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Question

Calculate solubility of Ca3(PO4)2(Ksp=1015) in presence of 0.1 M CaCl2 solution.

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Solution

[Ca2+]=101M
Ca3(PO3)23Ca2++2PO34
Ksp=[Ca2+]3[PO33]2
1015=103×4S2
S=10124=12×106=5×107
Solubility =5×107

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