Answer to Question #305776 in General Chemistry for James

Question

If the silver phosphate solubility product –
Ag3PO4 for pKs = 18.2; the concentration of silver in a
saturated solution will be:

Answer 1

Solution:

silver phosphate = Ag₃PO₄

pK_s(Ag₃PO₄) = 18.2

Convert pK_s to K_s:

pK_s = –log(K_s)

K_s = 10^–pKs = 10^–18.2 = 6.31×10^–19

Ag₃PO₄(s) ⇌ 3Ag⁺(aq) + PO₄^3–(aq)

___S__________3S_________S_______

Thus,

[Ag⁺] = 3S

[PO₄^3–] = S

The K_s expression for Ag₃PO₄(s) is:

K_s = [Ag⁺]³ × [PO₄^3–]

Therefore,

6.31×10^–19 = [3S]³ × [S] = 27S³ × S = 27S⁴

S⁴ = (6.31×10^–19) / (27) = 2.337×10^–20

S = (2.337×10^–20)^1/4 = 1.236×10^–5

[Ag⁺] = 3S = 3 × (1.236×10^–5) = 3.7×10^–5

[Ag⁺] = 3.7×10^–5 M

Answer: The concentration of silver (Ag⁺) in a saturated solution will be 3.7×10^–5 M