a) Solubility equilibrium of Sr3(PO4)2 is
Sr3(PO4)2(s) <-------> 3Sr2+(aq) + 2PO43-(aq)
Ksp = [Sr2+]3[PO43-]2 = 6.8 × 10-29
If solubility of Sr3(PO4)2 is represented by S
at saturated solution
[Sr2+] = 3S
[PO43-] = 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) + 2PO42-(aq)
Ksp = [Sr2+]3[PO43-]2 = 6.8 ×10-29
Initial concentration
[Sr2+] = 0.25
[PO43-] = 0
change in concentration
[Sr2+] = + 3x
[PO43-] = + 2x
equilibrium concentration
[Sr2+] = 0.25 + 3x
[PO43-] = 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
Daily Problem #33 Sr3(PO4)2 is an insoluble solid that has a Ksp value of 6.8 x...
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