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# 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.

7.58     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.

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By given abova data, we know the solubility of $Sr(OH)_2$ at 298 K = 19.23 g/L

So, concentration of $[Sr(OH)_2]$

= $19.23 / 121.63 M$ (Molecular weight of $Sr(OH)_2$ = 121.63 u)

= 0.1581 M

$Sr(OH)_2(aq) \rightarrow Sr^{2+}(aq) + 2 (OH^-)(aq)$

$\therefore Sr^{2+} = 0.1581M$
and the concentration of $[OH^-]$$= 2 \times 0.1581M = 0.3162 M$
Now

It is known that,
$K_w = [OH^-] [H^+]$

$[H^+] = \frac{10^{-14}}{0.3162}$
= $3.16 \times 10^{-14}$

Therefore $p^H = -\log(3.16 \times 10^{-14}) =13.5$

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