Question

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.

Answer 1

Solubility of Sr(OH)2 = 19.23 g/L
Then, concentration of Sr(OH)2 = $\frac{19.23}{121.63}$ M = 0.1581 M
Sr(OH)2(aq) $\rightarrow$ Sr2+(aq) + 2(OH-)(aq)
∴ [Sr2+] = 0.1581 M
[OH-] = 2 × 0.1581 M = 0.3126M
Now, Kw = [OH-] [H+] = $\frac{10^{-14}}{0.3126}$ ⇒ [H+] = 3.2 × 10-14
∴ pH = 13.495 ; 13.50