Asked by Leaf02
What is the molar solubility of Ba3(PO4)2. Ksp Ba3(PO4)2 = 1.3x10-29
Answers
Answered by
DrBob222
.....................Ba3(PO4)2 ==> 3Ba^2+ + 2PO4^3-
Initial..............solid......................0...............0
change...........solid-x...................3x.............2x
equilibrium......solid......................3x.............2x
Ksp = (Ba^2+)^3(PO4^3-)^2 = 1.2E-29
(3x)^3 * (2x)^2 = 1.3E-29
Solve for x = solubility of Ba3(PO4)2 in mols/L
Initial..............solid......................0...............0
change...........solid-x...................3x.............2x
equilibrium......solid......................3x.............2x
Ksp = (Ba^2+)^3(PO4^3-)^2 = 1.2E-29
(3x)^3 * (2x)^2 = 1.3E-29
Solve for x = solubility of Ba3(PO4)2 in mols/L
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