   Chapter 16, Problem 24E

Chapter
Section
Textbook Problem

# The solubility of the ionic compound M2X3, having a molar mass of 288 g/mol, is 3.60 × 10−7 g/L. Calculate the Ksp of the compound.

Interpretation Introduction

Interpretation: The mass and molar mass of M2X3 is given. The solubility product of M2X3 is to be calculated.

Concept introduction: The solubility product Ksp is the equilibrium constant that is applied when salt partially dissolve in a solvent. The solubility product of dissociation of AxBy is calculated as,

Ksp=[A]x[B]y

Explanation

Explanation

To determine: The solubility product of M2X3 .

Solubility of M2X3 is 1.25×109mol/L_ .

Given

Mass of M2X3 is 3.60×107g/L .

Molar mass of M2X3 is 288g/mol .

Formula

The solubility of M2X3 is calculated using the formula,

SolubilityofM2X3=MassofM2X3MolarmassofM2X3

Substitute the values of mass and molar mass of M2X3 in the above equation.

SolubilityofM2X3=MassofM2X3MolarmassofM2X3=3.60×107g/L×1288g/mol=1.25×109mol/L_

The concentration of M3+ is 2.5×109mol/L_ .

Solubility of M2X3 is 1.25×109mol/L .

Since, solid M2X3 is placed in contact with water. Therefore, compound present before the reaction is M2X3 and H2O . The dissociation reaction of M2X3 is,

M2X32M3+(aq)+3X2(aq)

Since, M2X3 does not dissolved initially, hence,

[M3+]initial=[X2]initial=0

The concentration at equilibrium can be calculated from the measured solubility of M2X3 . If 1.25×109mol/L of M2X3 is dissolved in 1.0L of solution, the change in solubility will be equal to 1.25×109mol/L . The reaction is,

M2X32M3+(aq)+3X2(aq)

Therefore,

1.25×109mol/LM2X32(1.25×109mol/L)M3++3(1.25×109mol/L)3X2

The equilibrium concentration of M3+ is written as,

[M3+]=[M3+]initial+changetoreachequilibrium

Substitute the value of [M3+]initial and change to reach equilibrium in the above equation

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