Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 4, Problem 4.131P

(a)

Interpretation Introduction

Interpretation:

The balanced net ionic equation for the precipitation reaction is to be determined.

Concept introduction:

There are three types of equations that are utilized to represent an ionic reaction:

1. Molecular equation

2. Total ionic equation

3. Net ionic equation

The molecular equation represents the reactants and products of the ionic reaction in undissociated form. In total ionic reaction, all the dissociated ions that are present in the reaction mixture are represented and in net ionic reaction, the useful ions that participate in the reaction are represented.

Precipitation reaction involves the reaction of two soluble ionic compounds to form an insoluble product. The insoluble product is known as a precipitate.

The reason for the precipitation reaction to occur is the formation of a product that is insoluble in nature.

Spectator ions are the ions that are not a part of the actual chemical change but are present in the reaction mixture to balance the charge on both sides of the reaction. They are represented in the total ionic reaction. These are the dissolved ions present in the reaction mixture.

(a)

Expert Solution
Check Mark

Answer to Problem 4.131P

The molecular equation of the following reaction is:

Na2C2O4(aq)+CaCl2(aq)CaC2O4(s)+2NaCl(aq)

The total ionic equation for the given reaction is:

2Na+(aq)+C2O42(aq)+Ca2+(aq)+2Cl(aq)CaC2O4(s)+2Na+(aq)+2Cl(aq)

The net ionic equation for the given reaction is:

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

Explanation of Solution

Sodium oxalate (Na2C2O4) reacts with calcium chloride to form calcium oxalate (CaC2O4) and sodium chloride. The molecular equation of the following reaction is:

Na2C2O4(aq)+CaCl2(aq)CaC2O4(s)+2NaCl(aq)

The total ionic equation for the given reaction is:

2Na+(aq)+C2O42(aq)+Ca2+(aq)+2Cl(aq)CaC2O4(s)+2Na+(aq)+2Cl(aq)

Na+ and Cl are the spectator ions that are present in the reaction mixture. Spectator ions are not present in the net ionic equation.

The net ionic equation for the given reaction is:

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

Conclusion

CaC2O4 is the solid insoluble substance that is formed in the reaction. Na+ and Cl are the spectator ions that are present in the reaction mixture.

(b)

Interpretation Introduction

Interpretation:

The balanced net ionic equation for the titration reaction is to be determined.

Concept introduction:

There are three types of equations that are utilized to represent an ionic reaction:

1. Molecular equation

2. Total ionic equation

3. Net ionic equation

The molecular equation represents the reactants and products of the ionic reaction in undissociated form. In total ionic reaction, all the dissociated ions that are present in the reaction mixture are represented and in net ionic reaction, the useful ions that participate in the reaction are represented.

Precipitation reaction involves the reaction of two soluble ionic compounds to form an insoluble product. The insoluble product is known as a precipitate.

The reason for the precipitation reaction to occur is the formation of a product that is insoluble in nature.

Spectator ions are the ions that are not a part of the actual chemical change but are present in the reaction mixture to balance the charge on both sides of the reaction. They are represented in the total ionic reaction. These are the dissolved ions present in the reaction mixture.

(b)

Expert Solution
Check Mark

Answer to Problem 4.131P

The net ionic equation of the following reaction is:

2MnO4(aq)+5H2C2O4(aq)+6H+(aq)2Mn2+(aq)+10CO2(g)+8H2O(l)

Explanation of Solution

Potassium permanganate (KMnO4) is a soluble salt and can be written in the dissociated form and oxalic acid is weak acid and is present in the undissociated form. KMnO4 oxidizes the oxalic acid to CO2 and gets reduced to form Mn2+. The molecular equation of the following reaction is:

K+(aq)+MnO4(aq)+H2C2O4(aq)Mn2+(aq)+CO2(g)+K+(aq)

K+ is the spectator ion that is present in the reaction mixture.

The reaction takes place in an acidic medium so H+ and H2O are used to balance the reaction. The net ionic equation of the following reaction is:

2MnO4(aq)+5H2C2O4(aq)+6H+(aq)2Mn2+(aq)+10CO2(g)+8H2O(l)

Conclusion

Potassium permanganate (KMnO4) is a soluble salt and written in dissociated form. Oxalic acid is a weak acid and present in dissociated form.

(c)

Interpretation Introduction

Interpretation:

The oxidizing agent in the reaction is to be identified.

Concept introduction:

A redox reaction is a type of reaction that involves the change in oxidation number of a molecule, atom or ion changes due to the transfer of an electron from one species to another.

An oxidizing agent is a substance that oxidizes another species and itself gets reduced in a chemical reaction. A reducing agent is the one that reduces another species and itself gets oxidized in a chemical reaction.

(c)

Expert Solution
Check Mark

Answer to Problem 4.131P

The oxidizing agent in the reaction is MnO4.

Explanation of Solution

The given redox reaction is:

2MnO4(aq)+5H2C2O4(aq)+6H+(aq)2Mn2+(aq)+10CO2(g)+8H2O(l)

The oxidation number of oxygen is 2.

The expression to calculate the oxidation number of manganese in MnO4 is:

[4(oxidation number of oxygen)+(oxidation number of mangenese)]=1 (1)

Rearrange equation (1) for the oxidation number of manganese.

Oxidation number of mangenese=[(1)4(oxidation number of oxygen)] (2)

Substitute 2 for the oxidation number of oxygen in equation (2).

Oxidation number of mangenese=[(1)4(2)]=[1+8]=+7

The oxidation state of Mn2+ is +2 that is equal to the charge present on the Mn2+.

The oxidation state of manganese in MnO4 is +7 and it decreases to +2 in Mn2+. MnO4 undergoes a reduction in the reaction, therefore, it acts as the oxidizing agent.

Conclusion

An oxidizing agent is a substance that oxidizes another species and itself gets reduced in a chemical reaction. MnO4 undergoes reduction and acts as the oxidizing agent in the reaction

(d)

Interpretation Introduction

Interpretation:

The reducing agent in the reaction is to be identified.

Concept introduction:

A redox reaction is a type of reaction that involves the change in oxidation number of a molecule, atom or ion changes due to the transfer of an electron from one species to another.

An oxidizing agent is a substance that oxidizes another species and itself gets reduced in a chemical reaction. A reducing agent is the one that reduces another species and itself gets oxidized in a chemical reaction.

(d)

Expert Solution
Check Mark

Answer to Problem 4.131P

The reducing agent in the reaction is H2C2O4.

Explanation of Solution

The given redox reaction is:

2MnO4(aq)+5H2C2O4(aq)+6H+(aq)2Mn2+(aq)+10CO2(g)+8H2O(l)

The expression to calculate the oxidation number of carbon in H2C2O4 is:

[2(oxidation number of hydrogen)+2(oxidation number of carbon)+4(oxidation number of oxygen)]=0 (3)

Rearrange equation (3) for the oxidation number of carbon.

Oxidation number of carbon=[2(oxidation number of hydrogen)4(oxidation number of oxygen)2] (4)

Substitute 2 for oxidation number of oxygen and +1 for the oxidation state of hydrogen in equation (4).

Oxidation number of carbon=[2(+1)4(2)2]=[2+82]=[+62]=+3

The expression to calculate the oxidation number of carbon in CO2 is:

[2(oxidation number of oxygen)+(oxidation number of carbon)]=0 (5)

Rearrange equation (5) for the oxidation number of carbon.

Oxidation number of carbon=[2(oxidation number of oxygen)] (6)

Substitute 2 for the oxidation number of oxygen in equation (4).

Oxidation number of carbon=[2(2)]=+4

The oxidation state of carbon in H2C2O4 is +3 and it increases to +4 in CO2. H2C2O4 undergoes oxidation in the reaction, therefore, it acts as the reducing agent.

Conclusion

A reducing agent is the one that reduces another species and itself gets oxidized in a chemical reaction. H2C2O4 undergoes oxidation and acts as the reducing agent in the reaction.

(e)

Interpretation Introduction

Interpretation:

The mass percent of CaCl2 in the original sample is to be determined.

Concept introduction:

Precipitation reaction involves the reaction of two soluble ionic compounds to form an insoluble product. The insoluble product is known as a precipitate.

The reason for the precipitation reaction to occur is the formation of a product that is insoluble in nature.

(e)

Expert Solution
Check Mark

Answer to Problem 4.131P

The mass percent of CaCl2 in the original sample is 55.06%.

Explanation of Solution

The given redox reaction is:

2MnO4(aq)+5H2C2O4(aq)+6H+(aq)2Mn2+(aq)+10CO2(g)+8H2O(l)

The formula to calculate the moles of MnO4 is,

MolesofMnO4=(molarity of MnO4(mol/L))(Volume ofMnO4(L)) (7)

Substitute 0.1019mol/L for the molarity of MnO4 and 37.68mL for the volume of MnO4 in the equation (7).

MolesofMnO4=(0.1019mol/L)(37.68mL)(1L1000mL)=0.0038396mol

The formula to calculate the moles of H2C2O4 is,

MolesofH2C2O4=(moles of MnO4(mol))(5molH2C2O42molMnO4) (8)

Substitute 0.0038396mol for the moles of MnO4 in the equation (8).

MolesofH2C2O4=(0.0038396mol)(5molH2C2O42molMnO4)=0.009599mol

The formula to calculate the mass of CaCl2 is,

MassofCaCl2=[(moles ofH2C2O4(mol))(1molCaCl21molH2C2O4)molecularmassofCaCl2(g/mol)]                              (9)

Substitute 0.009599mol for moles of H2C2O4 and 110.98g/mol for molecular mass of CaCl2 in the equation (2).

MassofCaCl2=[(0.009599mol)(1molCaCl21molH2C2O4)(110.98g/mol)]=1.06530g

The expression to calculate the mass percent of CaCl2 is,

mass %ofCaCl2=(massof CaCl2(g)mass of sample(g))(100) (10)

Substitute 1.06530g for the mass of CaCl2 and 1.9348g for the mass of sample in the equation (10).

Mass %ofCaCl2=(1.06530g1.9348g)(100)=55.05995%55.06%

Conclusion

The mass percent of CaCl2 in the original sample is 55.06%.

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Chapter 4 Solutions

Chemistry: The Molecular Nature of Matter and Change

Ch. 4.1 - A chemist dilutes 60.0 mL of 4.50 M potassium...Ch. 4.1 - Prob. 4.6BFPCh. 4.1 - Prob. 4.7AFPCh. 4.1 - Prob. 4.7BFPCh. 4.3 - Prob. 4.8AFPCh. 4.3 - Prob. 4.8BFPCh. 4.3 - Prob. 4.9AFPCh. 4.3 - Molecular views of the reactant solutions for a...Ch. 4.3 - It is desirable to remove calcium ion from hard...Ch. 4.3 - To lift fingerprints from a crime scene, a...Ch. 4.3 - Despite the toxicity of lead, many of its...Ch. 4.3 - Mercury and its compounds have uses from fillings...Ch. 4.4 - How many OH−(aq) ions are present in 451 mL of...Ch. 4.4 - Prob. 4.12BFPCh. 4.4 - Prob. 4.13AFPCh. 4.4 - Prob. 4.13BFPCh. 4.4 - Prob. 4.14AFPCh. 4.4 - Prob. 4.14BFPCh. 4.4 - Another active ingredient in some antacids is...Ch. 4.4 - Prob. 4.15BFPCh. 4.4 - What volume of 0.1292 M Ba(OH)2 would neutralize...Ch. 4.4 - Calculate the molarity of a solution of KOH if...Ch. 4.5 - Prob. 4.17AFPCh. 4.5 - Prob. 4.17BFPCh. 4.5 - Prob. 4.18AFPCh. 4.5 - Prob. 4.18BFPCh. 4.5 - Prob. 4.19AFPCh. 4.5 - Prob. 4.19BFPCh. 4.6 - Prob. 4.20AFPCh. 4.6 - Prob. 4.20BFPCh. 4 - Prob. 4.1PCh. 4 - What types of substances are most likely to be...Ch. 4 - Prob. 4.3PCh. 4 - Prob. 4.4PCh. 4 - Which of the following scenes best represents how...Ch. 4 - Prob. 4.6PCh. 4 - Prob. 4.7PCh. 4 - Prob. 4.8PCh. 4 - Prob. 4.9PCh. 4 - Prob. 4.10PCh. 4 - A mathematical equation useful for dilution...Ch. 4 - Prob. 4.12PCh. 4 - Prob. 4.13PCh. 4 - Prob. 4.14PCh. 4 - Prob. 4.15PCh. 4 - Does an aqueous solution of each of the following...Ch. 4 - Prob. 4.17PCh. 4 - Prob. 4.18PCh. 4 - Prob. 4.19PCh. 4 - Prob. 4.20PCh. 4 - Prob. 4.21PCh. 4 - Calculate each of the following quantities: Mass...Ch. 4 - Prob. 4.23PCh. 4 - Prob. 4.24PCh. 4 - Prob. 4.25PCh. 4 - Prob. 4.26PCh. 4 - Prob. 4.27PCh. 4 - Prob. 4.28PCh. 4 - Calculate each of the following quantities: Volume...Ch. 4 - Prob. 4.30PCh. 4 - Concentrated sulfuric acid (18.3 M) has a density...Ch. 4 - Prob. 4.32PCh. 4 - Muriatic acid, an industrial grade of concentrated...Ch. 4 - Prob. 4.34PCh. 4 - Prob. 4.35PCh. 4 - Prob. 4.36PCh. 4 - Write two sets of equations (both molecular and...Ch. 4 - Why do some pairs of ions precipitate and others...Ch. 4 - Use Table 4.1 to determine which of the following...Ch. 4 - The beakers represent the aqueous reaction of...Ch. 4 - Complete the following precipitation reactions...Ch. 4 - Prob. 4.42PCh. 4 - Prob. 4.43PCh. 4 - Prob. 4.44PCh. 4 - Prob. 4.45PCh. 4 - Prob. 4.46PCh. 4 - Prob. 4.47PCh. 4 - If 25.0 mL of silver nitrate solution reacts with...Ch. 4 - Prob. 4.49PCh. 4 - Prob. 4.50PCh. 4 - With ions shown as spheres and solvent molecules...Ch. 4 - The precipitation reaction between 25.0 mL of a...Ch. 4 - A 1.50-g sample of an unknown alkali-metal...Ch. 4 - Prob. 4.54PCh. 4 - The mass percent of Cl− in a seawater sample is...Ch. 4 - Prob. 4.56PCh. 4 - Prob. 4.57PCh. 4 - Write a general equation for a neutralization...Ch. 4 - Prob. 4.59PCh. 4 - (a) Name three common weak acids. 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