General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 17, Problem 17.108QP

(a)

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

(a)

Expert Solution
Check Mark

Answer to Problem 17.108QP

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)

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.

(b)

Expert Solution
Check Mark

Answer to Problem 17.108QP

Because General Chemistry, Chapter 17, Problem 17.108QP , 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)

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

(c)

Expert Solution
Check Mark

Answer to Problem 17.108QP

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 General Chemistry, Chapter 17, Problem 17.108QP , additional homework tip  2.  By calculating the pOH using concentration of hydroxide ion, the pH value is determined.

(d)

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

(d)

Expert Solution
Check Mark

Answer to Problem 17.108QP

[Mg2+] = 1.9×10-8MGeneral Chemistry, Chapter 17, Problem 17.108QP , 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)

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.

(e)

Expert Solution
Check Mark

Answer to Problem 17.108QP

General Chemistry, Chapter 17, Problem 17.108QP , 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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Chapter 17 Solutions

General Chemistry

Ch. 17.5 - Prob. 2PECh. 17.5 - Prob. 3PECh. 17.6 - Prob. 1PECh. 17.6 - Prob. 1RCCh. 17.7 - Prob. 1PECh. 17.7 - Prob. 1RCCh. 17 - Prob. 17.1QPCh. 17 - Prob. 17.2QPCh. 17 - Prob. 17.3QPCh. 17 - 17.4 The pKbs for the bases X−, Y−, and Z− are...Ch. 17 - 17.5 Specify which of these systems can be...Ch. 17 - 17.6 Specify which of these systems can be...Ch. 17 - 17.7 The pH of a bicarbonate–carbonic acid buffer...Ch. 17 - Prob. 17.8QPCh. 17 - 17.9 Calculate the pH of the buffer system 0.15 M...Ch. 17 - 17.10 What is the pH of the buffer 0.10 M...Ch. 17 - 17.11 The pH of a sodium acetate–acetic acid...Ch. 17 - 17.12 The pH of blood plasma is 7.40. Assuming the...Ch. 17 - Prob. 17.13QPCh. 17 - Prob. 17.14QPCh. 17 - 17.16 A student wishes to prepare a buffer...Ch. 17 - Prob. 17.17QPCh. 17 - Prob. 17.18QPCh. 17 - Prob. 17.19QPCh. 17 - 17.20 A 5.00-g quantity of a diprotic acid is...Ch. 17 - Prob. 17.21QPCh. 17 - Prob. 17.22QPCh. 17 - 17.23 The diagrams shown here represent solutions...Ch. 17 - 16.38 The diagrams shown here represent solutions...Ch. 17 - 17.25 Explain how an acid-base indicator works in...Ch. 17 - 17.26 What are the criteria for choosing an...Ch. 17 - 17.27 The amount of indicator used in an acid-base...Ch. 17 - 17.28 A student carried out an acid-base titration...Ch. 17 - 17.29 Referring to Table 17.1, specify which...Ch. 17 - 17.30 The ionization constant Ka of an indicator...Ch. 17 - 17.31 Define solubility, molar solubility, and...Ch. 17 - 17.32 Why do we usually not quote the Ksp values...Ch. 17 - 17.33 Write balanced equations and solubility...Ch. 17 - Prob. 17.34QPCh. 17 - Prob. 17.35QPCh. 17 - 17.36 Silver chloride has a larger Ksp than silver...Ch. 17 - Prob. 17.38QPCh. 17 - 17.39 The molar solubility of MnCO3 is 4.2 × 10−6...Ch. 17 - Prob. 17.40QPCh. 17 - Prob. 17.41QPCh. 17 - 17.42 Using data from Table 17.2, calculate the...Ch. 17 - 17.43 What is the pH of a saturated zinc hydroxide...Ch. 17 - 17.44 The pH of a saturated solution of a metal...Ch. 17 - Prob. 17.45QPCh. 17 - 17.46 A volume of 75 mL of 0.060 M NaF is mixed...Ch. 17 - 17.47 How does a common ion affect solubility? Use...Ch. 17 - Prob. 17.48QPCh. 17 - Prob. 17.49QPCh. 17 - Prob. 17.50QPCh. 17 - Prob. 17.51QPCh. 17 - 17.52 Calculate the molar solubility of BaSO4 (a)...Ch. 17 - Prob. 17.55QPCh. 17 - Prob. 17.56QPCh. 17 - 17.57 If 2.50 g of CuSO4 are dissolved in 9.0 ×...Ch. 17 - 17.58 Calculate the concentrations of Cd2+, , and...Ch. 17 - Prob. 17.59QPCh. 17 - Prob. 17.60QPCh. 17 - Prob. 17.61QPCh. 17 - Prob. 17.62QPCh. 17 - Prob. 17.63QPCh. 17 - 16.88 In a group 1 analysis, a student adds HCl...Ch. 17 - 17.65 Both KCl and NH4Cl are white solids. Suggest...Ch. 17 - 17.66 Describe a simple test that would enable you...Ch. 17 - Prob. 17.67QPCh. 17 - Prob. 17.68QPCh. 17 - Prob. 17.69QPCh. 17 - 17.70 The pKa of the indicator methyl orange is...Ch. 17 - Prob. 17.71QPCh. 17 - Prob. 17.72QPCh. 17 - 17.73 The two curves shown here represent the...Ch. 17 - 17.74 The two curves shown here represent the...Ch. 17 - Prob. 17.75QPCh. 17 - 17.76 A solution is made by mixing exactly 500 mL...Ch. 17 - Prob. 17.77QPCh. 17 - Prob. 17.78QPCh. 17 - 17.79 For which of these reactions is the...Ch. 17 - Prob. 17.80QPCh. 17 - Prob. 17.81QPCh. 17 - Prob. 17.82QPCh. 17 - Prob. 17.83QPCh. 17 - 17.84 Find the approximate pH range suitable for...Ch. 17 - Prob. 17.85QPCh. 17 - 17.86 Which of these substances will be more...Ch. 17 - Prob. 17.87QPCh. 17 - Prob. 17.88QPCh. 17 - Prob. 17.89QPCh. 17 - Prob. 17.90QPCh. 17 - Prob. 17.91QPCh. 17 - 17.92 When a KI solution was added to a solution...Ch. 17 - Prob. 17.93QPCh. 17 - Prob. 17.94QPCh. 17 - Prob. 17.95QPCh. 17 - 17.96 Solid NaI is slowly added to a solution that...Ch. 17 - Prob. 17.97QPCh. 17 - 17.98 (a) Assuming complete dissociation and no...Ch. 17 - 17.99 Acid-base reactions usually go to...Ch. 17 - 17.100 Calculate x, the number of molecules of...Ch. 17 - Prob. 17.101QPCh. 17 - 17.102 What reagents would you employ to separate...Ch. 17 - 17.103 CaSO4 (Ksp = 2.4 × 10−5) has a larger Ksp...Ch. 17 - 17.104 How many milliliters of 1.0 M NaOH must be...Ch. 17 - Prob. 17.105QPCh. 17 - Prob. 17.106QPCh. 17 - Prob. 17.107QPCh. 17 - Prob. 17.108QPCh. 17 - Prob. 17.109QPCh. 17 - Prob. 17.111SPCh. 17 - Prob. 17.112SPCh. 17 - Prob. 17.113SPCh. 17 - Prob. 17.114SPCh. 17 - Prob. 17.115SPCh. 17 - Prob. 17.116SPCh. 17 - 17.117 The titration curve shown here represents...Ch. 17 - Prob. 17.118SP
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