Introductory Chemistry: An Active Learning Approach
Introductory Chemistry: An Active Learning Approach
6th Edition
ISBN: 9781305079250
Author: Mark S. Cracolice, Ed Peters
Publisher: Cengage Learning
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Chapter 18, Problem 67E
Interpretation Introduction

Interpretation:

The amount of calcium oxalate that dissolves in 2.5×102mL of 0.22MNa2C2O4 if the Ksp for CaC2O4 is 2.4×109, is to be calculated.

Concept introduction:

Solubility product is the product of concentrations of ions raised to the power of their stoichiometric coefficients at equilibrium. It is represented by Ksp. It is applicable for saturated solutions.

Expert Solution & Answer
Check Mark

Answer to Problem 67E

The amount of calcium oxalate that will dissolve in 2.5×102mL of 0.22MNa2C2O4, is 3.5×107g.

Explanation of Solution

The equilibrium equation for calcium oxalate is shown below.

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

The solubility product constant for calcium oxalate is calculated by the formula shown below.

Ksp=[Ca2+][C2O42]…(1)

Where,

[Ca2+] is the molar concentration of calcium ion.

[ C2O42 ] is the molar concentration of oxalate ion.

The given value of Ksp=2.4×109.

The equilibrium equation for sodium oxalate is shown below.

Na2C2O4(s)2Na+(aq)+C2O42(aq)

In the above reaction, the concentration of Na2C2O4 is equal to the concentration of Na+ and C2O42. Since the concentration of Na2C2O4 is 0.22M, so, the concentration of C2O42 is also 0.22M.

Substitute the values of Ksp and concentration of C2O42 in equation (1).

2.4×109=[Ca2+]×0.22M[Ca2+]=2.4×1090.22M=1.09×108M

The concentration of [Ca2+] is 1.09×108M.

According to the equilibrium equation for calcium oxalate shown above, the concentration of CaC2O4 must be equal to the concentration of [Ca2+]. So, the concentration of CaC2O4 is 1.09×108M.

The molar mass of CaC2O4 is 128.10g/mol.

The volume of the solution is 2.5×102mL.

The conversion of mL into L is done as shown below.

1000mL=1L

So, the conversion of 2.5×102mL into L is shown below.

1L1000mL×2.5×102mL=0.25L

The volume of the solution is 0.25L.

The number of moles of solute is calculated as shown below.

Moles=MassofsoluteMolarmassofsolute…(2)

The molarity of the solution is calculated as shown below.

Molarity=NumberofmolesofsoluteVolumeofsolutioninL …(3)

Substitute equation (2) in equation (3) as shown below.

Molarity=MassofsoluteMolarmassofsoluteVolumeofsolutioninL

Substitute the values of molarity, molar mass and volume of solution in above equation as shown below.

Molarity=MassofsoluteMolarmassofsoluteVolumeofsolutioninL1.09×108M=Massofsolute128.10g/mol0.25L

Therefore, the mass of the solute is calculated as shown below.

Massofsolute=1.09×108M×128.10g/mol×0.25L=3.49×107g3.5×107g

Conclusion

The amount of calcium oxalate required is 3.5×107g.

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Chapter 18 Solutions

Introductory Chemistry: An Active Learning Approach

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