Chemistry
Chemistry
4th Edition
ISBN: 9780078021527
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 17, Problem 96AP

Water containing Ca 2+ and Mg 2+ ions is called hard water and is unsuitable for some household and industrial use because these ions react with soap to form insoluble salts, or curds. One way to remove the Ca 2+ ions from hard water is by adding washing soda (Na 2 CO 3 - 10H 3 O) . (a) The molar solubility of CaCO 3  is 9 .3  ×  10 -5 if. What is its molar solubility in a 0.050 M Na 2 CO 3 solution? (b) Why are Mg 2+ ions not removed by this procedure? (c) The Mg 2+ ions are removed as Mg ( OH ) 2 by adding slaked lime [ Ca ( OH ) 2 ] to the water to produce a saturated solution. Calculate the pH of a saturated Ca(OH), solution. (d) What is the concentration of Mg 2+ ions at this pH? (e) In general, which ion ( Ca 2+ or Mg 2+ ) would you remove first? Why?

Expert Solution & Answer
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Interpretation Introduction

Interpretation: The molar solubility of CaCO3 in 0.050 M Na2CO3 solution is to be calculated. The reason behind the non-removal of Mg2+ by this procedure is to be explained. The pH of the saturated Ca(OH)2 is to be calculated. The concentration of Mg2+ ions in the calculated pH is to be determined. The ion that will be removedout of the given ions is to be determined.

Concept introduction:

When a compound that contains an ion that is common with a dissolved substance is added to a solution at equilibrium, the equilibrium shifts to the left. This is known as the common ion effect.

Molar solubility is defined as the moles of solute dissolved in one liter of solution, and its unit is mol/L.

At a given temperature, the product of molar concentrations of the ions of salt present in the solution is known as the solubility product of the salt. It is represented by Ksp.

Higher is the value of solubility product of a salt, more is its solubility.

The presence of common ions in the solution decreases the solubility of a given compound.

For a general reaction: AB(s)A+(aq)+B(aq).

The solubility product can be calculated by the expression as: Ksp= [A+][B]

Here, Ksp is the solubility product constant and sp stands for solubility product.

The expression of pOH is represented as:

pOH=log[OH]

The relation between pH and pOH can be expressed as: pH+pOH=14

Answer to Problem 96AP

Solution:

(a) 1.7×107 M

(b)It is fairly soluble.

(c)12.40

(d) 1.9×108 M

(e) Ca2+

Explanation of Solution

a)The molar solubility of CaCO3 is 9.3×105 M, it’s molar solubility in 0.050 M Na2CO3.

This problem is based on the common ion effect.

Consider the dissociation of Na2CO3 as:

Na2CO3(s)2Na+(aq)+CO32(aq)

Let s be the solubility of CaCO3 in the Na2CO3 solution.

Consider the dissociation of CaCO3 as:

                       CaCO3(s)Ca2+(aq)+CO32(aq)Initial(M)                              0.00             0.050Change(M)                            +s                 +sEquilibrium(M)                       s                0.050+s

According to the above dissociation reaction, the expression for Ksp will be as:

Ksp=[Ca2+][CO32]

The solubility product constant of CaCO3 is 8.7×109.

Substitute 8.7×109 for Ksp, s for [Ca2+], and (0.050+s) for [CO32] in the above expression as:

8.7×109=s(0.050+s)

Since the value of Ksp is very small, the value of (0.050+s) will be approximately equal to 0.050.

8.7×109=0.050ss=8.7×1090.050=1.7×107 M

As the addition of washing soda to permanent hard water is done, most of the Ca2+ will be removed due to the common ion effect.

Therefore, the molar solubility of CaCO3 in 0.050 M Na2CO3 solution is 1.7×107 M.

b) Mg2+ Ions not removed by this procedure

The solubility product constant of MgCO3 is 4.0×105. From its Ksp value, it can be said that it is soluble in water. As a result, Mg2+ cannot be removed by the common ion effect method.

c) The pH of a saturated Ca(OH)2 solution.

The solubility product constant of Ca(OH)2 is 8.0×106. Write its dissociation reaction as:

Ca(OH)2Ca2++2OH

Let s be the solubility of Ca(OH)2.

So, according to the above reaction, the expression for Ksp will be as:

Ksp=(s)(2s)2=4s3

Substitute 8.0×106 for Ksp in the above expression as:

8.0×106=4s3s3=8.0×1064=2.0×106s=(2.0×106)1/3=0.0126 M

Now, the concentration of OH will be as:

[OH]=2s=2(0.0126 M)=0.0252 M

Calculate the pOH of the solution as:

pOH=log[OH]

Substitute 0.0252 M for [OH] as:

pOH=log[0.0252 M]=1.60

The relation between pH and pOH is:

pH+pOH=14

Substitute 1.60 for pOH as:

pH+1.60=14pH=12.40

Therefore, the pH of the Ca(OH)2 solution is 12.40.

d)The concentration of Mg2+ ions at this pH

From the previous part of the question, the concentration of OH is 0.0252 M, which is quite high. At this concentration, most of the Mg2+ will be removed in the form of Mg(OH)2. The concentration of the remaining Mg2+ is only due to following reaction, which takes place at equilibrium:

Mg(OH)2Mg2++2OH

According to the above reaction, the expression for Ksp will be as:

Ksp=[Mg2+][OH]2

The calculated concentration of OH is 0.0252 M.

Substitute 0.0252 M for [OH] and 1.2×1011 for Ksp in the above reaction as:

1.2×1011=[Mg2+][0.0252 M]2[Mg2+]=1.2×1011(0.0252 M)2=1.9×108 M

Therefore, the concentration of Mg2+ is 1.9×108 M.

e) The i on (Ca2+ or Mg2+) that would be removed first.

Out of Ca2+ & Mg2+, Ca2+ will be removed first because it is present in alarger quantity.

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Chapter 17 Solutions

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