Determine the value of Ksp for AgIO3 by constructing an ICE table, writing the solubility constant expression and solving the expression.

Question

It is found that up to 0.0980 g of AgIO3 dissolves in a 2.00 L of aqueous solution at a certain temperature.

Answer 1

AgIO₃(s) <==> Ag⁺(aq) + IO₃^-(aq)

Ksp = [Ag⁺][IO₃^-]

So now we need to find [Ag⁺] and [IO₃^-]

molar mass AgIO₃ = 282.8 g / mole

[AgIO₃] = 0.0980 g x 1 mol / 282.8 g = 3.465x10^-4 moles / 2.00 L = 1.733x10^-4 M

[Ag⁺] = [IO₃^-] = 1.733x10^-4 M

Ksp = (1.733x10^-4)(1.733x10^-4)

Ksp = 3.00x10^-8