# The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution.

Question:

The solubility of $\mathrm{Sr}(\mathrm{OH})_{2}$ at $298 \mathrm{~K}$ is $19.23 \mathrm{~g} / \mathrm{L}$ of solution. Calculate the concentrations of strontium and hydroxyl ions and the $\mathrm{pH}$ of the solution.

Solution:

Solubility of $\mathrm{Sr}(\mathrm{OH})_{2}=19.23 \mathrm{~g} / \mathrm{L}$

Then, concentration of $\mathrm{Sr}(\mathrm{OH})_{2}$

$=\frac{19.23}{121.63} \mathrm{M}$

$=0.1581 \mathrm{M}$

$\mathrm{Sr}(\mathrm{OH})_{2(a q)} \longrightarrow \mathrm{Sr}^{2+}{ }_{(a q)}+2\left(\mathrm{OH}^{-}\right)_{(a q)}$

$\therefore\left[\mathrm{Sr}^{2+}\right]=0.1581 \mathrm{M}$

$\left[\mathrm{OH}^{-}\right]=2 \times 0.1581 \mathrm{M}=0.3126 \mathrm{M}$

Now,

$K_{\mathrm{w}}=\left[\mathrm{OH}^{-}\right]\left[\mathrm{H}^{+}\right]$

$\frac{10^{-14}}{0.3126}=\left[\mathrm{H}^{+}\right]$

$\Rightarrow\left[\mathrm{H}^{+}\right]=3.2 \times 10^{-14}$

$\therefore \mathrm{pH}=13.495 ; 13.50$