At a certain temperature, the solubility of strontium arsenate, sr3(aso4)2, is 0.0560 g/l. what is the ksp of this salt at this temperature?

The solution for this problem is:
Get into moles first. .0560 grams over 540.8 grams per mole = 1.04 x l0^-4 moles 
Sr3(As04)2 = 3 Sr++(aq) plus 2 As04^-3(aq) 
Ksp = (Sr++)^3(As04^-3)^2 
(Sr++) = 3 X 1.04 x l0^-4= 3.11 x l0^-4 
(As04^-3) = 2 x 1.04 x l0^-4= 2.07 x l0^-4 
Ksp = (1.04 x l0^-4)^3 (2.07 x l0^-4)^2 which equals 4.82 x 10^-20

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