Question

Calculate the solubility of Ca(OH)2 in a 0.469M CaCl2 solution at 31°C, given that the Ksp of Ca(OH)2 is 4.96 x 10-6 at that temperature. Give your answer in milliMolar units 2.927 mM 1.626 mM 2.764 mM 1.382 mM 1.463 mM 3.252 mM​

Answer 1

The solubility of Ca(OH)2 in a 0.469M CaCl2 solution at 31°C, at given Ksp of Ca(OH)2 is determined as 1.626 mM.

Dissociation reaction of Ca(OH)2

The dissociation reaction of Ca(OH)2 is given as follows;

Ca(OH)₂  ⇄ Ca²⁺  + 2OH⁻¹

                     x           2x

Concentration of Ca²⁺ = 0.469 M

Ksp = [x][2x]²

ksp = (0.469)(2x²)

ksp = 4(0.469)x²

ksp = 1.876x²

4.96 x 10⁻⁶ = 1.876x²

x² = (4.96 x 10⁻⁶)/(1.876)

x² = 2.643 x 10⁻⁶

x = √(2.643 x 10⁻⁶)

x = 1.626 x 10⁻³ M

x = 1.626 mM

Thus, the solubility of Ca(OH)2 in a 0.469M CaCl2 solution at 31°C, at given Ksp of Ca(OH)2 is determined as 1.626 mM.

Learn more about solubility here: brainly.com/question/23946616

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