Chemistry
Chemistry
12th Edition
ISBN: 9780078021510
Author: Raymond Chang Dr., Kenneth Goldsby Professor
Publisher: McGraw-Hill Education
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Chapter 16, Problem 16.130QP

Water containing Ca2+ and Mg2+ 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 Ca2+ ions from hard water is by adding washing soda (Na2CO3 · 10H2O). (a) The molar solubility of CaCO3 is 9.3 × 10−5 M. What is its molar solubility in a 0.050 M Na2CO3 solution? (b) Why are Mg2+ ions not removed by this procedure? (c) The Mg2+ 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)2 solution. (d) What is the concentration of Mg2+ ions at this pH? (e) In general, which ion (Ca2+ or Mg2+) would you remove first? Why?

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The molar solubility, pH and concentration of given solution has to be calculated.

Concept introduction:

  • Buffer solution is defined as a solution that oppose changes in pH while adding little amount of either an acid or a base. In general, addition of acid or base does not affect the pH in buffer solution but if it is more than amount of conjugate base or conjugate acid, then buffer loses its buffering capacity.
  • Buffer solution is a combination of a weak acid and its conjugate base or vice-versa in an aqueous solution.
  • Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.
  • When the concentration of one of the ions of a chemical solution got higher, it reacts with counter charged ions and precipitated out as salt till the ion product equals solubility product is called common ion effect.

To calculate: the molar solubility of CaCO3 in Na2CO3

Answer to Problem 16.130QP

s = 1.7×10-7M

Explanation of Solution

ThedissociationofNa2CO3Na2CO3(s) H2O2Na+(aq)    +    CO32-(aq)2(0.050M)       0.050MConsider s be the molar solubility of CaCO3inNa2CO3CaCO3(s) H2OCa2+(aq)    +    CO32-(aq) Initial concentration(M):                             0.00                0.050 Changeinconcentration (M):                              +s                    +s  Equilibriumconcentration (M):                             +s                 0.050+sKsp = [Ca2+][CO32-]  8.7×10-9 = s(0.050+s) Small and neglect it, 0.050+s0.050 8.7×10-9 = 0.050s s = 1.7×10-7M

The molar solubility of CaCO3 in Na2CO3 is calculated using the solubility product expression of CaCO3.  By substituting the concentrations of ions and doing simple mathematical operations, the molar solubility of CaCO3 is determined.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The molar solubility, pH and concentration of given solution has to be calculated.

Concept introduction:

  • Buffer solution is defined as a solution that oppose changes in pH while adding little amount of either an acid or a base. In general, addition of acid or base does not affect the pH in buffer solution but if it is more than amount of conjugate base or conjugate acid, then buffer loses its buffering capacity.
  • Buffer solution is a combination of a weak acid and its conjugate base or vice-versa in an aqueous solution.
  • Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.
  • When the concentration of one of the ions of a chemical solution got higher, it reacts with counter charged ions and precipitated out as salt till the ion product equals solubility product is called common ion effect.

To explain: the reason of magnesium ion cannot be removed as above.

Answer to Problem 16.130QP

Because Chemistry, Chapter 16, Problem 16.130QP , additional homework tip  1 are moderately soluble (Ksp=4.0×10-5).

Explanation of Solution

Because Mg2+ions are moderately soluble (Ksp=4.0×10-5).  Hence, it cannot be removed.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The molar solubility, pH and concentration of given solution has to be calculated.

Concept introduction:

  • Buffer solution is defined as a solution that oppose changes in pH while adding little amount of either an acid or a base. In general, addition of acid or base does not affect the pH in buffer solution but if it is more than amount of conjugate base or conjugate acid, then buffer loses its buffering capacity.
  • Buffer solution is a combination of a weak acid and its conjugate base or vice-versa in an aqueous solution.
  • Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.
  • When the concentration of one of the ions of a chemical solution got higher, it reacts with counter charged ions and precipitated out as salt till the ion product equals solubility product is called common ion effect.

To calculate the pH of Ca(OH)2

Answer to Problem 16.130QP

pH = 12.40

Explanation of Solution

KspofCa(OH)2is 8.0×10-6                       Ca(OH)2      Ca2++2OH-At equilibrium:                                   s          2s                                        Ksp = 8.0×106 = [Ca2+][OH-]2                                        4s3 = 8.0×10-6                                          s  = 0.0126M                            [OH-] = 2s = 0.0252M                                pOH = -log(0.0252) = 1.60                                    pH = 12.40

The molar solubility of Ca(OH)2 is calculated using the solubility product expression of Chemistry, Chapter 16, Problem 16.130QP , additional homework tip  2.  By calculating the pOH using concentration of hydroxide ion, the pH value is determined.

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The molar solubility, pH and concentration of given solution has to be calculated.

Concept introduction:

  • Buffer solution is defined as a solution that oppose changes in pH while adding little amount of either an acid or a base. In general, addition of acid or base does not affect the pH in buffer solution but if it is more than amount of conjugate base or conjugate acid, then buffer loses its buffering capacity.
  • Buffer solution is a combination of a weak acid and its conjugate base or vice-versa in an aqueous solution.
  • Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.
  • When the concentration of one of the ions of a chemical solution got higher, it reacts with counter charged ions and precipitated out as salt till the ion product equals solubility product is called common ion effect.

To calculate the strength of Mg2+ ions

Answer to Problem 16.130QP

[Mg2+] = 1.9×10-8MChemistry, Chapter 16, Problem 16.130QP , additional homework tip  3

Explanation of Solution

Higher concentration of hydroxide ion removes most of Mg(OH)2.  But there is only small amount remaining due to below equilibrium

 Mg(OH)2(s)  Mg2+(aq)+2OH-(aq)       Ksp = [Mg2+][OH-]2  1.2×1011 = [Mg2+](0.0252)2   [Mg2+] = 1.9×10-8M

The concentration of Mg2+ ions is calculated using the solubility product expression of Mg(OH)2.  By substituting the concentrations of ion and Ksp  and doing simple mathematical operations, concentration of Mg2+ ions is determined.

(e)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The molar solubility, pH and concentration of given solution has to be calculated.

Concept introduction:

  • Buffer solution is defined as a solution that oppose changes in pH while adding little amount of either an acid or a base. In general, addition of acid or base does not affect the pH in buffer solution but if it is more than amount of conjugate base or conjugate acid, then buffer loses its buffering capacity.
  • Buffer solution is a combination of a weak acid and its conjugate base or vice-versa in an aqueous solution.
  • Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.
  • When the concentration of one of the ions of a chemical solution got higher, it reacts with counter charged ions and precipitated out as salt till the ion product equals solubility product is called common ion effect.

To explain: which one ions can be removed first.

Answer to Problem 16.130QP

Chemistry, Chapter 16, Problem 16.130QP , additional homework tip  4Because calcium ion is present in huge amount

Explanation of Solution

As calcium ion is present in huge amount. it can be removed first.

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