 , 13.50`