At a particular temperature, the K _sp for Ag₃PO₄ is 1.3×10^{- 20} M . If the equilibrium concentration of PO₄^{3 -} is 1.3×10^{- 8} M , what is the silver ion concentration?

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the given reaction is:

K = [Ag+][PO43-]/[Ag3PO4]

Let x be the concentration of Ag+ ions at equilibrium.

According to the given information,

K = 1.3×10^-20 M

[PO43-] = 1.3×10^-8 M

[Ag3PO4] = 0 (since Ag3PO4 is a solid)

Substituting these values in the equilibrium constant expression and simplifying,

1.3×10^-20 = x^3/[Ag3PO4]

x = (1.3×10^-20 ×[Ag3PO4])^(1/3)

At equilibrium, the solubility product of Ag3PO4, Ksp = [Ag+][PO43-]^3

Substituting the given values of Ksp and [PO43-],

1.3×10^-20 = x^3/(1.3×10^-8)^3

x = (1.3×10^-20 ×(1.3×10^-8)^3)^(1/3)

x = 1.0×10^-4 M

Therefore, the concentration of Ag+ ions at equilibrium is 1.0×10^-4 M.

The correct answer is choice B.

At a Particular Temperature, the K sp for Ag3PO4 Is

At a Particular Temperature, the K _sp for Ag₃PO₄ Is