chapter 5 selected solutions

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CHAPTER 5

INTRODUCTION TO REACTIONS

IN AQUEOUS SOLUTIONS

PRACTICE EXAMPLES

1A In determining total Cl , we recall the definition of molarity: moles of solute per liter of solution

from NaCl Clmol NaCl

L soln

mol Cl

mol NaClM Cl, =

0.438

1

1

1= 0.438

22

2

0.0512 mol MgCl 2 mol Clfrom MgCl , Cl = = 0.102 M Cl

1 L soln 1 mol MgCl

2Cl total = Cl from NaCl+ Cl from MgCl = 0.438 M + 0.102 M = 0.540 M Cl

1B (a)-1.5 mg F

L

1 g F

1000 mg F

1 mol F

18.998 g F

= 7.9 10-5 M F-

(b) 1.00 106 L5 -7.9 10 mol F

1L

2

1 mol CaF

2 mol F 2

2

78.075 g CaF

1 mol CaF

1 kg

1000 g= 3.1 kg CaF2

2A In each case, we use the solubility rules to determine whether either product is insoluble.The ions in each product compound are determined by simply switching the partners of the

reactant compounds. The designation (aq) on each reactant indicates that it is soluble.

(a) Possible products are potassium chloride, KCl, which is soluble, and aluminum hydroxide,

3

Al OH , which is not. Net ionic equation: 3+3

Al aq + 3 OH aq Al OH s

(b) Possible products are iron(III) sulfate, 2 4 3Fe SO , and potassium bromide, KBr, both

of which are soluble. No reaction occurs.

(c) Possible products are calcium nitrate, Ca(NO3)2, which is soluble, and lead(II) iodide,

PbI2 , which is insoluble. The net ionic equation is: 2+

2Pb aq + 2 I aq PbI s

2B (a) Possible products are sodium chloride, NaCl, which is soluble, and aluminum phosphate,

AlPO4 , which is insoluble. Net ionic equation: 33+

4 4Al aq + PO aq AlPO s

(b) Possible products are aluminum chloride, AlCl3 , which is soluble, and barium sulfate,BaSO4 , which is insoluble. Net ionic equation:

22+4 4Ba aq + SO aq BaSO s


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(c) Possible products are ammonium nitrate, NH NO4 3 , which is soluble, and lead (II) carbonate

PbCO3 , which is insoluble. Net ionic equation: 22+

3 3Pb aq + CO aq PbCO s

3A Propionic acid is a weak acid, not dissociated completely in aqueous solution. Ammonia similarlyis a weak base. The acid and base react to form a salt solution of ammonium propionate.

3 3 5 2 4 3 5 2NH aq + HC H O aq NH aq + C H O aq

3B Since acetic acid is a weak acid, it is not dissociated completely in aqueous solution (exceptat infinite dilution); it is misleading to write it in ionic form. The products of this reactionare the gas carbon dioxide, the covalent compound water, and the ionic solute calciumacetate. Only the latter exists as ions in aqueous solution.

2+3 2 3 2 2 2 2 3 2CaCO s + 2 HC H O aq CO g + H O l + Ca aq + 2 C H O aq

4A (a) This is a metathesis or double displacement reaction. Elements do not change oxidationstates during this reaction. It is not an oxidationreduction reaction.

(b) The presence of O2(g) as a product indicates that this is an oxidationreduction reaction.Oxygen is oxidized from O.S. = -2 in NO3

- to O.S. = 0 in O2(g). Nitrogen is reducedfrom O.S. = +5 in NO3

- to O.S. = +4 in NO2.

4B Vanadium is oxidized from O.S. = +4 in VO2+ to an O.S. = +5 in VO2+ while manganese is

reduced from O.S. = +7 in MnO4- to O.S. = +2 in Mn2+.

5A Aluminum is oxidized (from an O.S. of 0 to an O.S. of +3), while hydrogen is reduced(from an O.S. of +1 to an O.S. of 0).

3+: Al s Al aq + 3 e 2Oxidation

Reduction: + 22 H aq + 2 e H g 3

+ 3+ 2: 2 Al s + 6 H aq 2 Al aq + 3 H gNet equation

5B Bromide is oxidized (from 1 to 0), while chlorine is reduced (from 0 to 1). 2: 2 Br aq Br l + 2 eOxidation

Reduction: 2Cl g + 2 e 2 Cl aq

2 2: 2 Br aq + Cl g Br l + 2 Cl aqNet equation

6A Step 1: Write the two skeleton half reactions.

2+ 2+ 3+4MnO aq Mn aq Fe aq Fe aqand

Step 2: Balance each skeleton half reaction for O (with H O2 ) and for H atoms (withH+ ).

+ 2+ 2+ 3+4 2MnO aq 8H aq Mn aq 4 H O(l) Fe aq Fe aqand


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Step 3: Balance electric charge by adding electrons.

+ 2+ 2+ 3+4 2MnO aq 8 H aq 5e Mn aq 4 H O(l) Fe aq Fe aq eand

Step 4: Combine the two half reactions

2+ 3+Fe aq Fe aq + e 5

+ 2+

4 2MnO aq +8 H aq + 5 e Mn aq + 4 H O(l)

+ 2+ 2+ 3+4 2MnO aq +8 H aq + 5 Fe aq Mn aq + 4 H O(l) + 5 Fe aq

6B Step 1: Uranium is oxidized and chromium is reduced in this reaction. The skeleton

half-equations are: 2+ 22+ 3

2 2 7UO aq UO aq Cr O (aq) Cr (aq)and

Step 2: First, balance the chromium skeleton half-equation for chromium atoms:

2 3+

2 7Cr O aq 2 Cr aq

Next, balance oxygen atoms with water molecules in each half-equation:

2 22+ 3

2 2 2 7 2UO aq + H O(l) UO aq Cr O (aq) 2Cr (aq) 7H O(l)and

Then, balance hydrogen atoms with hydrogen ions in each half-equation:

22+ +

2 2

2 32 7 2

UO aq + H O(l) UO aq + 2 H aq

Cr O (aq) 14H (aq) 2Cr (aq) 7H O(l)

Step 3: Balance the charge of each half-equation with electrons.

22+ +2 2UO aq + H O(l) UO aq + 2 H aq + 2 e

2 + 3+

2 7 2Cr O aq +14 H aq + 6 e 2 Cr aq + 7 H O(l)

Step 4: Multiply the uranium half-equation by 3 and add the chromium half-equation to it.

22+ +2 2UO aq + H O(l) UO aq + 2 H aq + 2 e 3

2 + 3+2 7 2Cr O aq +14 H aq + 6 e 2 Cr aq + 7 H O(l) 2+ 2- + 2+ 3+ +

2 7 2 2 23 UO (aq)+Cr O (aq)+14 H (aq)+3 H O(l) 3 UO (aq)+2 Cr (aq)+7 H O(l)+6 H (aq)

Step 5: Simplify. Subtract 3 H O2 (l) and 6 H+ (aq) from each side of the equation.

2 22+ + 3+

2 7 2 23 UO aq + Cr O aq +8 H aq 3 UO aq + 2 Cr aq + 4 H O(l)

7A Step 1: Write the two skeleton half-equations.

S(s) SO aq OCl aq Cl aq3 2 ( ) ( ) ( )and

Step 2: Balance each skeleton half-equation for O (withH O2 ) and for H atoms (withH+ ).

2 +

2 3

2

3 H O(l) + S s SO aq + 6 HOCl (aq) 2H Cl (aq) H O(l)


2 +

2 3

2

3 H O(l) + S s SO aq + 6 H (aq) + 4 e

OCl (aq) 2H (aq) 2e Cl (aq) H O(l)


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Step 4: Change from an acidic medium to a basic one by adding OH to eliminate H+ .

2 +2 33H O(l) + S s + 6 OH (aq) SO aq + 6 H (aq) + 6 OH (aq) + 4 e

+ 2OCl aq + 2 H (aq) + 2 OH (aq) + 2 e Cl aq + H O(l) + 2 OH (aq)

Step 5: Simplify by removing the items present on both sides of each half-equation, and

combine the half-equations to obtain the net redox equation.

2

3 2{S s + 6 OH (aq) SO aq + 3 H O(l) + 4 e } 1

2{OCl aq + H O(l) + 2 e Cl aq + 2 OH (aq)} 2

2 -

2 3 2S s + 6 OH (aq) + 2 OCl aq 2H O(l) SO aq + 3 H O(l) + 2 Cl aq + 4OH

Simplify by removing the species present on both sides.

Net ionic equation: 2

3 2S s + 2 OH aq + 2 OCl aq SO aq + H O(l) + 2 Cl aq

7B Step 1: Write the two skeleton half-equations.

2 24 2 3 4MnO aq MnO s SO (aq) SO (aq) and

Step 2: Balance each skeleton half-equation for O (withH O2 ) and for H atoms (withH+ ).

+4 2 22 2

3 2 4

MnO aq + 4 H aq MnO s + 2 H O(l)

SO (aq) H O(l) SO (aq) 2H (aq)


+4 2 2MnO aq + 4 H aq + 3 e MnO s + 2 H O(l)

2 2 +

3 2 4SO aq + H O(l) SO aq + 2 H aq + 2 e

Step 4: Change from an acidic medium to a basic one by adding OH to eliminate +H .

+4 2 2MnO aq + 4 H aq + 4 OH aq + 3 e MnO s + H O(l) + 4 OH aq

2 2 +3 2 4SO aq + H O(l) + 2 OH aq SO aq + 2 H aq + 2OH aq + 2 e

Step 5: Simplify by removing species present on both sides of each half-equation, andcombine the half-equations to obtain the net redox equation.

4 2 2{MnO aq + 2 H O(l) + 3 e MnO s + 4 OH aq } 2

2 2

3 4 2{SO aq + 2 OH aq SO aq + H O(l) + 2 e } 3

2 -

4 3 22MnO aq + 3SO aq + 6 OH (aq) + 4 H O(l)

2

2 4 22 MnO s + 3SO aq + 3H O(l) +8 OH aq

Simplify by removing species present on both sides.

Net ionic equation:

2 24 3 2 2 42 MnO aq + 3SO aq + H O(l) 2 MnO s + 3SO aq + 2 OH aq


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8A Since the oxidation state of H is 0 in H2 (g) and is +1 in both NH3(g) and H2O(g), hydrogenis oxidized. A substance that is oxidized is called a reducing agent. In addition, theoxidation state of N in NO2 (g) is +4 , while it is 3 in NH3 ; the oxidation state of theelement N decreases during this reaction, meaning that NO2 (g) is reduced. The substancethat is reduced is called the oxidizing agent.

8B In 2

Au CN aq

, gold has an oxidation state of +1; Au has been oxidized and, thus,

Au(s) (oxidization state = 0), is the reducing agent. In OH- (aq), oxygen has an oxidationstate of -2; O has been reduced and thus, O2(g) (oxidation state = 0) is the oxidizing agent.

9A We first determine the amount of NaOH that reacts with 0.500 g KHP.

NaOH

1 mol KHP 1 mol OH 1 mol NaOHn = 0.5000 g KHP = 0.002448 mol NaOH

204.22 g KHP 1 mol KHP 1 mol OH

0.002448 mol NaOH 1000 mL[NaOH] = = 0.1019 M

24.03 mL soln 1 L

9B The net ionic equation when solid hydroxides react with a strong acid is OH- + H+ H2O.There are two sources of OH-: NaOH and Ca(OH)2. We compute the amount of OH

- fromeach source and add the results.

moles of OH from NaOH:

92.5 g NaOH 1 mol NaOH 1 mol OH= 0.235 g sample = 0.00543 mol OH

100.0 g sample 39.997 g NaOH 1 mol NaOH

2

2 2

2 2

moles of OH from Ca OH

7.5 g Ca OH 1 mol Ca OH 2 mol OH= 0.235 g sample = 0.00048 mol OH

100.0 g sample 74.093 g Ba OH 1 mol Ca OH

:

-2

total amount OH = 0.00543 mol from NaOH + 0.00048 mol from Ca OH = 0.00591 mol OH

+

+

0.00591 mol OH 1 mol H 1 mol HCl 1000 mL soln[HCl] = = 0.130 M

45.6 mL HCl soln 1 mol OH 1 mol H 1 L soln

10A First, determine the mass of iron that has reacted as Fe2+ with the titrant. The balancedchemical equation provides the essential conversion factor to answer this question.

Namely: 5 Fe aq MnO aq 8 H aq 5 Fe aq Mn aq 4 H O l2+ 43+ 2+

2

2+

4

2+

4

0.02140 mol MnO 5 mol Fe 55.847 g Femass Fe = 0.04125 L titrant = 0.246 g Fe

1 L titrant 1 mol Fe1 mol MnO

Then determine the % Fe in the ore. % Feg Fe

g ore= 65.4% Fe=

0.246

0.376100 %

10B The balanced equation provides us with the stoichiometric coefficients needed for the solution.

Namely: 2- 2+2 4 4 2 25 C O aq 2 MnO aq 16 H aq 10 CO g 2 Mn aq 8 H O l


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2

2 2 4 2 4 4

4 2 2 4 2

2 2 4 2 2 4 2 4

4

1 mol Na C O 1 mol C O 2 mol MnOamount MnO = 0.2482 g Na C O

134.00 g Na C O 1 mol Na C O 5 mol C O

= 0.0007409 mol MnO

4 4

44

0.0007409 mol MnO 1000 mL 1 mol KMnO

[KMnO ] = = 0.03129 M23.68 mL soln 1 L 1 mol MnO

KMnO4

INTEGRATIVE EXAMPLE

A. First, balance the equation. Break down the reaction of chlorate and ferrous ion as follows:

3

3 2

2 3

2 3

2

ClO +6H +6e Cl +3H O

6 Fe Fe

Net reaction: ClO 6Fe 6H Cl 6Fe 3H O

e

The reaction between Fe2+ and Ce4+ is already balanced. To calculate the moles of Fe2+ thatremains after the reaction with ClO3

-, determine the moles of Ce4+ that react with Fe2+:

mol Ce4+ = 0.01259 L 0.08362 M = 1.052710-3 mol = mol of excess Fe2+total mol of Fe2+ = 0.0500 L 0.09101 = 4.55110-3 mol

Therefore, the moles of Fe2+ reacted = 4.55110-3 - 1.052710-3 = 3.49810-3 mol. Todetermine the mass of KClO3, use the mole ratios in the balanced equation in conjunction

with the molar mass of KClO3.3 2 3 3 3

23 3

3

3

1 mol ClO 1 mol KClO 122.54 g KClO3.498 10 mol Fe

6 mol Fe 1 mol ClO 1 mol KClO

= 0.07144 g KClO

0.07144 g%KClO = 100% = 49.89%

0.1432 g

B. First, balance the equation. Break down the reaction of arsenous acid and permanganate asfollows:

- +

3 3 2 3 4

+ - 2+4 2

5 H AsO + H O H AsO + 2e + 2H

2 MnO + 8H 5e Mn + 4H O

Net reaction: + 2+3 3 4 3 4 25H AsO + 2MnO 6H 5H AsO + 2Mn + 3H O

moles of MnO4- = 0.02377 L 0.02144 M = 5.096310-4 mol


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To calculate the mass of As, use the mole ratios in the balanced equation in conjunction withthe molar mass of As:

4 3 34

4 3 3

5 mol H AsO 1 mol As 74.922 g As5.0963 10 mol MnO

2 mol MnO 1 mol H AsO 1 mol As

= 0.095456 g As

0.095456 gmass% As = 100% = 1.32%

7.25 g

EXERCISES

Strong Electrolytes, Weak Electrolytes, and Nonelectrolytes

1. (a) Because its formula begins with hydrogen, HC H O6 5 is an acid. It is not listed in

Table 5-1, so it is a weak acid. A weak acid is a weak electrolyte.

(b) Li SO2 4 is an ionic compound, that is, a salt. A salt is a strong electrolyte.

(c) MgI2 also is a salt, a strong electrolyte.

(d) 3 2 2CH CH O is a covalent compound whose formula does not begin with H.

Thus, it is neither an acid nor a salt. It also is not built around nitrogen,and thus it does not behave as a weak base. This is a nonelectrolyte.

(e) 2

Sr OH is a strong electrolyte, one of the strong bases listed in Table 5-2.

3. HCl is practically 100% dissociated into ions. The apparatus should light up brightly. Asolution of both HCl and HC2H3O2 will yield similar results. In strongly acidic solutions,the weak acid HC2H3O2 is molecular and does not contribute to the conductivity of thesolution. However, the strong acid HCl is practically dissociated into ions and is unaffectedby the presence of the weak acid HC2H3O2. The apparatus should light up brightly.

5. (a) Barium bromide: strong electrolyte (b) Propionic acid: weak electrolyte(c) Ammonia: weak electrolyte

Ion Concentrations

7. (a) K

mol KNO

L soln

mol K

mol KNOM K+ 3

+

3

+=0.238

1

1

1= 0.238

(b)

3 32

3 3

3 2

0.167 mol Ca NO 2 mol NONO = = 0.334 M NO

1 L soln 1 mol Ca NO


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(c)

3+2 43+ 3+3

2 4 3

0.083 mol Al SO 2 mol AlAl = = 0.166 M Al

1 L soln 1 mol Al SO

(d) Namol Na PO

L soln

mol Na

mol Na POM Na+ 3 4

+

3 4

+=0.209

1

3

1= 0.627

9.

Conversion pathway approach:

2 22 2

2 22 2

0.132 g Ba OH 8H O 1 mol Ba OH 8H O1000 mL 2 mol OHOH =

275 mL soln 1 L 315.5g Ba OH 8H O 1 mol Ba OH 8 H O

= 3.04 10 3 M OH

Stepwise approach:

22

22

22

22

22

2 -3 -2

0.132 g Ba OH 8H O 1000 mL

275 mL soln 1 L

1 mol Ba OH 8H O

315.5g Ba OH 8H O

2 mol OH

1 mol Ba OH 8 H O

= 0.480 g/L

0.00152 mol Ba OH 8H O0.480 g=

L L

0.00152 mol Ba OH 8H O= 3.04 10 M OH

L

11. (a)2+ 2+ 2+

2+ 4 2+

2+ 2+

14.2 mg Ca 1 g Ca 1 mol Ca[Ca ] = 3.54 10 M Ca

1 L solution 1000 mg Ca 40.078 g Ca

(b)+ + +

+ 3 +

+ +

32.8 mg K 1 g K 1000 mL solution 1 mol K [K ] = 8.39 10 M K

100 mL solution 1 L solution1000 mg K 39.0983 g K

(c)2+ 2+ 2+

2+ 3 2+

6 2+ 2+

225 g Zn 1 g Zn 1000 mL solution 1 mol Zn[Zn ] = 3.44 10 M Zn

1 mL solution 1 L solution1 10 g Zn 65.39 g Zn

13 In order to determine the solution with the largest concentration of K+, we begin byconverting each concentration to a common concentration unit, namely, molarity of K+.

++2 4

2 4

0.0850 M K SO 2 mol K0.17 M K

1 L solution 1 mol K SO

++1000 mL solution1.25 g KBr 1 mol KBr 1 mol K 0.105 M K

100 mL solution 1 L solution 119.0023 g KBr 1 mol KBr

+ + ++

+ +

1000 mL solution8.1 mg K 1 g K 1 mol K 0.207 M K

1 mL solution 1 L solution 1000 mg K 39.0983 g K

Clearly, the solution containing 8.1 mg K+ per mL gives the largest K+ of the three solutions.


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15. Determine the amount of I in the solution as it now exists, and the amount of I in thesolution of the desired concentration. The difference in these two amounts is the amount of

I that must be added. Convert this amount to a mass of MgI2 in grams.

1 L 0.1000 mol Imoles of I in final solution = 250.0 mL = 0.02500 mol I

1000 mL 1 L soln

1 L 0.0876 mol KI 1 mol Imoles of I in KI solution = 250.0 mL = 0.0219 mol I

1000 mL 1 L soln 1 mol KI

2 222

1 mol MgI 278.11 g MgI 1000 mgmass MgI required = 0.02500 0.0219 mol I

2 mol I 1 mol MgI 1 g

= 4.3 102 2 mg MgI

17. moles of chloride ion

2

2

0.625 mol KCl 1 mol Cl 0.385 mol MgCl 2 mol Cl= 0.225 L + 0.615 L

1 L soln 1 mol KCl 1 L soln 1 mol MgCl

= 0.141 + 0.474mol Cl mol Cl = 0.615 mol Cl-0.615 mol Cl

Cl = = 0.732 M0.225 L + 0.615 L

Predicting Precipitation Reactions

19. In each case, each available cation is paired with the available anions, one at a time, todetermine if a compound is produced that is insoluble, based on the solubility rules ofChapter 5. Then a net ionic equation is written to summarize this information.

(a) 2+

2Pb aq + 2 Br aq PbBr s

(b) No reaction occurs (all are spectator ions).

(c) 3+3

Fe aq + 3 OH aq Fe OH s

21.

Mixture Result (Net Ionic Equation)

(a) 3 2HI a + Zn NO (aq): No reaction occurs.

(b) 22+

4 2 3 3 3CuSO aq + Na CO aq : Cu aq + CO aq CuCO s

(c) 3

2+3 3 4 4 3 42 2Cu NO aq + Na PO aq : 3Cu aq + 2PO aq Cu PO s

23. (a) Add 2 4 4K SO aq ; BaSO s will form and MgSO4 will not precipitate.

2 2 4 4BaCl s + K SO aq BaSO s + 2 KCl aq


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(b) Add 2 2 3H O l ; Na CO s dissolves, but MgCO3 (s) will not dissolve (appreciably).

2water +

2 3 3 Na CO s 2 Na aq + CO aq

(c) Add KCl(aq); AgCl(s) will form, while Cu(NO3)2 (s) will dissolve.

3 3AgNO s + KCl aq AgCl s + KNO aq

25.

Mixture Net Ionic Equation

(a) 22+

3 2 4 4 42Sr NO aq + K SO aq : Sr aq +SO aq SrSO s

(b) 2+3 2 2Mg NO aq + NaOH aq : Mg aq + 2 OH aq Mg OH s

(c) 22+

2 2 4 4 4BaCl aq + K SO aq : Ba (aq) SO (aq) BaSO (s)

(upon filtering, KCl (aq) is obtained)

AcidBase Reactions

27. The type of reaction is given first, followed by the net ionic equation.

(a) Neutralization: 2 3 2 2 2 3 2OH aq + HC H O aq H O l + C H O aq

(b) No reaction occurs. This is the physical mixing of two acids.

(c) Gas evolution: + 2+2FeS s + 2 H aq H S g + Fe aq

(d) Gas evolution: +

3 2 3 2 2HCO aq + H aq "H CO aq " H O l + CO g

(e) Redox: + 2+ 2Mg s + 2 H aq Mg aq + H g

29. As a salt: +4 4NaHSO aq Na aq + HSO aq

As an acid: 2

4 2 4HSO aq + OH aq H O l + SO aq

31. Use (b) NH3(aq): NH3 affords the OH- ions necessary to form Mg(OH)2(s).

Applicable reactions: {NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)} 2

MgCl2(aq) Mg2+(aq) + 2 Cl-(aq)

Mg2+(aq) + 2 OH-(aq) Mg(OH)2(s)


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OxidationReduction (Redox) Equations

33. (a) The O.S. of H is +1, that of O is 2 , that of C is +4 , and that of Mg is +2 on eachside of this equation. This is not a redox equation.

(b) The O.S. of Cl is 0 on the left and 1 on the right side of this equation. The O.S. ofBr is 1 on the left and 0 on the right side of this equation. This is a redox reaction.

(c) The O.S. of Ag is 0 on the left and +1 on the right side of this equation. The O.S. ofN is +5 on the left and +4 on the right side of this equation. This is a redox reaction.

(d) On both sides of the equation the O.S. of O is 2 , that of Ag is +1, and that of Cr is+6 . Thus, this is not a redox equation.

35. (a) Reduction: 2 2+

3 2 3 22SO aq + 6 H aq + 4 e S O aq + 3 H O(l)

(b) Reduction: +3 2 22 NO aq +10 H aq +8 e N O g +5 H O l

(c) Oxidation: 4

Al s + 4 OH aq Al OH aq + 3 e

37. (a) Oxidation: { 22 I aq I s + 2 e } 5

Reduction: { + 2+

4 2MnO aq + 8 H aq + 5 e Mn aq + 4 H O l

} 2

Net: + 2+4 2 210 I aq + 2 MnO aq +16 H aq 5 I s + 2 Mn aq + 8 H O l

(b) Oxidation: { +2 4 2N H l N g + 4 H aq + 4 e } 3

Reduction: {

+

3 2

BrO aq + 6 H aq + 6 e Br aq + 3 H O l

} 2Net: 2 4 3 2 23 N H l + 2 BrO aq 3 N g + 2 Br aq + 6 H O l

(c) Oxidation: 2+ 3+Fe aq Fe aq + e

3 + 2+

4 2Reduction: VO aq + 6 H aq + e VO aq + 3 H O l

Net: 32+ + 3+ 2+

4 2Fe aq + VO aq + 6 H aq Fe aq + VO aq + 3 H O l

(d) Oxidation: { 22+ +

2 2UO aq + H O l UO aq + 2 H aq + 2 e } 3

Reduction: { +

3 2NO aq + 4 H aq + 3 e NO g + 2 H O l

} 2Net:

22+ +

3 2 23 UO aq + 2 NO aq + 2 H aq 3 UO aq + 2 NO g + H O l

39. (a) Oxidation: { 2 4 2MnO s + 4 OH aq MnO aq + 2 H O(l) + 3 e } 2

Reduction: 3 2ClO aq + 3 H O(l) + 6 e Cl aq + 6 OH aq

Net: 2 3 4 22 MnO s + ClO aq + 2 OH aq 2MnO aq + Cl aq + H O(l)


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(b) Oxidation: { 2

4 23Fe OH s + 5 OH aq FeO aq + 4 H O(l) + 3 e

} 2

Reduction: { -2OCl aq + H O(l) + 2 e Cl aq + 2OH aq } 3

Net: 2

4 232 Fe OH s + 3 OCl aq + 4 OH aq 2FeO aq + 3 Cl aq + 5 H O(l)

(c) Oxidation: { 2 3 2ClO (aq) + 2 OH aq ClO aq + H O l + e } 5

Reduction: 2 2ClO (aq) + 2 H O l + 5 e Cl (aq) + 4 OH aq

Net: 2 3 26 ClO (aq) + 6 OH aq 5ClO aq + Cl aq + 3 H O (l)

(d) Oxidation: (Ag (s) Ag+ (aq) + 1 e ) 3Reduction: 4 H2O(l) + CrO4

2- + 3 e Cr(OH)3(s) + 5 OH-

Net: 3 Ag(s) + CrO42- + 4 H2O(l) 3 Ag+(aq) + Cr(OH)3(s) + 5 OH

-

41. (a) Oxidation: 2 3 2Cl g +12 OH aq 2 ClO aq + 6 H O(l) +10 e

Reduction: { 2Cl g + 2 e 2 Cl aq

} 5

Net: 2 3 26 Cl g +12 OH aq 10 Cl aq + 2 ClO aq + 6 H O(l)

Or: 2 3 23 Cl g + 6 OH aq 5 Cl aq + ClO aq + 3 H O(l)

(b) Oxidation: 2 +

2 4 2 3S O aq + 2 H O(l) 2 HSO aq + 2 H aq + 2 e

Reduction: 2 2+

2 4 2 3 2S O aq + 2 H aq + 2 e S O aq + H O (l)

Net: 2 2

2 4 2 3 2 32 S O aq + H O(l) 2 HSO aq + S O aq

43. (a) Oxidation: { +2 2 3NO aq + H O l NO aq + 2 H aq + 2 e } 5

Reduction: { + 2+

4 2MnO aq +8 H aq + 5 e Mn aq + 4 H O l } 2

Net: + 2+2 4 3 25 NO aq + 2 MnO aq + 6 H aq 5 NO aq + 2 Mn aq + 3 H O l

(b) Oxidation: { Mn2+ (aq) + 4 OH- (aq) MnO2 (s) + 2 H2O (l) + 2 e- } 3

Reduction: { MnO4- (aq) + 2 H2O (l) + 3 e

- MnO2 (s) + 4 OH- (aq) } 2

Net: 3 Mn2+ (aq) + 2 MnO4- (aq) + 4 OH- (aq) 5 MnO2 (s) + 2 H2O (l)

(c) Oxidation: { +2 5 3C H OH CH CHO + 2 H aq + 2 e } 3

Reduction:

2 + 3+2 7 2Cr O aq +14 H aq + 6 e 2 Cr aq + 7 H O l

Net: 2 + 3+2 7 2 5 2 3Cr O aq + 8 H aq + 3 C H OH 2 Cr aq + 7 H O l + 3 CH CHO


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109

45. For the purpose of balancing its redox equation, each of the reactions is treated as if it takesplace in acidic aqueous solution.

(a) 2 H2O(g) + CH4(g) CO2(g) + 8 H+(g) + 8 e-

{2 e- + 2 H+(g) + NO(g) N2(g) + H2O(g) }4

CH4(g) + 4 NO(g) 2 N2(g) + CO2(g) + 2 H2O(g)

(b) {H2S(g) 1/8 S8(s)+ 2 H+(g) + 2 e- }2

4 e- + 4 H+(g) + SO2(g) 1/8 S8(s) + 2 H2O(g)2 H2S(g) + SO2(g) 3/8 S8(s) + 2 H2O(g) or16 H2S(g) + 8 SO2(g) 3 S8(s) + 16 H2O(g)

(c) {Cl2O(g) + 2 NH4+(aq) + 2 H+(aq) + 4 e- 2 NH4Cl(s) + H2O(l) } 3

{2 NH3(g) N2(g) + 6 e- + 6 H+(aq) }2

6 NH3(g) + 6 H+(aq) 6 NH4

+(aq)

10 NH3(g) +3 Cl

2O(g) 6 NH

4Cl(s) + 2N

2(g)+ 3 H

2O(l)

Oxidizing and Reducing Agents

47. The oxidizing agents experience a decrease in the oxidation state of one of their elements, whilethe reducing agents experience an increase in the oxidation state of one of their elements.

(a) 2

3SO aq

is the reducing agent; the O.S. of S = +4 in SO and32

= +6

in SO42

.

4MnO aq

is the oxidizing agent; the O.S. of Mn = +7 in 4MnO and + 2

inMn2+ .

(b) 2H g is the reducing agent; the O.S. of H = 0 in 2H g and = +1 in 2H O g .

2NO g is the oxidizing agent; the O.S. of N = +4 in 2 NO g and 3 in NH3 (g).

(c) 4

6Fe CN aq

is the reducing agent; the O.S. of Fe = +2 in

4

6Fe CN

and = +3 in 3

2 26Fe CN . H O aq

is the oxidizing agent; the O.S. of O = 1

in H O and2 2 = 2 in H O2 .

Neutralization and AcidBase Titrations

49. The problem is most easily solved with amounts in millimoles.

+

NaOH +

0.128 mmol HCl 1 mmol H 1 mmol OHV = 10.00 mL HCl aq

1 mL HCl aq 1 mmol HCl 1 mmol H


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110

1 mL NaOH aq1 mmol NaOH= 13.3 mL NaOH aq soln

0.0962 mmol NaOH1 mmol OH

51. The net reaction is 3 5 2 2 3 5 2OH aq + HC H O aq H O(l) + C H O aq .


3 5 2

3 5 2

base

0.3057 mmol HC H O 1 mmol KOH 1 mL baseV = 25.00 mL acid

1 mL acid 1 mmol HC H O 2.155 mmol KOH

= 3.546 mL KOH solution

Stepwise approach:

3 5 2

3 5 2

3 5 2

3 5 2

0.3057 mmol HC H O25.00 mL acid

1 mL acid

1 mmol KOH

1 mmol HC H O1 mL base

= 3.546 mL KOH solution2.155 mmol KOH

= 7.643 mmol HC H O

7.643 mmol HC H O = 7.643 mmol KOH

7.643 mmol KOH

53. 2NaOH aq + HCl aq NaCl aq + H O(l) is the titration reaction.

0.1085molHCl 1mol NaOH0.02834L

1Lsoln 1 mol HCl[NaOH] = = 0.1230 M NaOH

0.02500 L sample

55. The mass of acetylsalicylic acid is converted to the amount of NaOH, in millimoles, thatwill react with it.

9 7 4 9 7 4

9 7 4 9 7 4

0.32 g HC H O 1 mol HC H O 1 mol NaOH 1000 mmol NaOHNaOH =

23 mL NaOH aq 180.2 g HC H O 1 mol HC H O 1 mol NaOH

= 0.077 M NaOH

57. The equation for the reaction is 3 3 2HNO aq + KOH aq KNO aq + H O 1 .

This equation shows that equal numbers of moles are needed for a complete reaction.We compute the amount of each reactant.

mmol HNO mL acidmmol HNO

mL acidmmol HNO3

33= 25.00

0.132

1= 3.30

mmol KOH mL acidmmol KOH

mL basemmol KOH= 10.00

0.318

1= 3.18

There is more acid present than base. Thus, the resulting solution is acidic.


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111

59. 2 3 2 2 3 2base2 3 2

4.0 g HC H O 1 mol HC H O1.01 g vinegarV = 5.00 mL vinegar

1 mL 100.0 g vinegar 60.0 g HC H O

1

1

1

0.1000

1000

1= 34

2 3 2

mol NaOH

mol HC H O

L base

mol NaOH

mL

LmL base

61. Answer is (d): 120 % of necessary titrant added in titration of NH3

5 NH3+

5 HCl+

1 HCl

required forequivalence

point

20 % excess

5 NH4+ + 6 Cl- + H3O

+

(depicted in question's drawing )

Stoichiometry of OxidationReduction Reactions

63.


2 3 4 4

2 3

2 3 2 3 4

4

1mol As O 4 mol MnO 1mol KMnO0.1078 g As O

197.84 g As O 5 mol As O 1molMnO[ MnO ]=

1L22.15mL

1000mL

4

= 0.01968 M KMnO

Stepwise approach:

4

-4 -44

4

4

-42 32 3 2 3

2 3

-4 -442 3 4

2 3

4

[KMnO ]=

1molKMnO10 mol MnO 10

1molMnO

mol KMnO

L solution

1molAs O0.1078 g As O = 5.449 10 mol As O

197.84g As O

4molMnO5.449 10 mol As O = 4.359 10 mol MnO

5molAs O

4.359 = 4.359

-4

-24 4

4

4mol KMnO

1L22.15 mL 0.02215 L solution

1000mL

mol KMnO 4.359 10 mol KMnO[ KMnO ]= = = 1.968 10 M

L solution 0.02215 L solution


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112

65. First, we will determine the mass of Fe, then the percentage of iron in the ore.2 2+

2 72 2+

2 7

0.05051 mol Cr O1 L 6 mol Fe 55.85 g Femass Fe = 28.72 mL

1000 mL 1 L soln 1 mol Fe1 mol Cr O

mass Fe = 0.4861 g Fe % .Fe =0.4861g Fe

0.9132gore

Fe 100% 53 23%

67. First, balance the titration equation:

Oxidation: { 2

2 4 2C O aq 2 CO g + 2 e } 5

Reduction: { + 2+4 2MnO aq + 8 H aq + 5 e Mn aq + 4 H O l } 2

Net: 2 + 2+2 4 4 2 25 C O aq + 2 MnO aq +16 H aq 10 CO g + 2 Mn aq +8 H O l

2 2 4

4 4Na C O 2 2 4

2 2 4 4

24 2 4 2 2 4

24 4 2 4

25.8 mL satd soln KMnO 0.02140 mol KMnO1000 mLmass =1.00 L satd soln Na C O

1 L 5.00 mL satd soln Na C O 1000 mL KMnO

1 molMnO 5 molC O 1 mol Na C O 1341 molKMnO 2 molMnO 1 molC O

2 2 4

2 2 4

2 2 4

Na C O 2 2 4

.0 g Na C O1 mol Na C O

mass = 37.0 g Na C O

Integrative and Advanced Exercises

71. A possible product, based on solubility rules, is 243 )(POCa . We determine the % Ca in this

compound.

3 4 2

molar mass 3 40.078 g Ca 2 30.974 g P 8 15.999 g O

120.23 g Ca 61.948 g P 127.99 g O 310.17 g

120.23 g Ca% Ca 100% 38.763%

310.17 g Ca (PO )

Thus, 243 )(POCa is the predicted product. The net ionic equation follows.

(aq)H2(s))PO(Ca(aq)HPO2(aq)Ca3 2432

42

74. Let us first determine the mass of Mg in the sample analyzed.Conversion pathway approach:

2 2 72 2 7

2 2 7 2 2 7

1 mol Mg P O 2 mol Mg 24.305 gmass Mg 0.0549 g Mg P O 0.0120 g Mg

222.55 g Mg P O 1 mol Mg P O 1 mol Mg

6 0.0120 g Mg ppm Mg 10 g sample 108 ppm Mg110.520 g sample


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113

Stepwise approach:

-42 2 72 2 7 2 2 7

2 2 7

-4 -42 2 7

2 2 7

-4

1 mol Mg P O0.0549 g Mg P O = 2.47 10 mol Mg P O

222.55 g Mg P O

2 mol Mg2.47 10 mol Mg P O 4.93 10 mol Mg

1 mol Mg P O24.305 g

4.93 10 mol Mg 0.0120 g Mg1 mol Mg

6 0.0120 g Mg ppm Mg 10 g sample 108 ppm Mg110.520 g sample

75. Let Vrepresent the volume of added 0.248 M CaCl2 that must be added.

We know that [Cl ] = 0.250 M, but also,

V

V

L0.335

CaClmol1

Clmol2

solnL1

CaClmol248.0

KClmol1

Clmol1

solnL1

KClmol186.0L335.0

]Cl[ 2

2

L0874.0250.0496.0

0623.00838.0496.00623.00.2500.0838)335.0(250.0

VVVV

80. (a) [FeS2 + 8 H2O Fe3+ + 2 SO4

2 + 16 H+ + 15 e] 4[O2 + 4 H

+ + 4 e 2 H2O] 15

overall: 4 FeS2(s) + 15 O2(g) + 2 H2O(l) 4 Fe3+(aq) + 8 SO42

(aq) + 4 H+(aq)

(b) One kilogram of tailings contains 0.03 kg (30 g) of S. We have

moles of FeS2 =2

2

1 mol FeS1molS30 g S 0.468 mol FeS

32.07 g S 2 mol S

moles of H+ =+

+2

2

4 mol H0.468 mol FeS 0.467 mol H

4 mol FeS

moles of CaCO3 =+ 3

3+

1 mol CaCO0.467 mol H 0.234 mol CaCO

2 mol H

mass of CaCO3 =3

3 3

3

100.09 g CaCO0.234 mol CaCO 23.4 g CaCO

1 mol CaCO


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114

83. 3}e2(g)Cl(aq)Cl{2:Oxidation 2

OH7(aq)Cr2e6(aq)H14(aq)OCr:Reduction 232

72

(g)Cl3OH7(aq)Cr2(aq)H14(aq)OCr(aq)Cl6:Net 2232

72

We need to determine the amount of Cl2(g) produced from each of the reactants. The limitingreactant is the one that produces the lesser amount of Cl2..

reactantlimitingthefromproducedamountthe,Clmol629.0

OCrmol1

Clmol3

OCrKmol1

OCrmol1

OCrKg2.294

OCrKmol1

sampleg.100

OCrKg98.5g6.62Clamount

Clmol54.1

Clmol6

Clmol3

HClmol1

Clmol1

HClg46.36

HClmol1

solng.100

HClg1.30

mL1

g1.15mL325Clamount

2

2

72

2

722

2

72

722

7227222

2

22

Then we determine the mass of Cl2(g) produced.2

2 2

2

70.91 g Cl= 0.629 mol Cl = 44.6 g Cl

1 mol Cl

85. balanced)(not(g)ClONaCl(aq)(aq)NaClO(g)Cl 222

(g)ClO2NaCl(aq)2(aq)NaClO2(g)Cl 222

2 2 22

2 2

222

2

2.0 mol NaClO 2 mol ClO 67.45 g ClO3.785 Lamount ClO 1 gal

1 gal 1 L soln 2 mol NaClO 1 mol ClO

97 g ClO produced5.0 10 g ClO (g)

100 g ClO calculated

88. (a) First, balance the redox equations needed for the calculation.

Oxidation: {HSO3- (aq)+ H2O(l) SO4

2- (aq) + 3 H+ (aq) + 2 e- } 3

Reduction: {IO3- (aq) + 6 H+ (aq) + 6 e- I-(aq) + 3 H2O(l) } 1

Net: 3 HSO3- (aq) + IO3

- (aq) 3 SO42- (aq) + 3 H+ (aq) + I- (aq)

The solution volume of 5.00 L contains 29.0 g NaIO3. This represents

29.0 g/197.9g/mol NaIO3 = 0.147 mol NaIO3.

(b) From the above equation, we need 3 times that molar amount of NaHSO3, which is3(0.147 mol) = 0.441 mol NaHSO3; the molar mass of NaHSO3 is 104.06 g/mol.

The required mass then is 0.441(104.06) = 45.9 g.


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115

For the second process:

Oxidation: {2 I-(aq) I2(aq) + 2 e- } 5

Reduction: {2 IO3- (aq) + 12 H+ (aq) + 10 e- I2(aq) + 6 H2O(l) } 1

Net: 5 I

-

(aq) + IO3-

(aq) + 6 H

+

(aq) 3 I2(aq) + 3 H2O(l)

In Step 1, we produced 1 mol of I- for every mole of IO3- reactant; therefore we had

0.147 mol I-.

In step 2, we require 1/5 mol IO3- for every mol of I-.

We require only 1.00 L of the solution in the question instead of the 5.00 L in the firststep.

89. Mg(OH)2(aq) + 2 HCl(aq) MgCl2(aq) + 2 H2O(l) (1)

Al(OH)3(aq) + 3 HCl(aq) AlCl3(aq) + 3 H2O(l) (2)

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) (3)

initial moles of HCl =

0.500 mol0.0500 L = 0.0250 mol

1 L

moles of HCl that reacted with NaOH =

moles of HCl left over from reaction with active ingredients =

-30.377 mol NaOH 1 mol HCl0.0165 L = 6.22 10 mol

1 L 1 mol NaOH

moles of HCl that react with active ingredients =

-30.0250 mol - 6.22 10 mol = 0.0188 mol

32

# moles HCl that# moles HCl that+ = total moles of HCl reacted/used

react with Al(OH)react with Mg(OH)

# moles HCl that react with Mg(OH)2 =

22

2 2

1 mol Mg(OH) 2 mol HClX grams Mg(OH)58.32 g Mg(OH) 1 mol Mg(OH)

# moles HCl that react with Al(OH)3 =

33

3 3

1 mol Al(OH) 3 mol HCl0.500-X grams Al(OH)

78.00 g Al(OH) 1 mol Al(OH)


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116

2X 3(0.500 X)0.0188

58.32 78.00

X = 0.108, therefore the mass of Mg(OH)2 in the sample is 0.108 grams.

% Mg(OH)2 = (0.108/0.500) 100 = 21.6 %Al = 100 %Mg(OH)2 = 78.4

91.

3 19 16 4 19 16 4

3 19 16 4

19 16 4

1 mol CHI 1 mol C H O 308.33 g C H O1 mol AgI0.1386 g AgI

234.77 g AgI 3 mol AgI 1 mol CHI 1 mol C H O

0.06068 g C H O

% 19 16 4C H O =0.06068 g

100 0.4346 %13.96 g

93. (a) CaO(s) + H2O(l) Ca2+(aq) + 2 OH(aq)

H2PO4(aq) + 2 OH(aq) PO4

3(aq) + 2 H2O(l)

HPO4(aq) + OH(aq) PO4

3(aq) + H2O(l)

5 Ca2+(aq) + 3 PO43(aq) + OH(aq) Ca5(PO4)3OH(s)

(b)

33 24 4

34

2

1 mol PO10.0 10 g P 1 mol P 5 mol Ca

1.00 10 L L 30.97 g P 1 mol P 3 mol PO

1mol CaO 56.08 g CaO= 301.80 g CaO = 302 g = 0.302 kg

1 mol Ca 1 mol CaO

FEATURE PROBLEMS

94. From the volume of titrant, we can calculate both the amount in moles of NaC H5 5 and

(through its molar mass of 88.08 g/mol) the mass of NaC H5 5 in a sample. The remaining massin a sample is that of C H O4 8 (72.11 g/mol), whose amount in moles we calculate. The ratio of

the molar amount of C H O4 8 in the sample to the molar amount of NaC H5 5 is the value ofx.


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117


5 55 5

5 5

1 mol NaC H0.1001 mol HCl 1 mol NaOHmoles of NaC H = 0.01492 L

1 L soln 1 mol HCl 1 mol NaOH

= 0.001493 mol NaC H

5 54 8 5 5

5 5

4 8

88.08 g NaC Hmass of C H O = 0.242 g sample 0.001493 mol NaC H 1 mol NaC H

= 0.111 g C H O

x

4 84 8

4 8

5 5

1mol C H O0.110gC H O

72.11gC H O= = 1.03

0.001493 mol NaC H

Stepwise approach:

-3

-3 -3

-3 -35 55 5

-3

0.1001 mol HCl0.01492 L = 1.493 10 mol HCl1 L soln

1 mol NaOH1.493 10 mol HCl = 1.493 10 mol NaOH

1 mol HCl

1 mol NaC H1.493 10 mol NaOH 1.493 10 mol NaC H

1 mol NaOH

1.493 10 mol N

5 55 5 5 55 5

88.08 g NaC HaC H 0.1315 g NaC H

1 mol NaC H

4 8 5 5 4 8mass of C H O = 0.242 g sample 0.1315 g NaC H = 0.111 g C H O

-34 84 8 4 8

4 8

-34 8

5 5

1mol C H O0.111g C H O = 1.54 10 mol C H O

72.11gC H O

1.54 10 mol C H O= 1.03

0.001493 mol NaC H

For the second sample, parallel calculations give 0.001200 mol NaC H5 5 , 0.093 g C H4 8 ,

x = 1.1. There is rounding error in this second calculation because it is limited to twosignificant figures. The best answer is from the first runx ~1.03 or 1. The formula isNaC5H5(THF)1.

95. First, we balance the two equations.

Oxidation: +2 2 4 2H C O aq 2 CO g + 2 H aq + 2 e

Reduction: + 2+2 2MnO s + 4 H aq + 2 e Mn aq + 2 H O l

Net: + 2+2 2 4 2 2 2H C O aq + MnO s + 2 H aq 2 CO g + Mn aq + 2 H O l


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118

Oxidation: { +2 2 4 2H C O aq 2 CO g + 2 H aq + 2 e } 5

Reduction: { + 2+4 2MnO aq + 8 H aq + 5 e Mn aq + 4 H O l } 2

Net: + 2+2 2 4 4 2 25 H C O aq + 2 MnO aq + 6 H aq 10 CO g + 2 Mn aq +8 H O l

Next, we determine the mass of the excess oxalic acid.

4 4 2 2 42 2 4 2

4 4

0.1000mol KMnO 1mol MnO 5mol H C Omass H C O 2H O 0.03006 L

1L 1mol KMnO 2 mol MnO

2 2 4 2 2 2 4 2

2 2 4 2 2 4 2

1 mol H C O 2H O 126.07 g H C O 2H O

1 mol H C O 1 mol H C O 2H O

2 2 4 2= 0.9474 g H C O 2H O

The mass of H C O H O2 2 4 22 that reacted with MnO2 2 2 4 2=1.651 g 0.9474 g = 0.704 g H C O 2H O

2 2 4 2 2

2 2 2 4 2

2 2 4 2 2 2 4 2

2

1 mol H C O 1 mol MnO 86.9 g MnOmass MnO = 0.704 g H C O 2H O

126.07 g H C O 2H O 1 mol H C O 1 mol MnO

= 0.485 g MnO

% MnO0.485g MnO

0.533gsample100% 91.0% MnO2

22

97. The molecular formula for CH3CH2OH is C2H6O and for CH3COOH is C2H4O2.

The first step is to balance the oxidationreduction reaction.

Oxidation: [C2H6O + H2O C2H4O2 + 4 H+ + 4 e] 3

Reduction: [Cr2O72 + 14 H+ + 6e 2 Cr3+ + 7 H2O] 2

Overall: 3 C2H

6O + 2 Cr

2O

7

2 + 16 H+ 3 C2H

4O

2+ 4 Cr3+ + 11 H

2O

Before the breath test:

-42 2 70.75 mg K Cr O 1 g 1 mol 1000 mL = 8.498 10 M3 mL 1000 mg 294.19 g 1 L

= 810-4 M (to 1 sig fig)

For the breath sample:

BrAC = 2 60.05 g C H O 1 mL blood

100 mL blood 2100 mL breath =

72 62.3810 g C H O

mL breath

mass C2H6O =7

2 62.3810 g C H O

mL breath

500. mL breath = 1.19 104 g C2H6O


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119

Calculate the amount of K2Cr2O7 that reacts:

24 2 6 2 7 2 2 7

2 6 22 6 2 6 2 7

62 2 7

1 mol C H O 2 mol Cr O 1 mol K Cr O1.19 10 g C H O

46.068 g C H O 3 mol C H O 1mol Cr O

= 1.72 10 mol K Cr O

# mol K2Cr2O7 remaining = moles K2Cr2O7 before moles K2Cr2O7 that reacts

moles K2Cr2O7 before =-6

2 2 7

1 g 1 mol0.75 mg K Cr O = 2.5 10 mol

1000 mg 294.19 g

# mol K2Cr2O7 remaining = 2.5 106 mol 1.72 106 mol = 0.78 106 mol

concentration of K2Cr2O7 after the

breath test = 0.78 106 mol/0.003 L = 2.6 104 mol/L = 3 104 mol/L (to 1 sig fig)

102. The answer is (b).Conversion pathway approach:

2

2

0.0050 mol Ba(OH) 2 mol OH0.300 L = 0.0030 mol

1 L 1mol Ba(OH)

Stepwise approach:

-322

-32

2

0.0050 mol Ba(OH)0.300 L = 1.5 10 mol Ba(OH)

1 L

2 mol OH1.5 10 mol Ba(OH) = 0.0030 mol

1mol Ba(OH)

103. The answer is (d), because H2SO4 is a strong diprotic acid and theoretically yields0.20 mol of H+ for every 0.10 mol of H2SO4.

104. The answer is (c). Based on the solubility guidelines in Table 5-1, carbonates (CO32-) are

insoluble.

105. The answer is (a). Reaction with ZnO gives ZnCl2 (soluble) and H2O. There is no reactionwith NaBr and Na2SO4, since all species are aqueous. By the process of elimination, (a) isthe answer.

106.

3 2 3 2

- 22

Balanced equation: 2 KI + Pb(NO ) 2KNO + PbI

Net ionic equation: 2I + Pb PbI (s)


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120

107.

3

2 3 2 2

22 2

Balanced equation: Na CO + 2HCl 2NaCl + H O + CO

Net ionic equation: CO + 2H H O (l) + CO (g)

108.

(a)

4

3 4 3 2 3 3 4 2

2+ 3

3 4 2

Balanced equation: 2 Na PO + 3 Zn(NO ) 6NaNO + Zn (PO )

Net ionic equation: 3 Zn + PO Zn (PO ) (s)

(b)3 2 2 3

2+2

Balanced equation: 2 NaOH + Cu(NO ) Cu(OH) + 2 NaNO

Net ionic equation: Cu + 2 OH Cu(OH) (s)

(c)2 2 3 3

2+ 23 3

Balanced equation: NiCl + Na CO NiCO + 2 NaCl

Net ionic equation: Ni + CO NiCO (s)

109. (a) Species oxidized: N in NO(b) Species reduced: O2(c) Oxidizing agent: O2(d) Reducing agent: NO(e) Gains electrons: O2(f) Loses electrons: NO

110. The answer is (b). The charges need to be balanced on both sides. Using a coefficient of 4,the charges on both sides of the reaction becomes +12.

111. The answer is (d), 5 ClO- to 1 I2. The work to balance the half-reactions is shown below:

+2

2 2 3

Reduction: 5ClO + 2H + 2e Cl + H O

Oxidation: I + 6H O 2IO + 10e + 12H

To combine the above reactions, the oxidation reaction should be multiplied by 5. Thecombined equation is:

2 2 3Combined: 5ClO + I + H O 5Cl + 2IO + 2H

112. The answer is (a). The balanced half-reaction is as follows:+ - 4

2 2NpO + 4 H + e Np + 2H O

113. (a) False. Based on solubility rules, BaCl2 dissolves well in water. Therefore, it is astrong electrolyte.(b) True. Since H- is a base, H2O is by necessity an acid. It also reduces H

- (-1) to H2 (0).(c) False. The product of such a reaction would be NaCl and H2CO3, neither of which

precipitates out.(d) False. HF is among the strongest of weak acids. It is not a strong acid, because it

doesnt completely dissociate.


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(e) True. For every mole of Mg(NO3)2, there are 3 moles of ions, in contrast to 2 moles ofions for NaNO3.

114. (a) No. Oxidation states of C, H or O do not change throughout the reaction.(b) Yes. Li is oxidized to Li+ and H in H2O is reduced from +1 to 0 in H2.

(c) Yes. Ag is oxidized and Pt is reduced.(d) No. Oxidation states of Cl, Ca, H, and O remain unchanged.

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