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

given, solubility of Sr(OH)2 is 19.23g/L

molecular weight of Sr(OH)2 is 121.6 g/mol

         we know,

 molarity = solubility/molecular weight

              = 19.23 g/L/121.6g/mol

              =0.1581 M

Sr(OH)2   ⇔ Sr²⁺   +             2OH⁻

0.1581M    0.1581M         2×0.1581M = 0.3162M

[Sr²⁺ ] = 0.1581M and [OH⁻] = 0.3162M

we know, ionic product(Kw) = [H⁺][OH⁻]  = 10⁻¹⁴

                       [H⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/0.3162  = 3.16 × 10⁻¹⁴

now Arrhenius formula,

    pH = -log[H⁺] = -log(3.16 × 10⁻¹⁴) = 14 -0.4997 = 13.5003