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Chapter 14: Problem 50

Calculate the solubility of a substance MN that ionizes to form \(\mathrm{M}^{2+}\) and \(\mathrm{N}^{2-}\) ions given that \(K_{s p}=8.1 \times 10^{-6} .\)

Short Answer

Expert verified
The molar solubility 'x' of the substance MN is approximately \(2.84 \times 10^{-3}\) mol/L.

Step by step solution

01

Write the chemical equilibrium

When MN dissolves, it dissociates into M^2+ and N^2- ions. This reaction can be presented as follows: MN(s) ⇌ M^2+(aq) + N^2-(aq).
02

Setting up the solubility product expression

The equilibrium expression for this reaction in terms of the molar solubility (x) can be formulated as follows based on the exponents of the ions in the balanced equation: \(K_{sp} = [M^{2+}][N^{2-}] = x * x = x^2.\)
03

Calculate molar solubility

Substitute the given value of \(K_{sp}\) into the solubility product equation and solve for 'x' to get the molar solubility: \(x^2 = 8.1 \times 10^{-6}\). Solving this gives \(x = \sqrt{8.1 \times 10^{-6}}.\)

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