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

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

1 a

2 d

3 a

4 c

explanation: