Determine the number of grams of strontium hydroxide that can be dissolved in 421.07 mL of distilled water to make a saturated solution of Sr(OH)2 (Ksp = 3.2 x 10-4)

Question

Determine the number of grams of strontium hydroxide that can be dissolved in 421.07 mL of distilled water to make a saturated solution of Sr(OH)₂ (K_sp = 3.2 x 10^-4)

Answer 1

Sr(OH)₂(s) ==> Sr²⁺(aq) + 2OH^-(aq)

Ksp = [Sr²⁺][OH^-]²

If we let [Sr²⁺] = x, then [OH^-] = 2x

3.2x10^-4 = (x)(2x)²

3.2x10^-4 = 4x³

x = 8.9x10^-3 M This is the molar solubility of Sr(OH)₂

For 421.07 ml we find moles that can dissolve as follows:

421.07 ml x 1 L / 1000 ml = 0.42107 L x 8.9x10^-3 mol/L = 3.75x10^-3 moles Sr(OH)₂

grams Sr(OH)₂ soluble in 421 mls = 3.75x10^-3 moles x 121.6 g/mol = 0.46 g Sr(OH)₂ (to 2 sig. figs.)