Finding the Ksp value of Ca(OH)2

Question

A student determined the solubility of Ca(OH)2 in 0.200 M CaCl2 as 0.0026 mol/L. Calculate the Ksp value for Ca(OH)2 using these data.

I found this to be =7.03E-8, but it says that I have entered my results to the the wrong precision.

Answer 1

Ca(OH)₂(s) <===> Ca²⁺(aq) + 2OH-(aq)

Ksp = [Ca²⁺][OH-]²

Since there is a common ion present (Ca2+) provided by the added CaCl2, we know the [Ca2+] = 0.2 M

The molar solubility is given as 0.0026 M and since you get 2 OH- from each Ca(OH)2, the molar solubility of OH- will be 2 x 0.0026 M = 0.0052 M

Plugging these values into the Ksp expression, we obtain...

Ksp = [0.200][0.0052]²

Ksp = 5.4x10^-6 (2 sig. figs. based on the value of 0.0026 M)

Answer 2

CaCl₂ ↔ Ca²⁺ + 2Cl^-

[Ca²⁺] [Cl^-]

Initial: 0 0

Change: +x +2x

Equilibrium: +x +2x

K_sp = [Ca²⁺][Cl^-]² = (x)(2x)² = 4x³ = 4(.0026)³ = 7.0x10^-8

Maybe they only want two significant figures.