Sr(OH)_{2 }→ Sr^{2+} + 2 OH^{-}

$\left[{\mathrm{OH}}^{-}\right]\mathbf{}\mathbf{=}\mathbf{}(1.6x{10}^{-3}\frac{\overline{)\mathrm{mol}\mathrm{Sr}{\left(\mathrm{OH}\right)}_{2}}}{1L})\mathbf{}\left(\frac{2\mathrm{mol}{\mathrm{OH}}^{-}}{1\overline{)\mathrm{mol}\mathrm{Sr}{\left(\mathrm{OH}\right)}_{2}}}\right)$

[OH-] = 3.2x10^{-3 }M

$\overline{){\mathbf{pOH}}{\mathbf{\hspace{0.17em}}}{\mathbf{=}}{\mathbf{}}{\mathbf{-}}{\mathbf{log}}\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}}$

pOH = -log (3.2x10^{-3 }M)

Calculate pH for 1.6x10^{-3} M Sr(OH)_{2}.

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