This is a bear of a problem. If the course is a freshman class then I think the author of the problem did not think it through. If the class is AP freshman class or advanced class I wonder if the use of H3PO4 instead of some other acid was the intended acid. The solubility of Ca3(PO4)2 is increased by the use of an acid but since H3PO4 is a weak acid I think the place to start is to determine the (H^+). That means using k1 and ignoring k2 and k3.
..................H3PO4 ==> H^+ + H2PO4^-
I....................0.1 M.........0.............0
C.....................-x..............x............x
E..................0.1-x............x.............x
k1 = (H^+)(H2PO4^-)/(H3PO4)
Substitute the E line into k1 and solve for x. You should use the quadratic formula and not use 0.1-x = 0.1. Keep the value of x and proceed.
.....................Ca3(PO4)2 ==> 3Ca^2+ + 2PO4^3-
I.....................solid.....................0..............0
C...................solid-y...................y...............y
E....................solid.....................y................y
Ksp = 2.0E-29 = (Ca^2+)^3(PO4^3-)^2
You have Ksp. (Ca^2+) = y; (PO4^3-) is obtained from the following:
After you have determined (H+) from H3PO4 then
k3 for H3PO4 = (H^+)(PO4^3-)/(HPO4^-). You know (H^+), (HPO4^-) = k2 and you know k3.Solve for (PO4^3-) from the H3PO4 and call that z and substitute into the following equation.
Ksp = (3y)^3*(2y + z)^2 = 2.8E29
Solve this for y = solubility Ca3(PO4)2. Good luck!
Ca3(PO4)2 has a Ksp = 2.0 x 10-29. What is its molar solubility in 0.1 M H3PO4?
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