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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.

Answers (1)

best_answer

Given the above 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

Posted by

manish

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