The Ksp of calcium phosphate Ca3(PO4)2 is 1.3 × 10-32 at 25°C. What is the solubility of Ca3(PO4)2 in moles/Liter in a 0.20 Molar solution of Na3PO4?

A.

2.3 × 10-11 moles/Liter

B.

4.3 × 10-13 moles/Liter

C.

2.3 × 10-32 moles/Liter

D.

1.3 × 10-32 moles/Liter

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.

b is the correct answer

mark me brainliest and hope this