   Chapter 17, Problem 94GQ

Chapter
Section
Textbook Problem

The solubility product constant for calcium oxalate is estimated to be 4 × 10−9. What is its solubility in grams per liter?

Interpretation Introduction

Interpretation:

Solubility of salt CaC2O4 has to be calculated in gL1 units using given Ksp value.

Concept introduction:

The solubility of a salt is defined as the maximium amount of salt that can be dissolved in definite amount of solvent. It is expressed in moles per liter or grams per liter. Solubility in terms of moles per liter is called molar solubility and is defined as the number of moles of solute (salt) dissolved in per liter of solution.

Solubility product constant Ksp is an equilibrium constant and is defined as the product of the equilibrium concentration of the ions of the salt raised to the power of their coefficients in the balanced chemical equation.

Expression for Ksp of a salt is given as,

AxBy(s)xAy+(aq)+yBx(aq)Ksp=[Ay+]x[Bx]y

Relation between Ksp and s is derived as follows,

EquationAxByxAy++yBxInitial(M)00Change(M)+xs+ysEquilibrium(M)xsys

Ksp=[Ay+]x[Bx]yKsp=(xs)x(ys)y=xxyy(s)x+y

Rearrange the expression for s.

(s)x+y=Kspxxyy=(Kspxxyy)1/(x+y)

Here,

• x is the coefficient of cation A+y
• y is the coefficient of anion Bx
• s is the molar solubility

Ksp is calculated by using molar solubility of the salt.

Explanation

Solubility of salt CaC2O4 in gL1 units is to be calculated as follows;

Given:

The value of solubility product Ksp of CaC2O4 is 4×109.

When CaC2O4 dissolved in water it dissociates as,

CaC2O4(s) Ca2+(aq)+ C2O42(aq)

The ICE table is as follows,

EquationCaC2O4Ca2++C2O42Initial (M)00Change (M)+s+sEquilibrium (M) ss

Expression of Ksp for CaC2O4 is as follows,

Ksp=[Ca2+][C2O42]

Substitute s for [Ca2+] and s for [C2O42].

Ksp=(s)(s)

Here,

• Ksp is solubility product constant
• s is solubility

Ksp=s2s2=Ksp

Rearrange for s,

s=Ksp

Substitute 4×109 for Ksp

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