Answer to Question #289949 in General Chemistry for Victoria

Question

Determine the solubility of calcium hydroxide in molL-1 and gL-1 (0.01-0.10 molL-1 is

classified as sparingly soluble).

Answer 1

Ca ( OH )₂ = Ca²⁺ + 2 OH^- .

Given Solubility of Ca ( OH )₂ = 0.01 mol/ l .

Molar mass of Ca( OH )₂ = 40 + 34 = 74 .

Solubility of Ca ( OH )₂ = ( 0.01 × 74 ) g / L .

= 0.74 g / L .

Solubility prodict = 4 ( 0.01 )³ . = 4 × 10^-6 .