Question

The solubility product, Ksp, of Ca (PO4 is 2.0 x 10-32. Calculate its solubility (in g/L) (a) (4 points) in pure water (b) (4 points) in a 0.150 M solution of CaCh

Answer 1

a) S = 1.7099 E-11 g/L

b) S = 8.88 E-31 g/L

Step-by-step explanation:

a) in water:

• Ca(PO4)2 ↔ Ca²+ + 2PO4²-

solubility: S S 2S

Ksp = [ Ca²+ ] * [ PO4²-]² = ( S ) * ( 2S )² = 4S³ = 2.0 E-32

⇒ S = ∛ (2.0 E-32 / 4) = 1.7099 E-11 g/L

b) 0.150 M sln CaCH:

• CaCH ↔ Ca²+ + CH²-

0.15 0.15 0.15

Ksp = ( S ) * ( 2S + 0.15 )² ......... we despise solubility as adding

⇒ Ksp = ( S ) * ( 0.15 )² = 2.0 E-32

⇒ S = 2.0 E-32 / 0.0225 = 8.88 E-31 g/L