   Chapter 16, Problem 104CP

Chapter
Section
Textbook Problem

# Calcium oxalate (CaC2O4) is relatively insoluble in water (Ksp = 2 × 10−9). However, calcium oxalate is more soluble in acidic solution. How much more soluble is calcium oxalate in 0.10 M H+ than in pure water? In pure water, ignore the basic properties of C2O42−.

Interpretation Introduction

Interpretation: The solubility of calcium oxalate in 0.10MH+ and in pure water is to be compared.

Explanation

Explanation

To determine: how much more soluble is CaC2O4 in 0.10MH+ than in pure water.

The solubility of CaC2O4 in pure water is 4×10-5mol/L_ .

The reaction for the solubility of CaC2O4 in pure water is,

CaC2O4(s)Ca+2(aq)+C2O42(aq)

The number of moles of CaC2O4 to be dissolved to reach equilibrium is considered to be x .

Initialconcentration(mol/L)[CaC2O4]=x[Ca+2]=0[C2O42]=0toreachequilibriumxmoles/LdissolvesEquilibriumconcentration(mol/L)[CaC2O4]=0[Ca+2]=x[C2O42]=x

The molar solubility of CaC2O4 is calculated by the formula,

Ksp=[Ca+2][C2O42]

Where,

• Ksp is the solubility product of CaC2O4 .

The value of Ksp is 2.9×109 .

Substitute the equilibrium concentrations into the above formula.

2.9×109=[x][x]x2=2.9×109x=4×10-5mol/L_

Therefore, the solubility of CaC2O4 in pure water is 4×10-5mol/L_ .

The equilibrium constant for solubility of CaC2O4 in 0.10MH+ is 4.8×10-4_ .

The CaC2O4 dissolves according to the reaction,

CaC2O4(s)Ca+2(aq)+C2O42(aq)K=2.9×109

In acidic solution, CaC2O4 dissolves according to the reaction,

C2O42+H+HC2O4K1=1.6×104 (1)

HC2O4+H+H2C2O4K2=15 (2)

The overall reaction for the solubility of CaC2O4 in 0.10MH+ is obtained by adding equation(1), equation(2) and equation(3). The resultant equation is,

CaC2O4(s)+2H+Ca+2+H2C2O4

The solubility product of the resultant equation is calculated as,

Ksp=K×K1×K2

Substitute the values of K , K1 and K2 in the above equation

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