Question

The K_sp for silver(I) phosphate is 1.8 x 10^–18. Calculate the molar solubility of silver(I) phosphate.

Group of answer choices

1.8 x 10^–1 M

1.6 x 10^–5 M

2.1 x 10^–5 M

7.2 x 10^–1 M

3.7 x 10^–5 M

Answer 1

Solution:-

Ag₃PO₄ (s)   $\rightleftharpoons$ 3 Ag⁺ (aq) + PO₄^3- (aq)

Let, the solubility of Ag​​​​​​​_{3​​​​​​​}PO₄ be "s" M.

[Ag​​​​​​​⁺] = 3s

[PO₄^3-] = s

Thus, the solubility product,

Ksp = [Ag​​​​​​​⁺]³ [PO₄^3-]

(1.8 × 10^-18) = (3s)³ (s) = 27s⁴

or, s = (1.8 × 10^-18 / 27)^1/4 = 1.6 × 10^-5 M

Hence, molar solubility of silver(I) phosphate is = 1.6 × 10^-5 M

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