   # The solubility of the ionic compound M 2 X 3 , having a molar mass of 288 g/mol, is 3.60 × 10 −7 g/L. Calculate the K sp of the compound. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 15, Problem 24E
Textbook Problem
154 views

## 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 of Solution

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

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts
Nutrition authorities often recommend. drinking water, plain or lightly flavored, to quench thirst staying hydr...

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

What is the difference between hydrocarbons and other organic molecules?

Biology: The Dynamic Science (MindTap Course List)

Match the term listed in Column A with its definition from Column B

Nutrition Through the Life Cycle (MindTap Course List)

An accelerating voltage of 2.50103 V is applied to an electron gun, producing a beam of electrons originally tr...

Physics for Scientists and Engineers, Technology Update (No access codes included)

Name two sources that contribute material to the ISM.

Foundations of Astronomy (MindTap Course List) 