Question

Daily Problem #33 Sr3(PO4)2 is an insoluble solid that has a Ksp value of 6.8 x 10-29 a. Calculate the molar solubility of Sr3(PO4), in pure water. b. Calculate the molar solubility of Sr3(PO4)2 in 0.25 M SrCl.

Answer 1

a) Solubility equilibrium of Sr₃(PO₄)₂ is

Sr₃(PO₄)₂(s) <-------> 3Sr²⁺(aq) + 2PO₄^3-(aq)

K_sp = [Sr²⁺]³[PO₄^3-]² = 6.8 × 10^-29

If solubility of Sr^{​​​​​​}₃(PO₄)₂ is represented by S

at saturated solution

[Sr²⁺] = 3S

[PO₄^3-] = 2S

So,

( 3S)³( 2S)² = 6.8 ×10^-29

108S⁵ = 6.8 ×10^-29

S⁵ = 6.296 × 10^-31

S = 9.12×10^-7

Therefore,

Molar solubility of Sr₃(PO₄)₂ in pure water = 9.12 ×10^-7 M

b)

Sr₃(PO₄)₂(s) <------> 3Sr²⁺(aq) + 2PO₄^2-(aq)

K_sp = [Sr^{​​​​​​2+}]³[PO₄^3-]² = 6.8 ×10^-29

Initial concentration

[Sr²⁺] = 0.25

[PO₄^3-] = 0

change in concentration

[Sr²⁺] = + 3x

[PO₄^3-] = + 2x

equilibrium concentration

[Sr²⁺] = 0.25 + 3x

[PO₄^3-] = 2x

so,

(0.25 + 3x)³( 2x)² = 6.8 ×10^-29

solving for x

x = 3.30 ×10^-14

Therefore,

molar solubility of Sr₃(PO4)₂ in 0.25M SrCl2 = 3.30×10^-14 M