Molar mass of $\mathrm{Sr(OH)_2}$ = 87.6 + 34 = 121.6 $\text{g mol}^{-1}$ Solubility of $\mathrm{Sr(OH)_2}$ in moles $\text{L}^{-1}$ = $\frac{19.23\,\mathrm{g\,L^{-1}}}{121.6\,\mathrm{g\,mol^{-1}}}$ = 0.1581 M Assuming complete dissociation, $\mathrm{Sr(OH)_2}$ → $\mathrm{Sr}^{2+} + 2\,\mathrm{OH}^-$ ∴ $[\mathrm{Sr}^{2+}]$ = 0.1581 M, $[\mathrm{OH}^-]$ = 2 × 0.1581 = 0.3162 M pOH = –log (0.3162) = 0.5, ∴ pH = 14 – 0.5 = 13.5