Question

The Ksp of cobalt(II) hydroxide, Co(OH)2, is 5.92 × 10-15. Calculate the molar solubility of this compound.

Answer 1

The molar solubility of cobalt(II) hydroxide is approximately 1.44 × 10-5 M.

Step-by-step explanation:

The molar solubility of a compound can be calculated using the solubility product constant (Ksp). The Ksp of cobalt(II) hydroxide, Co(OH)2, is provided as 5.92 × 10-15. Since Co(OH)2 dissociates into Co²⁺ ions and OH⁻ ions, and the stoichiometry of the reaction is 1:2, the equilibrium concentration of Co²⁺ ions and OH⁻ ions will be 2x and x, respectively. Therefore, we can set up the expression for the Ksp: Ksp = [Co²⁺][OH⁻]² = (2x)(x)² = 5.92 × 10-15.

Simplifying this equation, we get x³ = 2.96 × 10-15. Taking the cube root of both sides, we find that the molar solubility of Co(OH)2 is approximately 1.44 × 10-5 M.

Answer 2

The Ksp of cobalt(II) hydroxide is equal to [OH-]2[Co2+]=5.92*10-15. And [OH-]=2[Co2+]=2*[Co(OH)2](dissolved). So the molar solubility of this compound is 1.14*10-5 M.