Answer:
THE SOLUBILITY OF Ca3(PO4)2 is 2.3 * 10^-11 M
Explanation:
Step 1: write out the solubility equation and the solubility product formula:
Ca3(PO4)2(s) ---------> 3Ca^2+(aq) + 2PO4^3- (aq)
The solubility product formula for the equation will therefore be;
Ksp = [Ca2+]^3 [PO4^3-]^2
Step2 : obtain the molar solubility;
Let the molar solubility be represented as S
Ksp = [3S]^3 [2S]^2
Ksp = 1.3 *10 ^-32
Solubility of PO4^2- = 0.2 since Na3PO4 will dissociate completely and give off PO4^2- as it is a strong electrolyte.
Also, the amount of PO4^2- in the system obtained from Ca3(PO4)2 is smaller than that from Na3PO4 so its value can be ignored when calculating the molar solubility.
Ksp = 93S)^3 * (0.2 + 2S)^2
1.3 *10^-32 = (3S)^3 (0.2)^2
1.3 *10^-32 = 27 (S)^2 * 4 *10^-2
27 S^3 = 1.3*10^-32 / 4*10^-2
27 S^3 = 5.2 * 10 ^-30
S^3 = 5.2 *10^-30 / 27
S^3 = 0.19259 *10 ^-30
S = 2.3 * 10^-11 M
So therefore, the molar solubility of Ca3(PO4)2 is 2.3*10^-11 M.
