Mg(OH)₂ ⇄ Mg²⁺ + 2 OH⁻
Ksp = [Mg²⁺] [OH⁻]²
6.0 x 10⁻¹⁰ = 0.10 x [OH⁻]²
[OH⁻] = 7.746 x 10⁻⁵ M
when Mg(OH)₂ 1st precipitates, [OH⁻] = 7.746 * 10⁻⁵ M
Fe(OH)₂ <—> Fe²⁺ + 2OH⁻
Ksp = [Fe²⁺] [OH⁻]²
7.9 x 10⁻¹⁶ = [Fe²⁺] x (7.746 x 10⁻⁵)²
[Fe²⁺] = 1.32 x 10⁻⁷ M
Answer: 1.32 x 10⁻⁷ M
Answer:
mass of 1 mole of 127 grams, KCI, 552mL = 51.1099987g
Explanation:
