The molar solubility of calcium phosphate in a 0.288 M potassium phosphate solution is ... ?

ksp calcium phosphate - -> 1.0 * 10^-25

3.55 * 10⁻⁹ M

Explanation:

Potassium phosphate is a strong electrolyte that dissociates according to the following equation.

K₃PO₄ (aq) → 3 K⁺ (aq) + PO₄⁻ (aq)

If the concentration of K₃PO₄ is 0.288 M, the concentration of PO₄⁻ will be 0.288 M.

We can find the molar solubility (S) of calcium phosphate using an ICE chart.

Ca₃ (PO₄) ₂ ⇄ 3 Ca²⁺ + 2 PO₄³⁻

I 0 0.288

C + 3S + 2S

E 3S 0.288 + 2S

The solubility product (Ksp) is

Ksp = 1.0 * 10⁻²⁵ = [Ca²⁺]³.[PO₄³⁻]² = (3S) ³ (0.288 + 2S) ²

Since 2S << 0.288, we can neglect it to simplify the calculations.

Ksp = 1.0 * 10⁻²⁵ = (3S) ³ (0.288) ²

S = 3.55 * 10⁻⁹ M

5.116x10⁻⁹ M

Explanation:

First, let's know the molecular formula of calcium phosphate. Calcium is a metal of group 2, so it loses 2 electrons in an ionic bond and forms the cation Ca⁺². Phosphate is the ion PO₄⁻³, so to form the compound, first comes the cation, then the anion and the charges must change between then:

Ca₃ (PO₄) ₂

The chemical equation for its dillution is:

Ca₃ (PO₄) ₂ (s) ⇄ 3Ca⁺² (aq) + 2PO₄⁻³ (aq)

In the solution, there is ions phosphate from potassium phosphate. Potassium is from group 1, so it loses 1 electron and forms the ion K⁺, so: K₃PO₄. It means that in 1 mol of the salt, will be 1 mol of phosphate. Then, the initial concentration of phosphate is 0.288M. So:

Ca₃ (PO₄) ₂ (s) ⇄ 3Ca⁺² (aq) + 2PO₄⁻³ (aq)

M 0 0.288 Initial

-S + 3S + 2S Reacted

M-x 3S 0.288 + 2S Equilibrium

S is the molar solubility. In the expression of the equilibrium constant (Kps) the solids don't participate. And the concentration must be raised to the coefficient, so:

Kps = [Ca⁺²]³x[PO₄⁻³]²

1.0x10⁻²⁵ = (3S) ³x (0.288 + 2S) ²

1.0x10⁻²⁵ = (9S³) x (0.083 + 1.152S + 4S²)

1.0x10⁻²⁵ = 0.747S³ + 10.368S⁴ + 36S⁵

Which is an equation of the 5th grade, the equation must be solved by try and error or by a computer.

S = 5.116x10⁻⁹ M