First, we have to change Sr3(ASO4)2 from 0.059 g/L to moles:
and when the molar mass of Sr3(ASO4)2 = 87.6*3 + 74.9*2 + 16*8
= 540.6 g/mol
Molarity (X) = 0.059 g/L * 1 mol / 540.6 g/mol
= 1.1 x 10^-4 moles
When the solubility equation is:
Sr3(ASO4)2 ↔ 3Sr++(aq) + 2ASO-3(aq)
so, Ksp = [Sr++]^3[ASO-3]^2
from ICE table and the balanced equation:
[Sr++] = 3 X
= 3* 1.1 x 10^-4 = 3.3x10^-4 M
[ASO-3] = 2X
= 2* 1.1 x 10^-4 = 2.2 x 10^-4 M
So, by substitution:
∴Ksp = (3.3 x 10^-4)^3 * (2.2 x 10^-4) ^2
= 1.74 x 10^-18
