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.

Explanation

Molarity of solution `=("Mass of Sr"(OH_(2))//"Molar Mass")/("Volume of solution in litres")`
`=((19.23)//(87.6 + 34g) "mol"^(-1)L)/(1L)`
`Sr(OH)_(2) overset((aq))(to) underset(0.1581 M)(Sr^(2) (aq)) + underset(2 xx 1.1581 = 0.3162 M)(2OH^(-) (aq))`
`[Sr^(2+)]=0.1581 M, [OH^(-)] = 0.3162 M`
`pOH =- log [OH^(-)] =- log (3.162 xx10^(-1))`
`=- (log 3.162 - log 10) =(1- log 3.162) = (1-0.5) =0.5`
`pH =14 - pOH =14 -0.5 =13.5`