Question

At 25 ∘C only 0.0180 mol of the generic salt AB is soluble (at equilibrium) in 1.00 L of water.What is the value for Ksp for this salt at 25 ∘C?

Answer 1

Ksp = 3.24 x 10⁻⁴

Step-by-step explanation:

The dissociation equilibrium for a generic salt AB is:

AB(s) ⇄ A⁺(aq) + B⁻(aq)

s s

For instance, the expression for the Ksp constant is:

Ksp = [A⁺] [B⁻] = s x s = s²

According to the problem, 0.0180 mol of the salt is soluble in 1.00 L os water. That means that the solubility of the salt (s) is equal to 0.0180 mol per liter.

s = moles of solute/L of solution = 0.0180 mol/L

Thus, we calculate Ksp from the s value as follows:

Ksp = s² = (0.0180)² = 3.24 x 10⁻⁴