   Chapter 16, Problem 21E

Chapter
Section
Textbook Problem

# Use the following data to calculate the Ksp value for each solid.a. The solubility of CaC2O4 is 4.8 × 10−5 mol/L.b. The solubility of BiI3 is 1.32 × 10−5 mol/L.

(a)

Interpretation Introduction

Interpretation: The solubility of CaC2O4 and BiI3 is given. The solubility product of CaC2O4 and BiI3 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 CaC2O4 .

The concentration of Ca2+ is 4.8×105mol/L_ .

Given

Solubility of CaC2O4 is 4.8×105mol/L .

Since, solid CaC2O4 is placed in contact with water. Therefore, compound present before the reaction is CaC2O4 and H2O . The dissociation reaction of CaC2O4 is,

CaC2O4(s)Ca2+(aq)+C2O42(aq)

Since, CaC2O4 does not dissolved initially, hence,

[Ca2+]initial=[C2O42]initial=0

The concentration at equilibrium can be calculated from the measured solubility of CaC2O4 . If 4.8×105mol of CaC2O4 is dissolved in 1.0L of solution, the change in solubility will be equal to 4.8×105mol/L . The reaction is,

CaC2O4(s)Ca2+(aq)+C2O42(aq)

Therefore,

4.8×105mol/LCaC2O44.8×105mol/LCa2++4.8×105mol/LC2O4

The equilibrium concentration of [Ca2+] is written as,

[Ca2+]=[Ca2+]initial+changetoreachequilibrium

Substitute the value of [Ca2+]initial and change to reach equilibrium in the above equation.

[Ca2+]=[Ca2+]initial+changetoreachequilibrium=0+4.8×105mol/L=4.8×105mol/L_

The concentration of C2O42 is 4

(b)

Interpretation Introduction

Interpretation: The solubility of CaC2O4 and BiI3 is given. The solubility product of CaC2O4 and BiI3 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

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