The solubility of $Sr(OH)_2$ at $298 \, K$ is $19.23 \, g/L$ of solution. Calculate the concentrations of strontium and hydroxyl ions and the $pH$ of the solution.

(N/A) Molar mass of $Sr(OH)_2 = 87.62 + 2 \times (16.00 + 1.01) = 121.64 \, g/mol$.
Concentration of $Sr(OH)_2 = \frac{19.23 \, g/L}{121.64 \, g/mol} = 0.1581 \, M$.
Dissociation reaction: $Sr(OH)_{2(aq)} \to Sr^{2+}_{(aq)} + 2OH^{-}_{(aq)}$.
Concentration of strontium ions: $[Sr^{2+}] = 0.1581 \, M$.
Concentration of hydroxyl ions: $[OH^{-}] = 2 \times 0.1581 \, M = 0.3162 \, M$.
To find $pH$: $pOH = -\log[OH^{-}] = -\log(0.3162) \approx 0.50$.
$pH = 14 - pOH = 14 - 0.50 = 13.50$.