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

Molar mass of Sr `(OH)_(2) = 87.6+34=121.6 g "mol" ^(-1)`
Solubility of Sr `(OH)_(2) ` in moles `L^(-1)=(19.23g L^(-1))/(121.6g "mol"^(-1))=0.1581M`
Assuming complete dissociation , Sr`(OH)_(2)rarrSr^(2)+2OH^(-)`
`:. [Sr^(2+)]=0.1581 M, [OH^(-)]=2xx0.1581=0.3162`M
`pOH=-log 0.3162=0.5, :. pH = 14-0.5=13.5`