Petrucci Chap 21 Answers

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CHAPTER 21 CHEMISTRY OF THE MAIN-GROUP ELEMENTS I:

GROUPS 1, 2, 13, AND 14

PRACTICE EXAMPLES 1A (E) From Figure 21-2, the route from sodium chloride to sodium nitrate begins with

electrolysis of NaCl(aq) to form NaOH(aq). 2NaCl aq + 2 H2O l electrolysis 2 NaOH aq + H2 g + Cl2 g followed by addition of NO2 g to NaOH(aq). 2 3 22 NaOH aq + 3 NO g 2 NaNO aq + NO g + H O(l)

1B (E) From Figure 21-2, we see that the route from sodium chloride to sodium thiosulfate

begins with the electrolysis of NaCl(aq) to produce NaOH(aq), 2 NaCl aq + 2 H2O l electrolysis 2 NaOH aq + H2 g + Cl2 g and continues through the reaction of SO2 g with the NaOH(aq) in an acid-base reaction

[ SO2 g is an acid anhydride] to produce 2 3 2 2 3 2Na SO aq : 2 NaOH aq + SO g Na SO aq + H O(l) and (3) concludes with

the addition of S to the boiling solution: boil2 3 2 2 3Na SO aq +S(s) Na S O aq . 2A (M) The first reaction indicates that 0.1 mol of NaNO2 reacts with 0.3 mol of Na to yield

0.2 mol of compound X and 0.05 mol of N2. Since this reaction liberates nitrogen gas, it is very likely that compound X is an oxide of sodium. The two possibilities include Na2O and Na2O2. Na2O2 will not react with oxygen, but Na2O will. Therefore, compound X is most likely Na2O and compound Y is Na2O2. The balanced chemical equations for two processes are:

NaNO2 (s)+3Na(s) 2Na2O(s)+ 12 N2 (g)

Na2O(s)+12

O2 (g) Na2O2 (s)

2B (M) Since compound X is used in plaster of Paris it must contain calcium. Calcium reacts

with carbon to form calcium carbide, according to the following chemical equation: Ca(s)+2C(s) CaC2 (s)

Furthermore, compound X or calcium carbide reacts with nitrogen gas to form compound Y or calcium cyanamide: CaC2 (s)+N2 (g) CaCN2 (s)+C(s)

CN22- anion is isoelectronic with CO2 and it therefore contains 16 electrons. The structure of this anion is:

N C N

Chapter 21: Chemistry of the Main-Group Elements I: Groups 1, 2, 13, and 14

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3A (M) The first two reactions, are those from Example 21-3, used to produce 2 3B O .

2 4 7 2 2 4 2 4 23

2 3 23

Na B O 10H O(s) + H SO (l) 4B OH (s) + Na SO (s) + 5H O(l)

2B OH (s) B O (s) + 3H O(g)

The next reaction is conversion to 3BCl with heat, carbon, and chlorine.

2 3 2 3 22 B O s + 3C s + 6Cl g 4 BCl g + 3CO g 4LiAlH is used as a reducing agent to produce diborane. 3 4 2 6 34 BCl g + 3LiAlH (s) 2 B H g + 3LiCl(s) + 3AlCl (s)

3B (M) Na2B4O7.10 H2O(s) + H2SO4(aq) 4 B(OH)3(s) + Na2SO4(aq) + 5 H2O(l)

2 B(OH)3(s) B2O3(s) + 3 H2O(l) B2O3(s) + 3 CaF2(s) + 3 H2SO4(l) 2 BF3(g) + 3 CaSO4(s) + 3 H2O(g)

INTEGRATIVE EXAMPLE

A (M) NaCN and Al(NO3)3 dissociate in water according to the following equations:

NaCN(aq) Na+(aq)+CN-(aq)Al(NO3)3(aq) Al3+(aq)+3NO3- (aq)

NaCN is a salt of strong base (NaOH) and weak acid (HCN) and its solution is therefore basic. We proceed by first determining the [OH]- concentration in a solution which is 1.0M in NaCN:

CN- +H2O HCN+OH- 1.0 / / 1.0-x x x

Kb KwKa (HCN )

1.0 1014

6.2 1010 1.6 105

Kb [HCN ][OH ]

[CN ] x x

1.0 x x2

1.0 x 1.6 105

Assume, x


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B (M) When 2 2BeCl 4H O is heated, it decomposes to Be(OH)2(s), H2O(g), and HCl(g), as discussed in Section 21-3. 2 2BeCl 4H O comprises [Be(H2O)4]2+ and Cl ions. Because of the high polarizing power of Be2+, it is difficult to remove the coordinated water molecules by heating the solid and the acidity of the coordinated H2O molecules is enhanced. When 2 2BeCl 4H O is dissolved in water, [Be(H2O)4]2+ ions react with water, and [Be(H2O)3(OH)]+ and H3O+ ions are produced. Hence, a solution of 2 2BeCl 4H O is expected to be acidic. When 2 2CaCl 6H O is heated, CaCl2(s) and H2O(g) are produced. Because the charge density and polarizing power of Ca2+ is much less than that of Be2+, it is much easier to drive off the coordinated water molecules by heating the solid. When 2 2CaCl 6H O is dissolved in water, Ca2+(aq) and Cl(aq) are produced. However, the charge density of the Ca2+ ion is too low to affect the acidity of water molecules in the hydration sphere of the Ca2+ ion. Hence, a solution of 2 2CaCl 6H O has a neutral pH.

EXERCISES Group 1: The Alkali Metals 1. (E) (a) 22 Cs(s) + Cl (g) 2 CsCl(s) (b) 2 2 22 Na(s) + O (g) Na O (s) (c) 2 3 2 2 Li CO s Li O s + CO g (d) 2 4 2Na SO s + 4 C s Na S s + 4 CO g (e) 2 2K s + O g KO s 2. (a)

2 Rb s + 2 H2O l 2RbOH aq + H2 g

(b) 3 2 3 2 22 KHCO s K CO aq + H O(l) + CO g (c) 2 2 22 Li s + O g Li O s (d) 2 4 2 42 KCl s + H SO aq K SO aq + 2 HCl g (e) 2 2LiH s + H O l LiOH aq + H g 3. (E) Both LiCl and KCl are soluble in water, but Li PO3 4 is not very soluble. Hence the

addition of K PO (aq)3 4 to a solution of the white solid will produce a precipitate if the white solid is LiCl, but no precipitate if the white solid is KCl. The best method is a flame test; lithium gives a red color to a flame, while the potassium flame test is violet.


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4. (E) When heated, Li CO2 3 decomposes to 2CO (g) and 2 2 3Li O(l). K CO (s) simply melts when heated. The evolution of CO (g)2 bubbles should be sufficient indication of the difference in behavior. The flame test affords a violet flame if K+ is present, and a red flame if Li+ is present.

5. (M) First we note that sodium carbonate ionizes virtually completely when dissolved in H2O and

thus is described as highly soluble in water. This is made evident by the fact that there is no Ksp value for Na2CO3. By contrast, the existence of Ksp values for MgCO3 and Li2CO3 shows that these compounds have lower solubilities in water than Na2CO3. To decide whether MgCO3 is more or less soluble than Li2CO3, we must calculate the molar solubility for each salt and then compare the two values. Clearly, the salt that has the larger molar solubility will be more soluble in water. The molar solubilities can be found using the respective Ksp expressions for the two salts.

1. MgCO3 -8

sp = 3.5 10K Mg2+(aq) + CO32-(aq) Let s = molar solubility of MgCO3 excess s s s s2 = 3.510-8 s = 1.910-4 M

2. Li2CO3 -2

sp = 2.5 10K 2 Li+(aq) + CO32-(aq) Let s = molar solubility of Li2CO3 excess s 2s s

(2s)2(s) = 2.510-8 4s3 = 2.510-8 s = -8

32.510

4 = 0.18 M

We can conclude that Li2CO3 is more soluble than MgCO3. Thus, the expected order of increasing solubility in water is: MgCO3 < Li2CO3 < Na2CO3.

6. (M) We know sodium metal was produced at the cathode from the reduction of sodium ion, Na+. Thus, hydroxide must have been involved in oxidation at the anode. The hydrogen in hydroxide ion already is in its highest oxidation state and thus cannot be oxidized. This leaves oxidation of the hydroxide ion to elemental oxygen as the remaining reaction.

Cathode, reduction : {Na e Na(l)} 4+

Anode, oxidation : 4OH O (g) + 2 H O(g) + 4e2 2

Net : 4 4 4Na OH Na(l) + O (g) + 2 H O(g)+ 2 2 7. (M) (a) H (g)2 and Cl (g)2 are produced during the electrolysis of NaCl(aq). The electrode reactions are:

Anode, oxidation: 2Cl (aq) Cl (g) + 2e2

Cathode, reduction: 2 22H O(l) + 2e H (g) + 2OH (aq)

We can compute the amount of OH produced at the cathode.


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mol OH 2.50min 60s1min

0.810C1s

1mole

96,500C 2molOH

2mole 1.26 103 molOH

Then we compute the [OH ] and, from that, the pH of the solution.

[OH mol OHLsoln

M

] ..

.126 100872

145 103

3 pOH = log(1.45 10 3 2 839) . pH=14.000 2.839 11.161

(b) As long as NaCl is in excess and the volume of the solution is nearly constant, the solution pH only depends on the number of electrons transferred.

8. (M) (a) total energy = 33600s 1C/s 1J3.0V 0.50A h 5.4 10 J1hr 1A 1V C

We obtained the first conversion factor for time as follows.

50 1 50 106 6

. . W1 10 W

1 WJ / s

1WJ

1s

time = 3 961s 1hr 1day 1y5.4 10 J 1.1 10 s 34 y

5 10 J 3600s 24hr 365days

(b) The capacity of the battery is determined by the mass of Li present.

1 C/s 3600 s 1 mol e 1 mol Li 6.941 g Limass Li = 0.50 A h = 0.13 g Li1 A 1 h 96485 C 1 mol Li1 mol e

9. (M) (a) We first compute the mass of NaHCO3 that should be produced from 1.00 ton

NaCl, assuming that all of the Na in the NaCl ends up in the NaHCO3. We use the unit, ton-mole, to simplify the calculations.

3

3 33

3

1 ton-mol NaCl 1ton-mol Namass NaHCO 1.00 ton NaCl58.4 ton NaCl 1ton-mol NaCl

1tol-mol NaHCO 84.0 ton NaHCO = 1.44 ton NaHCO1ton-mol Na 1 ton mol NaHCO

% yield ton NaHCO produced ton NaHCO expected

yield= 1.031.44

100% = 71.5%33

(b) NH3 is used in the principal step of the Solvay process to produce a solution in which NaHCO3 is formed and from which it will precipitate. The filtrate contains NH Cl,4 from which NH3 is recovered by treatment with Ca(OH)2 . Thus, NH3 is simply used


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during the Solvay process to produce the proper conditions for the desired reactions. Any net consumption of NH3 is the result of unavoidable losses during production.

10. (M) (a) 22 4 4Ca(OH) (s) SO (aq) CaSO (s) 2OH (aq)

(b) We sum two solubility reactions and combine their values of Ksp 22Ca(OH) (s) Ca (aq) 2OH (aq)

Ksp 55 10 6. 2 2

4 4Ca (aq) SO (aq) CaSO (s) 1 1 9 1 10 6/ / .Ksp

24 42Ca OH s + SO aq CaSO s + 2 OH aq Keq = 5.5 109.1 10 = 0.606

6

Because Keq is close to 1.00, we conclude that the reaction lies neither very far to the right (it does not go to completion) nor to the left.

(c) Reaction: 22 4 4Ca(OH) (s) SO (aq) CaSO (s) 2OH (aq)

Initial: 1.00 M 0 M Changes: -x M +2x M Equil: (1.00x) M 2x M

K xx

x x x x= = 0.60 = 41.00

4 = 0.60 0.60 4 + 0.60 0.60 = 02

42

22 2

OH

SO

x b b aca

= 42

= 0.60 0.36 + 9.608

= 0.322 M

SO M OH M42 = 1.00 = 0.68 = 2 = 0.64 x x

11. (M) Use Gfo values to calculate Go for the reaction and then the equilibrium constant K at 298 K.

Na2O2 (s) Na2O(s)+

12

O2 (g)

Gfo (kJmol-1) -449.63 -379.09

Go 379.09 (449.63) 70.54kJmol-1Go RT ln K8.314JK-1mol-1 298K ln K 70.54 1000Jmol-1

ln K 70.54 1000Jmol-1

8.314JK-1mol-1 298K 28.47 K e28.47 4.32 1013


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Since K pO212 pO2 K 2 1.87 1025 . The equilibrium constant and partial pressure of

oxygen are both very small at 298 K. Therefore, Na2O2(s) is thermodynamically stable with respect to Na2O(s) and O2(g) at 298 K. 12. (M) Use Gfo values to calculate Go for the reaction and then the equilibrium constant K at 298 K.

2KO2 (s) K2O(s)+

32

O2 (g)

Gfo (kJmol-1) -240.59 -322.09

Go 322.09 (2 240.59) 159.1kJmol-1Go RT ln K8.314JK-1mol-1 298K ln K 159.11000Jmol-1

ln K 159.11000Jmol-1

8.314JK-1mol-1 298K 64 K e64 1.604 1028

Since K pO232 pO2 K

23 2.952 1019 . The equilibrium constant and partial pressure

of oxygen are both very small at 298 K. Therefore, KO2(s) is thermodynamically stable with respect to K2O(s) and O2(g) at 298 K. Group 2: The Alkaline Earth Metals 13. (M)

2 22,electrolysis

2 2 2 3 22

3 2 2 4 4 22

The reactions are as follows. Ca OH s + 2 HCl aq CaCl aq + 2H O(l)

CaCl l Ca l + Cl g Ca OH s + CO g CaCO s + H O g

CaCO s CaO s + CO g Ca OH s + H SO aq CaSO s + 2 H O(l)

3 4 4 22Ca OH s + H PO aq CaHPO aq + 2H O(l) . Actually CaO s is the industrial starting material from which 2Ca(OH) is made. 2 2CaO(s) + H O(l) Ca(OH) (s)

14. (M)

Once we return to Mg(OH)2 from MgSO4, the other substances can be made by the indicated pathways. The return reaction is: 4 2 2 4MgSO (aq) + 2 NaOH(aq) Mg(OH) (s) + Na SO (aq). Then the other reactions are

CaO CaCO3CO2 Ca(OH)2 HCl CaCl2 Ca

H2SO4 CaSO4H3PO4CaHPO4

electrolysis

MgO MgCO3

CO2 Mg(OH)2 HCl MgCl2 Mg

H2SO4 MgSO4H3PO4MgHPO4

electrolysis N2 Mg3N2


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2 2 2Mg(OH) (s) + 2 HCl(aq) MgCl (aq) +2 H O(l) , electrolysis

2 2 2 2 3 2MgCl (l) Mg(l) + Cl (g) ; Mg(OH) (s) +CO (g) MgCO (s) + H O(g)

3 2 2 2 4 4 2MgCO (s) MgO(s) +CO (g); Mg(OH) (s) + H SO (aq) MgSO (s) + 2H O(l)

2 3 4 4 2Mg(OH) (s) + H PO (aq) MgHPO (aq) + 2H O(l)

15. (M) The reactions involved are: Mg2+(aq) + Ca2+(aq) + 2OH(aq) Mg(OH)2(s) + Ca2+(aq) Mg(OH)2(s) + 2 H+(aq) + 2 Cl(aq) Mg2+(aq) + 2 H2O(l) + 2 Cl(aq) Mg2+(aq) + 2 Cl(aq) MgCl2(s) MgCl2(s) (electrolysis) Mg(l) + Cl2(g) Mg(l) Mg(s) Mg2+ + 2 Cl(aq) Mg(s) + Cl2(g) (Overall reaction) As can be seen, the process does not violate the principle of conservation of charge.

16. (M) (a) MgO vs. BaO: MgO would have the higher melting point because, although Mg2+ and Ba2+ have the same charge, Mg2+ is a smaller ion. Smaller ions have a larger electrostatic attraction to anions (here, in both cases, the anion is O2), which is due to the smaller charge separation (Coulomb's law).

(b) MgF2 vs. MgCl2 solubility in water. F is smaller than Cl, hence, electrostatic attraction between Mg2+ and the halide is greater in F than Cl. If we assume that hydration of the ions is similar, we expect that MgF2 is less soluble than MgCl2. (Note: Ksp given for MgF2, which is sparingly soluble, while no Ksp value is given for readily soluble MgCl2.)

17. (M) (a) 2 2BeF (s) + Mg(s) Be(s) + MgF s (b) Ba s + Br2 l BaBr2 s (c) UO2 s + 2 Ca s U s + 2 CaO s (d) 3 3 2MgCO CaCO s MgO s + CaO s + 2 CO g (e) 2 H3PO4 aq + 3 CaO s Ca3 PO4 2 s + 3 H2O l

18. (M) (a) heat3 2 22Mg HCO s MgO s + 2 CO g + H O(l) (b) electrolysis2 2BaCl l Ba l + Cl g (c) Sr s + 2 HBr aq SrBr2 aq + H2 g (d) 2 4 4 22H SO aq + Ca OH aq CaSO s + 2H O l (e) 314 2 4 2 22 2CaSO 2 H O s CaSO H O s + H O g 19. (D) Let us compute the value of the equilibrium constant for each reaction by combining

the two solubility product constants. Large values of equilibrium constants indicate that the reaction is displaced far to the right. Values of K that are much smaller than 1 indicate that the reaction is displaced far to the left.

(a) 2+ 2 104 4 spBaSO s Ba aq +SO aq 1.1 10K


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2+ 2 93 3 spBa aq + CO aq BaCO s 1/ = 1/ (5.1 10 )K 102 2 24 3 3 4 91.1 10BaSO s + CO aq BaCO s +SO aq = = 2.2 105.1 10K

Thus, the equilibrium lies slightly to the left.

(b) 2+ 3 253 4 4 sp2Mg PO s 3 Mg aq + 2 PO aq = 2.1 10K 2+ 2 3 8 33 3 sp3{Mg aq + CO aq MgCO s } 1/ ( ) = 1/ (3.5 10 )K

2 33 4 3 3 42

253

38

Mg PO s + 3CO aq 3 MgCO s + 2 PO aq

2.1 10= = 4.9 10 Thus, the equilibrium lies to the left.3.5 10

K

(c) 2+ 6sp2Ca OH s Ca aq + 2 OH aq = 5.5 10K 2+ 92 spCa aq + 2 F aq CaF s 1/ 1/ (5.3 10 )K

2 2Ca(OH) s + 2F aq CaF s + 2 OH aq 6

39

5.5 10= = 1.0 105.3 10

K

Thus, the equilibrium lies to the right.

20. (M) We expect the reaction to occur to a significant extent in the forward direction if its

equilibrium constant is >> 1. (a) 2+ 2 93 3 spBaCO s Ba aq + CO aq = 5.1 10K 2+ 2 52 3 2 2 3 2 a2 HC H O aq 2H aq + 2C H O aq = 1.8 10K

2

+ 2 113 3 aH aq + CO aq HCO aq 1/ = 1/ (4.7 10 )K

1

+ 73 2 3 aH aq + HCO aq H CO aq 1/ = 1/ (4.2 10 )K

3 2 3 2 2 3 2 2 32BaCO s + 2 HC H O aq Ba C H O aq + H CO aq K =

5.1109 1.8 105 24.7 1011 4.2 107 = 8.4 10

2

Thus, no significant reaction occurs (the equilibrium mixture contains appreciable amounts of all four species involved in the reaction.

(b) 2+ 6sp2Ca OH s Ca aq + 2 OH aq = 5.5 10K 2+ 2 54 3 2 b2 NH aq + 2 OH aq 2 NH aq + 2 H O(l) 1/ = 1/ 1.8 10K

+ 2+4 3 22

64

25

Ca OH s + 2 NH aq Ca aq + 2 NH aq + 2 H O(l)

5.5 10= = 1.7 10 Thus, this reaction would occur to a significant extent.1.8 10

K


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(c) 2+ 62 spBaF s Ba aq + 2 F aq = 1.0 10K

2+ 2 43 2 a2 H O aq + 2 F aq 2 HF aq + 2 H O(l) 1/ = 1/ 6.6 10K

6

2+2 3 2 24

1.0 10BaF s + 2H O aq Ba aq + 2HF aq + 2H O(l); = = 2.36.6 10

K

This reaction would occur to some extent, but certainly not to completion. 21. (M) The SO42 ion is a large polarizable ion. A cation with a high polarizing power

will polarize the SO42 ion and kinetically assist the decomposition to SO3. Because Be2+ has the largest charge density of the group 2 cations, we expect Be2+ to be the most polarizing and thus, BeSO4 will be the least stable with respect to decomposition.

22. (M) The CO32 ion is a large polarizable ion. A cation with a high polarizing power

will polarize the CO32 ion and kinetically assist the decomposition to CO2. A cation with a low polarizing power will not polarize the anion as much; thus, the decomposition of CO32 to CO2 will not occur as readily in the presence of such a cation. Because Ba2+ has the smallest charge density of the group 2 cations, we expect Ba2+ to be the least polarizing; thus, BaSO4 will be the most stable with respect to decomposition.

Group 13: The Boron Family 23. (M) (a) 4 10B H contains a total of 4 3 +10 1 = 22 valence electrons or 11 pairs. Ten of

these pairs could be allocated to form 10 BH bonds, leaving but one pair to bond the four B atoms together, which is clearly an electron deficient situation.

(b) In our analysis in part (a), we noted that the four B atoms had but one electron pair to bond them together. To bond these four atoms into a chain requires three electron pairs. Since each electron pair in a bridging bond replaces two normal bonds, there must be at least two bridging bonds in the 4 10B H molecules. By analogy with 2 6B H , we might write the structure below left. But this structure uses only a total of 20 electrons. (The bridge bonds are shown as dots, normal bondselectron pairsas dashes.) In the structure at right below, we have retained some of the form of 2 6B H , and produced a compound with the formula 4 10B H and 11 electron pairs. (The experimentally determined structure of 4 10B H consists of a four-membered ring of alternating B and H atoms, held together by bridging bonds. Two of the B atoms have two H atoms bonded to each of them by normal covalent bonds. The other two B atoms have one H atom covalently bonded to each. One final BB bond joins these last two B atoms, across the diameter of the ring.). See the diagram that follows:


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(c) 4 10C H contains a total of 4 4 +10 1 = 26 valence electrons or 13 pairs. A plausible

Lewis structure follows. (Note that each atom possess an octet of electrons.)

24. (E) (a) (b) 25. (M) (a) 2BBr3(l) + 3H2(g) 2B(s) + 6HBr(g)

(b) i) B2O3(s) + 3 C(s) 3 CO(g) + 2 B(s) ii) 2 B(s) + 3 F2(g) 2 BF3(g) (c) 2 B(s) + 3 N2O(g) 3 N2(g) + B2O3(s)

26. (M) Each boron atom has an oxidation number of +3. The hydroxyl oxygens are each 2, while the bridging oxygens are each 1. Finally, the hydroxyl H atoms are all in the +1 oxidation state. The oxidation numbers for the all the constituent atoms add up to the charge on the perborate ion, namely, 2.

27. (M) (a) 2 Al s + 6 HCl aq 2 AlCl3 aq + 3 H2 g

(b) +2 242 NaOH aq + 2 Al s + 6 H O l 2 Na aq + 2 Al OH aq + 3 H g

(c) Oxidation: 3+{Al s Al aq + 3 e } 2 Reduction: 2 +4 2 2{SO aq + 4 H aq + 2e SO g + 2 H O(l)} 3 _________________________________________________________________________________________________________________________________

Net: 2 + 3+4 2 22 Al s + 3 SO aq +12 H aq 2 Al aq + 3 SO aq + 6 H O(l) 28. (M) (a) 2 Al(s) + 3 Br2(l) 2 AlBr3(s)

(b) 2 Al s + Cr2O3 s heat 2Cr l + Al2O3 s

H C C C C

H H H

H

H H H H

H

F B F

F

F

F B

F

FN

H

H

C

H

H

C

H

H

H_ +

H H H H H

HB

HB

HB

HB

H

H H H

HB

HB

HB

H

BH

H

H


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(c) 2 3Fe O s + OH aq no reaction 2 3 2 4Al O s + 2 OH aq + 3 H O(l) 2 Al OH aq

29. (M) One method of analyzing this reaction is to envision the HCO3

ion as a combination of CO2 and OH

. Then the OH reacts with Al3+ and forms Al OH 3 . This method of envisioning HCO3

does have its basis in reality. After all,

2 3 2 2 3 2H CO = H O + CO + OH HCO + H O

Al3+ aq + 3 HCO3 aq Al OH 3 s + 3 CO2 g

Another method is to consider the reaction as, first, the hydrolysis of hydrated aluminum ion to produce 3 Al OH s and an acidic solution, followed by the reaction of the acid with bicarbonate ion.

3+ +2 2 2 36 3 3Al H O aq + 3H O(l) Al OH H O s + 3H O aq

+3 3 2 23H O aq + 3HCO aq 6H O(l) + 3CO g

This gives the same net reaction:

3+2 3 2 2 26 3 3Al H O aq + 3HCO aq Al OH H O s + 3CO g + 3H O(l)

30. (E) The Al3+ aq ion hydrolyzes. 3+ +2 3Al aq + 3 H O(l) Al OH s + 3 H aq Subsequently, the hydrogen ion that is produced reacts with bicarbonate ion to liberate 2CO g .

+ 3 2 2H aq + HCO aq H O(l) + CO g 31. (M) Aluminum and its oxide are soluble in both acid and base.

+ 3+2

+ 3+2 3 2

2 Al s + 6 H aq 2 Al aq + 3 H g

Al O s + 6 H aq 2 Al aq + 3 H O(l)

2 242 Al s + 2 OH aq + 6 H O(l) 2 Al OH aq + 3 H g 2 3 2 4Al O s + 2 OH aq + 3 H O(l) 2 Al OH aq

Al(s) is resistant to corrosion only over the pH range 4.5 to 8.5. Thus, aluminum is inert only when the medium to which it is exposed is neither highly acidic nor highly basic.

32. (M) Both Al and Mg are attacked by acid and their ions are both precipitated by hydroxide

ion. + 3+ 22 Al s + 6 H aq 2 Al aq +3 H g + 2+ 2Mg s + 2 H aq Mg aq + H g 3+ 3Al aq +3 OH aq Al OH s 2+ 2Mg aq + 2 OH Mg OH s


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But, of these two solid hydroxides, only Al OH 3 s redissolves in excess OH aq . Al OH 3 s + OH aq Al OH 4 aq Thus, the analytical procedure consists of dissolving the sample in HCl(aq) and then treating the resulting solution with NaOH(aq) until a precipitate forms. If this precipitate dissolves completely in excess NaOH(aq), the sample is aluminum 2S. If at least some of the precipitate does not dissolve, the sample was magnesium.

33. (M) CO2 g is, of course, the anhydride of an acid. The reaction here is an acid-base

reaction. Al OH 4 aq + CO2 aq Al OH 3 s + HCO3 aq

HCl(aq), being a strong acid, cant be used because it will dissolve the Al(OH)3(s). 34. (M) (a) Oersted: 2 Al2O3 s + 3 C s + 6 Cl2 g 4 AlCl3 s + 3 CO2 g

(b) Whler: AlCl3 s + 3 K s Al s + 3 KCl s 35. (M) 2 KOH(aq) + 2 Al(s) + 6 H2O(l) 2 K[Al(OH)4](aq) + 3 H2(g)

2 K[Al(OH)4](aq) + 4 H2SO4(aq) K2SO4(aq) + Al2(SO4)3(aq) + 8 H2O(l) crystallize 2 KAl(SO4)2(s)

36. (E) HCO3 and CO32 solutions are basic (they produce an excess of OH). Al3+(aq) in the

presence of OH(aq) will precipitate as the hydroxide Al(OH)3(s). 37. (E) The structure of, and bonding in, digallane is similar to the structure of and bonding in diborane, B2H6.

Ga

H

H

Ga

H

H

H

H 38. (M) The structure of, and bonding in, GaH2Cl2 is similar to the structure of and bonding in Al2Cl6, except that the terminal Cls in Al2Cl6 are replaced by Hs.

Ga

Cl

Cl

Ga

H

H

H

H

Group 14: The Carbon Family 39. (E) In the sense that diamonds react imperceptibly slowly at room temperature (either with

oxygen to form carbon dioxide, or in its transformation to the more stable graphite), it is essentially true that diamonds last forever. However, at elevated temperatures, diamond will burn to form CO2(g) and thus the statement is false. Also, the transformation


1077

C(diamond) C(graphite) might occur more rapidly under other conditions. Eventually, of course, the conversion to graphite occurs.

40. (E) The graphite in the pencil lead is a good dry lubricant that will make slippery the

stickiness (or reluctance) in the lock and enable it to work smoothly. The key carries the graphite to the site that needs to be lubricated within the lock mechanism.

41. (M) (a) 3 SiO2(s) + 4 Al(s) 2 Al2O3(s) + 3 Si(s)

(b) K2CO3(s) + SiO2(s) CO2(g) + K2SiO3(s)

(c) Al4C3(s) + 12 H2O(l) 3 CH4(g) + 4Al(OH)3(s) 42. (M) (a) 2 KCN(aq) + AgNO3(aq) KAg(CN)2(aq) + KNO3(aq) or KCN(aq) + AgNO3(aq) Ag(CN)(s) + KNO3(aq)

(b) Si3H8(l) + 5 O2(g) 3 SiO2(s) + 4 H2O(l)

(c) N2(g) + CaC2(s) CaNCN(s) + C(s) 43. (M) A silane is a silicon-hydrogen compound, with the general formula 2 +2Si H .n n A silanol is

a compound in which one or more of the hydrogens of silane is replaced by an OH group. Then, the general formula becomes 2 +1Si H OHn n . In both of these classes of compounds, the number of silicon atoms, n, ranges from 1 to 6. Silicones are produced when silanols condense into chains, with the elimination of a water molecule between every two silanol molecules.

2 2 2 2 2HO Si H OH + HO Si H OH HO Si H O Si H OH + H On n n n n n n n 44. (M) Both the alkali metal carbonates and the alkali metal silicates are soluble in water.

Hence, they also are soluble in acids. However, the carbonates will produce gaseous carbon dioxide 2CO on reaction with acid, 2 +3 2 3 2 2CO (aq) + 2 H (aq) H CO H O(l) + CO (g) , while the silicates will produce silica 2SiO in an analogous reaction 4 +4 4 4 2 2SiO (aq) + 4H (aq) H SiO 2 H O(l) +SiO (s) . The silica is produced in many forms depending on reaction conditions: a colloidal dispersion, a gelatinous precipitate, or a semisolid gel. Silicates of cations other than those of alkali metals are insoluble in water, as are the analogous carbonates. However, the carbonates will dissolve in acids (witness the reaction of acid rain on limestone carvings and marble statues, both forms of 3CaCO ), while the silicates will not. (Silicate rocks are not significantly affected by acid rain.)

45. (M) (1) 4 8 2 22 CH g + S g 2 CS g + 4 H S g

(2) 2 2 4 2 2CS g + 3 Cl g CCl l + S Cl l

(3) 2 2 2 4 84 CS g +8 S Cl g 4 CCl l + 3 S (s)


1078

46. (M) (a) 3 2 33 3CH SiCl(l) + H O(l) CH Si OH(aq) + HCl(aq) 3 3 3 23 3 32 CH Si OH(aq) CH Si O Si CH (s) + H O(l)

(b) A silicone polymer does not form from 3 3CH Si Cl . Only a dimer is produced.

(c) The product that results from the treatment of 3 3CH SiCl is two long SiOSi chains, with 3CH groups on the outside, linked by SiOSi bridges. Part of one of these chains and the beginnings of the bridges are shown below.

47. (D) Muscovite or white mica has the formula KAl2(OH)2(AlSi3O10). Since they are not

segregated into O2 units in the formula, all of the oxygen atoms in the mineral must be in the 2 oxidation state. Potassium is obviously in the +1 oxidation state, as are the hydrogen atoms in the hydroxyl groups. Up to this point, we have -24 from the twelve oxygen atoms and +3 from the potassium and hydrogen atoms for a net number of -21 for the oxidation state. We still have three aluminum atoms and three silicon atoms to account for. In oxygen-rich salts such as mica, we would expect that the silicon and the aluminum atoms would be in their highest possible oxidation states, namely +4 and +3, respectively. Since the salt is neutral, the oxidation numbers for the silicon and aluminum atoms must add up to +21. This is precisely the total that is obtained if the silicon and aluminum atoms are in their highest possible oxidation states: (3 (+3) + 3 (+4) = +21). Consequently, the empirical formula for white muscovite is consistent with the expected oxidation state for each element present.

48. (D) Crysotile asbestos has the formula [Mg3Si2O5(OH)4]. Since they are not segregated to

O2 units in the formula, all of the oxygen atoms in the mineral must be in the 2 oxidation state. The three magnesium atoms are obviously in the +2 oxidation state, while hydrogen atoms in the hydroxyl groups have an oxidation state of +1. Up to this point, then, the sum of the oxidation numbers equals 8 (18 from O-atoms, +6 from Mg-atoms, and + 4 from H-atoms). Since the mineral is neutral, the two silicon atoms must have oxidation states that sum to +8 (8 + 8 = neutral mineral), therefore, each silicon would need to be in the +4 oxidation state. In oxygenrich salts such as asbestos, we would expect that the silicon atoms would be in their highest possible oxidation state, namely +4. Thus, the oxidation states for the element, in this mineral are precisely consistent with expectations.

49. (M) (a) 3 3 22PbO s + 2 HNO aq Pb NO s + H O(l)

(b) SnCO3 s SnO s + CO2 g

(c) PbO s + C s Pb l + CO g

O Si

O

CH3

O Si O

O

CH3

Si

O

CH3

O Si O

O

CH3

Si

O

CH3

O Si O

O

CH3


1079

(d) 2 Fe3+ aq + Sn2+ aq 2 Fe2+ aq + Sn4+ aq

(e) 2 22 PbS s + 3 O g 2 PbO s + 2 SO g 2 2 32 SO g + O g 2 SO g

SO3 g + PbO s PbSO4 s

Or perhaps simply: 2 4PbS s + 2O g PbSO s

Yet a third possibility: 2 3PbO s + SO s PbSO s , followed by 3 2 42 PbSO s + O s 2PbSO s

50. (M) (a) Treat tin(II) oxide with hydrochloric acid.

2 2SnO s + 2 HCl aq SnCl aq + H O(l)

(b) Attack tin with chlorine. Sn s + 2 Cl2 g SnCl4 s

(c) First we dissolve PbO2 s in HNO3 aq and then treat the resulting solution with K2CrO4 aq to precipitate PbCrO4(s)

2 3 3 2 222 PbO s + 4 HNO aq 2 Pb NO aq + O g + 2 H O(l) 3 2 4 4 32Pb NO aq + K CrO aq PbCrO s + 2 KNO aq 51. (M) We start by using the Nernst equation to determine whether the cell voltage still is

positive when the reaction has gone to completion.

(a) Oxidation: 2+ 3+Fe aq Fe aq + e } 2 = 0.771 VE Reduction: + 2+2 2PbO s + 4 H aq + 2 e Pb aq + 2 H O(l) = +1.455 VE

Net: 2+ + 3+ 2+2 22 Fe aq + PbO s + 4 H (aq) 2 Fe aq + Pb aq + 2 H O(l)

Ecell V V V = 0.771 +1.455 = +0.684

In this case, when the reaction has gone to completion,

Fe M Fe2+ 3+= 0.001 , = 0.999 M, and Pb2+ = 0.500 M.

Ecell = Eocell 3 2 2

2 2

0.0592 [Fe ] [Pb ]log2 [Fe ]

= 0.684 V

2

2

0.0592 [0.999] [0.500]log2 [0.001]

= 0.515 V. Thus, this reaction will go to completion. (b) Oxidation: 2 24 2 82 SO aq S O aq + 2 e = 2.01VE

Reduction: + 2+2 2PbO s + 4 H aq + 2 e Pb aq + 2 H O(l) = +1.455 VE 2 2+ 2+4 2 2 8 22 SO aq + PbO s + 4 H (aq) S O aq + Pb aq + 2 H O(l)


1080

Ecell V This reaction is not even spontaneous initially = 2.01+1.455 = 0.56 .

(c) Oxidation:

{Mn2+ 1 10-4 M +4H2O(l) MnO4 aq +8H+ aq +5 e-} 2 ; o = 1.51 VE

Reduction: + 2+2 2{PbO s +4 H aq +2 e Pb aq +2 H O(l)}5 o =+1.455 VE Net: -42+ + 2+2 4 21 10 M2Mn + 5PbO s + 4H 2MnO aq + 5Pb aq + 2H O(l)

ocell = 1.51+1.455 = 0.06 V.E The standard cell potential indicates that this reaction is

not spontaneous when all concentrations are 1 M. Since the concentration of a reactant Mn2+ is lower than 1.00 M, this reaction is even less spontaneous than the standard

cell potential indicates. 52. (M) A positive value of Ecell

indicates that a reaction should occur. (a) 2+ 4+Oxidation : Sn aq Sn aq + 2 e = 0.154 VE

2Reduction : I s + 2 e 2 I aq = +0.535 VE _________________________________________ 2+ 4+2Net : Sn aq + I s Sn aq + 2 I aq cell = +0.381 VE Yes, Sn2+ aq will reduce I2 to I .

(b) 2+ 4+Oxidation : Sn aq Sn aq + 2 e = 0.154 VE 2+Reduction : Fe aq + 2 e Fe s = 0.440 VE

________________________________________ 2+ 2+ 4+ cellNet : Sn aq + Fe aq Sn aq + Fe s = 0.594 VE No, Sn2+ aq will not reduce Fe2+ aq to Fe(s).

(c) 2+ 4+Oxidation : Sn aq Sn aq + 2 e = 0.154 VE 2+Reduction : Cu aq + 2 e Cu s = +0.337 VE

______________________________________________________ 2+ 2+ 4+ cellNet : Sn aq + Cu aq Sn aq + Cu s = +0.183 VE Yes, Sn2+ aq will reduce Cu2+ aq to Cu(s).

(d) 2+ 4+Oxidation : Sn aq Sn aq + 2 e = 0.154 VE 3+ 2+Reduction : {Fe aq + e Fe aq } 2 = +0.771 VE

___________________________________________________________ 2+ 3+ 4+ 2+ cellNet : Sn aq + 2 Fe aq Sn aq + 2 Fe aq = +0.617 VE Yes, Sn2+ aq will reduce Fe3+ aq to Fe2+ aq .

53. (M) As we move down a group, the lower oxidation state is generally favored. Thus, we expect PbCl2, with lead in the +2 oxidation state, to be the product.


1081

54. (M) As we move down a group, the lower oxidation state is generally favored. The higher oxidation state is favored for elements higher in the group. Thus, we expect GeF4, with lead in the +4 oxidation state, to be the product.

INTEGRATIVE AND ADVANCED EXERCISES

55. (M) What most likely happened is that a deliquescent solid has absorbed water vapor from the air over time and formed a saturated solution. This saturated solution can be used in any circumstance where the solid would be used to prepared an aqueous solution. Of course, the concentration will have to be determined by some means other than weighing, such as a gravimetric analysis or using a standard to react with the substance in a titration. Alternatively, the saturated solution can be heated in a drying oven to drive off the water and the solid may be recovered. Often, unfortunately, a solid with an unknown percent of water results from this treatment.

56. (D) (a) A solution of CO2(aq) has [CO32-] = Ka2[H2CO3] = 5.610-11 M. Since the Ksp = 2.810-9 for CaCO3, the [Ca2+] needed to form a precipitate from this solution can be computed.

[Ca2+] = -9

sp2+ -11

2.8 10= = 50. M[Ca ] 5.6 10

K

This is too high to reach by dissolving CaCl2 in solution. The reason a solid forms is that the OH- produced by the Ca(OH)2 neutralized some of the HCO3- from the ionization of CO2(aq), thereby increasing the [CO32-] above a value of 5.610-11 M.

(b) The equation for redissolving can be obtained by combining several equations.

CaCO3(s) Ca2+(aq) + CO32-(aq) Ksp = 2.810-9 CO32-(aq) + H3O+(aq) HCO3-(aq) + H2O(l) 1/Ka2 = 1/(5.610-11) CO2(aq) + 2 H2O(l) HCO3-(aq) + H3O+(aq) Ka1 = 4.210-7

CaCO3(s) + CO2(aq) + H2O(l) Ca2+(aq) + 2 HCO3-(aq) K =

Ksp Ka1Ka2

K =

(2.8 10-9 ) (4.2 10-7 )(5.6 10-11) = 2.110

-5 = [Ca2+ ][HCO3

-]2

[CO2]

If CaCO3(s) is precipitated from 0.005 M Ca2+(aq) and then redissolved, [Ca2+] = 0.005 M and [HCO3-] = 20.005 M = 0.010 M. We use these values in the above expression to compute the [CO2].

[CO2] = [Ca2+ ][HCO3

-]2

(2.110-5) = 0.002 M We repeat the calculation for saturated Ca(OH)2, in which [OH-] = 2 [OH-], after

first determining [Ca2+] in this solution.


1082

Ksp = [Ca2+][OH-]2 = 4[Ca2+] = 5.5 10-6 [Ca2+] = 6

35.5 10 = 0.011 M

4

[CO2] =

[Ca2+ ][HCO3-]2

(2.110-5) = (0.011)(0.022)2

2.110-5 = 0.25 M Thus, to redissolve the CaCO3 requires that the [CO2] = 0.25 M, assuming the solution is initially saturated with Ca(OH)2(aq).

(c) follow the solution for (b) above.

57. (M) There are two principal reasons why the electrolysis of NaCl(l) is used to produce sodium commercially rather than the electrolysis of NaOH(l). First, NaCl is readily available whereas NaOH is produced from the electrolysis of NaCl(aq). Thus the raw material NaCl is much cheaper than is NaOH. Second, but less important, when sodium is produced from the electrolysis of NaCl(l), a by-product is Cl2(g), whereas O2(g) is a by-product of the production of NaOH(l). Since O2(g) can be produced more cheaply by the fractional distillation of liquid air, Cl2(g) can be sold for a higher price than O2(g). Thus, the two reasons for producing sodium from NaCl(l) rather than NaOH(l) are a much cheaper raw material and a more profitable by-product.

58. (M) The triiodide ion is linear (AX2E3). The Li+ ion has a high charge density and significant polarizing power. The Li+ ion will polarize the I3 to a significant extent and presumably assists the decomposition of I3 to I2 and I.

I I I

59. (M) Use ofG values to calculate oG for the reaction and then K. Because 122OK P , the value of K can then be used to calculate

2OP .

Li2O2 (s) Li2O(s)+

12

O2 (g)

Gfo[kJ / mol] -419.02 -466.40

Go 466.40 (419.02) 47.38kJmol-1Go RT ln K8.314JK-1mol-1 1000K ln K 47.38 1000Jmol-1

ln K 47.38 1000Jmol-1

8.314JK-1mol-1 1000K 5.70 K e5.70 298

K PO212 PO2 K 2 2982 8.8 104

60. (M) Use o o o1hydr. hydr. hydr.TS H G to calculate ohydr.S values:


1083

Shydr.o (Li ) 1298 522 (481) 0.14J/KmolShydr.o (Na ) 1298 407 (375) 0.11J/KmolShydr.o (K ) 1298 324 (304) 0.07J/KmolShydr.o (Rb ) 1298 299 (281) 0.06J/KmolShydr.o (Cs ) 1298 274 (258) 0.05J/Kmol

All the ohydr.S values are negative because, in the process, a gas is being converted into a liquid solution. However, ohydr.S is most negative for Li+. The ohydr.S values increase as we move down the group from Li+ to Cs+. As the charge density and polarizing power of the metal cation decreases, so too does the degree of order (or organization) in the hydration sphere. Thus, the entropy change should be most negative for Li+ and least negative for Cs+.

61. (M) (a) Use the Kapustinskii equation in the form 120, 200 34.51z zUr r r r

, with

r+ and r in picometers, to calculate U in kJ mol1. For LiO2, z+ = +1, z = 1, and = 2. (b) When we apply the BornFajansHaber cycle to the reaction Li(s) + O2(g) LiO2(s), we get ofH = of,Li(g)H + IE(1)Li 43 + U. Values of of,Li(g)H and IE(1)Li are given in Appendix D and in Table 21.2, respectively. (c) For the reaction 2 LiO2(s) Li2O(s) + 32 O2(g),

oH = 2

of,Li O(s)H 2 2of,LiO (s)H . (d) The decomposition reaction is exothermic

( oH < 0) and, with the assumption that entropy effects are not important, the decomposition of LiO2(s) to Li2O(s) and O2(g) is thermodynamically favorable.

62. (M) (a) The mass of MgO(s) that would be produced can be determined with information

from the balanced chemical equation for the oxidation of Mg(s).

MgO(s) 2 )g(OMg(s) 2 2

MgO g 332.0MgO mol 1MgO g 30.40

Mg mol 2MgO mol 2

Mg g 305.24Mg mol 1Mg g 0.200MgO mass

(b) If the mass of product formed from the starting Mg differs from the 0.332 g MgO predicted, then the product could be a mixture of magnesium nitride and magnesium oxide. This scenario is not implausible since molecular nitrogen is readily available in the atmosphere. (s)NMg(g)NMg(s) 3 232

mass Mg3N2 =0.200 g Mg 1 mol Mg24.305 g Mg

1 mol Mg3N23 mol Mg

100.93 g Mg3N21 mol Mg3N2

=0.277 g Mg3N2


1084

We can use a technique similar to determining the percent abundance of an isotope. Let f = the mass fraction of MgO in the product. The (1.000 f) is the mass fraction of Mg3N2 in the product. product mass = mass MgO f + [mass Mg3N2 (1.000 f)] 0.315 g product = 0.332 f + [0.277 (1.000 f)] = 0.277 + f(0.332 0.277)

69.0277.0332.0277.0315.0

f The product is 69% by mass MgO.

63. (M) Reaction (21.4) is K(g) NaCl(l) Na(l)KCl(l) C850 It is practical when the reactant element is a liquid and the product element is a gas at the same temperature.

(a) Since Li has a higher boiling point (1347C) than K (773.9C), a reaction similar to (21.4) is not a feasible way of producing Li metal from LiCl.

(b) On one hand, Cs has a lower boiling point (678.5C) than K (773.9C), and thus a reaction similar to (21.4) is a feasible method of producing Cs metal from CsCl. However, the ionization energy of Na is considerably larger than that of Cs, making it difficult to transfer an electron from Na to Cs+.

64. (M) (a) 2 Al(s) + Fe2O3(s) 2 Fe(s) + Al2O3(s) H = -852 kJ (b) 4 Al(s) + 3 MnO2(s) 2 Al2O3(s) + 3 Mn(s) H = -1792 kJ (c) 2 Al(s) + 3 MgO(s) Al2O3(s) + 3 Mg(s) H = +129 kJ

65. (M) Li+(aq) + e- Li(s) G = -293.3 kJ = -nFE = -1 mol e- (96,485 C/mol e-)E E = -3.040 V (this value is the same as the one that appears in Table 21.2.)

66. (M) The partial pressure of H2 is 748 mmHg 21 mmHg = 727 mmHg

amount H2 P VR T

727 mm Hg (1 atm/760 mmHg) 0.104 L0.08206Latm K-1 mol-1 296 K =4.10 10

3 mol H2

The electrode reactions are the following.

)(H aq)(OH 2e 2O(l)H 2 :Cathode e 2g)(Claq)(Cl 2 :Anode 222 g Thus 2 mol OH(aq) are produced per mole of H2(g); 8.20 103 mol OH(aq) are produced.

M 0.0328M 1028.3soln L 250.0

OH mol1020.8]OH[ 2

3

Then we compute the ion product and compare its value to the value of Ksp for Mg(OH)2. 2sp

114222sp Mg(OH)for 108.11037.2)0328.0)(220.0(][OH]Mg[ KQ

Thus, Mg(OH)2 should precipitate.


1085

67. (D) (a) We determine the amount of Ca2+ associated with each anion in 106 g of the water.

mol 44.1

HCO mol 2Ca mol 1

HCO g 02.61

HCO mol 1HCO g 176)(HCO Caamount

mol 592.0SO mol 1Ca mol 1

SO g 06.96SO mol 1

SO g 9.56)(SO Caamount

3

2

3

333

2

24

2

24

242

42

42

Then we determine the total mass of Ca2+, numerically equal to the ppm Ca2+.

222

222 Ca ppm 4.81Ca g 4.81

Ca mol 1Ca g 08.40Ca mol )44.1592.0(Ca mass

(b) The reactions for the removal of HCO3(aq) begin with the formation of hydroxide ion resulting from dissolving CaO(s). Hydroxide ion reacts with bicarbonate ion to form carbonate ion, which then combines with calcium ion to form the CaCO3(s) precipitate.

aq)(OH 2(aq)CaO(l)HCaO(s) 22

s)(CaCO aq)(COaq)(Ca

aq)(COO(l)Haq)(HCOaq)(OH

32

32

2323

For each 1.000 106 g of water, we need to remove 176 g HCO3(aq) with the added CaO(s).

CaO g 7.48CaO mol 1CaO g 08.56

OHmol 2CaO mol 1

HCO mol 1OH mol 1

HCO g 02.61HCO mol 1

waterg 10000.1HCO g 176

waterg 10602CaO mass33

36

33

(c) Here we determine the total amount of Ca2+ in 602 kg of water, and that added as CaO.

Ca2+ =

602 kg1 106 kg

(0.592+1.44) mol Ca

2+ + 48.7 g CaO 1 mol CaO56.08 g CaO

1 mol Ca2+

1 mol CaO

= 2.09 mol Ca

2+

The amount of Ca2+ that has precipitated equals the amount of HCO3 in solution, since each mole of HCO3 is transformed into 1 mole of CO32, which reacts with and then precipitates one mole of Ca2+. Thus the amount of Ca2+ that has precipitated is (0.602)(1.44) = 0.867 mol Ca2+ as CaCO3(s). Then we can determine the concentration of Ca2+ remaining in solution.

M 1003.2 waterL 1 waterg 10

waterg 10Ca mol 867.0 totalCa mol 2.09]Ca[ 3

3

6

222

To consider the Ca2+ removed, its concentration should be decreased to 0.1% (0.001) of its initial value, or 2.03 106 M. We use the Ksp expression for CaCO3 to determine the needed [CO32]

2 2 9 6sp 3

92

3 6

[Ca ][CO ] 2.8 10 1.46 10 M

2.8 10 [CO ] 0.0014 M2.03 10

K


1086

This carbonate ion concentration can readily be achieved by adding solid Na2CO3.

(d) The amount of CO32 needed is that which ends up in the precipitate plus that needed to attain the 0.0014 M concentration in the 602 kg = 602 L of water.

n

CO3-2 = 602 L 0.00203 mol Ca

2+

1 L water 1 mol CO3

2-

1 mol Ca2+

+

0.0014 mol CO32-

1 L water

= 2.1 mol CO32-

This CO32 comes from the added Na2CO3(s).

322

32

32

3

322332 CONa g 102.2CONamol 1

CONa g 99.105CO mol 1

CONa mol 1CO mol 0.2CONa mass

2

68. (M) (a) In the cell reaction, three moles of electrons are required to reduce each mole of Al 3+.

mass Al 8.00 h 3600 s

1 h 1.00 10

5 C1 s

1 mol e

96,485 C 1 mol Al

3 mol e 26.98 g Al

1 mol Al 2.68105 g Al

The 38% efficiency is not considered in this calculation. All of the electrons produced must pass through the electrolytic cell; they simply require a higher than optimum voltage to do so, leading to resistance heating (which consumes some of the electrical energy).

(b) total energy 4.5 V 8.00 h 3600 s1 h

1.00 105 C

1 s 1 J

1 V C 1 kJ

1000 J 1.3107 kJ

mass coal = 1.3107 kJ electricity 1 kJ heat0.35 kJ electricity

1 g C32.8 kJ

1 g coal0.85 g C

1 metric ton106g

mass coal = 1.3 metric tons of coal

69. (M) We begin by rewriting Equation 21.25 for the electrolysis of Al2O3(s) with n = 12 e-.

3 C(s) + 2 Al2O3(s) 4 Al(s) + 3 CO2(g) G = 4(0 kJ/mol) + 3(-394 kJ) [3(0 kJ/mol) + 2(-1520 kJ/mol)] = +1858 kJ G = 1.858 106 J = -nFE = -12 mole e-(96,485 C/mol e-)E E = -1.605 V Note: this is just an estimate because G values are at 298 K, whereas

the reaction occurs at a temperature that is much higher than 298 K.

If the oxidation of C(s) to CO2(g) did not occur, then the cell reaction would just be the reverse of the formation reaction of Al2O3 with n = 6 e-.

E = Eo = G

o

nF= 1.520 10

6 J6mole- 96485C/mole- -2.626V

70. (M) It would not be unreasonable to predict that Raoults law holds under these circumstances. This is predicated on the assumption, of course, that Pb(NO3)2 is completely ionized in aqueous solution.


1087

Pwater xwater Pwater xwater PwaterPwater 97% 0.97

Thus, there are 97 mol H2O in every 100 mol solution. The remaining 3 moles are 1 mol Pb2+ and 2 mol NO3. Thus there is 1 mol Pb(NO3)2 for every 97 mol H2O. Compute the mass of Pb(NO3)2 in 100 g H2O.

massPb(NO3 )2 = 100.0 g H2O

1 mol H2O

18.02 g H2O

1 mol Pb(NO3 )297 mol H

2O

331.2 g Pb(NO3 )21 mol Pb(NO

3)

2

= 19 g Pb(NO3)

2

If we did not assume complete ionization of Pb(NO3)2, we would obtain 3 moles of un-ionized Pb(NO3)2 in solution for every 97 moles of H2O. Then there would be 57 g Pb(NO3)2 dissolved in 100 g H2O. A handbook gives the solubility as 56 g Pb(NO3)2/100 g H2O, indicating only partial dissociation or, more probably, extensive re-association into ion pairs, triplets, quadruplets, etc.

71. (M) Considerable energy is required to produce the Pb4+ cationfour ionization steps, one for the removal of each electron. It therefore needs quite a large lattice energy to compensate for its energy of production. Both Br and I are large anions, and therefore the Pb4+Br and Pb4+I interionic distances are long. But lattice energy depends on the charge of the cation (which is quite large) multiplied by the charge of the anion (which is reasonably small) divided by the square of the interionic distance (long, as we have said). Thus, we predict a small lattice energy that is insufficient to stabilize the Pb4+ cation. We would predict, however, that PbF4 and PbCl4 (both with small anions) and PbO2 and PbS2 (both with small, highly charged anions) would be stable compounds. In addition, notice that Eo {Pb4+|Pb2+} = 1.5 V (from Table 21-6) is sufficient to oxidize Br to Br2 and I to I2, since Eo {Br2|Br} = +1.065 V and Eo {I2|I} = +0.535 V. Thus, even if PbI4 or PbBr4 could be prepared they would be thermodynamically unstable.

72. (M) Na2B4O7.10 H2O(s) + 6 CaF2(s) + 8 H2SO4(aq) 4 BF3(aq) + 6 CaSO4(aq) + 17 H2O(l) + 2 NaHSO4(aq)

73. (D) (a) 1.00 M NH4Cl is a somewhat acidic solution due to hydrolysis of NH4+(aq); MgCO3 should be most soluble in this solution. The remaining two solutions are buffer solutions, and the 0.100 M NH31.00 M NH4Cl buffer is more acidic of the two. In this solution, MgCO3 has intermediate solubility. MgCO3 is least soluble in the most alkaline solution, namely, 1.00 M NH31.00 M NH4Cl.

2 2 83 3 sp

54 3 2

2 113 3 3 2 2

MgCO (s) Mg (aq) CO (aq) 3.5 10

NH (aq) OH (aq) NH (aq) H O(l) 1/ 1/1.8 10

CO (aq) H O (aq) HCO (aq) H O(l) 1/ 1/ 4.7 10b

K

K

K

14

2 3

23 4 3 3

2 H O(l) H O (aq) OH (aq) 1.0 10

MgCO (s) NH (aq) Mg HCO ( ) NH (aq)

wK

aq


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8 14

sp w 75 11

b 2

K K 3.5 10 1.0 10K 4.1 10K K 1.8 10 4.7 10

In 1.00 M NH4Cl

23 4 3 3Reaction: MgCO (s) NH (aq) Mg (aq) HCO (aq) NH (aq)

Initial: 1.00 M 0 M 0 M 0 MChanges: M M M x x x x

2 33 3

4

3 7 33

MEquil: (1.00 )M M M M

[Mg ][HCO ][NH ] [NH ] 1.00 1.00

4.1 10 7.4 10 M molar solubility of MgCO in1.0

x x x xx x x xK

x

x

40 M NH Cl( 1.00 M, thus the approximation was valid)x

(b) In 1.00 M NH31.00 M NH4Cl

23 4 3 3 Reaction: MgCO (s) NH (aq) Mg (aq) HCO (aq) NH (aq)

Initial: 1.00 M 0 M 0 M 1.00 M Changes: M M M M Equil: (1.00

x x x x

2 23 3

4

)M M M (1.00 )M[Mg ][HCO ][NH ] (1.00 ) 1.00

[NH ] 1.00 1.00

x x x xx x x xK

x

7 43 3 4 4.1 10 6.4 10 M molar solubility of MgCO in 1.00 M NH -1.00 M NH Clx

( 1.00 M, thus the approximation was valid)x

(c) In 0.100 M NH31.00 M NH4Cl 2

3 4 3 3 Reaction: MgCO (s) NH (aq) Mg (aq) HCO (aq) NH (aq) Initial: 1.00 M 0 M 0 M 0.100 M

Changes: M M M M Equil: (1.00 )M M M (0.100 )M

x x x xx x x x

2 23 3

4

73

3 3 4

[Mg ][HCO ][NH ] (0.100 ) 0.100 1.00 1.00[NH ]

4.1 10 2.0 10 M solubility of MgCO in 0.100 M NH -1.00 M NH Cl0.100

( is 2 % of 0.100 M, so

x x x xKx

x

x

the approximation was valid)


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74. (M) If the value of the equilibrium constant for the reaction is quite large, then the conversion will be almost complete. We write the net ionic equation and then add the two solubility reactions to obtain the net ionic equation for the conversion. The equilibrium constants are multiplied together to give the equilibrium constant for the conversion.

22 3 3

2 62 sp

2 2 93 3 sp

22 3

Net ionic equation: Ca(OH) (s) CO (aq) CaCO (s) 2 OH (aq)

Ca(OH) (s) Ca (aq) 2 OH (aq) 5.5 10

Ca (aq) CO (aq) CaCO (s) 1/ 1/ 2.8 10

Ca(OH) (s) CO (aq

K

K

6

33 ovedrall 9

5.5 10) CaCO (s) 2 OH (aq) 2.0 102.8 10

K

Because Koverall > 1000, the conversion is substantially complete.

75. (M) The density of each metal depends on two factors: its atomic size and the masses of the individual atoms. Density is thus, proportional to the atomic mass divided by the cube of the atomic radius. We calculate this ratio for each of the elements.

3 3 3

3 3

atomic mass atomic mass atomic massLi: Na: K: (atomic radius) (atomic radius) (atomic radius)

6.941 (g/mol) 22.99 (g/mol) 1.98 3.57 [1.52 (A)] [1.86 (A)]

339.10 (g/mol) 3.34[2.27 (A)]

We see that the calculated ratios follow the same pattern as the densities: Na K Li, thus explaining why Na has a higher density than both Li and K.

76. (M) We would expect MgO(s) to have a larger value of lattice energy than MgS(s) because of the smaller interionic distance in MgO(s).

Lattice energy of MgO: Enthalpy of formation: MgO(s)(g)OMg(s) 2 kJ1002.6 2f H Sublimation of Mg(s): (g)MMg(s) g kJ146 subH Ionization of Mg(g)

Ionization of Mg(g):

e(g)MMg(g) g e(g)M(g)Mg 2g

kJ7.7371 I kJ14512 I

21 Dissociation O2(g): O(g)(g)O221 kJ7.4824.974DE 21 O(g) electron affinity: (g)OeO(g) kJ411EA1 O(g) electron affinity: (g)Oe(g)O 2- kJ744EA2 Lattice energy: MgO(s)(g)O(g)M 22 g L.E = -3789 kJ(see below)


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L.E = -602 kJ 146 kJ 737.7 kJ 1451 kJ 248.7 kJ + 141 kJ 744 kJ = -3789 kJ

Lattice energy of MgS: Enthalpy of formation: MgS(s)S(g)Mg(s) kJ1046.3 2f H Sublimation of Mg(s): (g)MMg(s) g kJ146 subH Ionization of Mg(g)

Ionization of Mg(g):

e(g)MMg(g) g e(g)M(g)Mg 2g

kJ7.7371 I kJ14512 I

21 Dissociation of S2(g): S(g)(g)Srhombic21 kJ8.7826.557DE 21 S(g) electron affinity: (g)SeS(g) kJ4.200EA1 S(g) electron affinity: (g)Se(g)S 2- kJ456EA2 Lattice energy: 2+ 2-Mg (g) + S (g) MgS(s) L.E = -3215 kJ

L.E = -346 kJ 146 kJ 737.7 kJ 1451 kJ 278.8 kJ + 200.4 kJ 456 kJ = -3215 kJ

77. (M) In one unit cell, there are 18

8 + 12

6 = 4 fulleride ions and 14

12 + 1 + 8 = 12

alkali metal ions. The ratio of cations to anions is 12:4 = 3:1, and so the fulleride ion is C603 and the empirical formula is M3C60.

FEATURE PROBLEMS 78. (D) Li(s) Li(g) 159.4 kJ; (UsingHrxn = Hfproducts Hfreactants in Appendix D) Li(g) Li+(g) + e 520.2 kJ; (Data given in Table 21.2, Chapter 21)

Li+(g) Li+(aq) 506 kJ; (Provided in the question) Li(s) Li+(aq) + e 174 kJ

1/2 H2(g) H(g) 218.0 kJ (UsingHrxn = Hfproducts Hfreactants in Appendix D) H(g) H+(g) + e 1312 kJ (Use RH(NA)Bohr theory = 2.179 1018 J(6.022 1023)) H+(g) H+(aq) 1079 kJ (Provided in the question) 1/2 H2(g) H+(aq) + e 451 kJ (a) Li(s) + H+(aq) Li+(aq) + 1/2 H2(g) H = 174 kJ 451 kJ = 277 kJ G

Eox = o 3277 10 J J= 2.87 = 2.87 V

C C1 mol e 96,485mol e

GnF


1091

In this reaction, Li(s) is being oxidized to Li+(aq). If we wish to compare this to the reduction potential for Li+(aq) being reduced to Li(s), we would have to reverse the reaction, which would result in the same answer only as a negative value. Alternatively, we can change the sign of the oxidation potential (Li Li+) to -2.87 V for the reduction potential for Li+ Li.

(it apears as -3.040 V in Appendix D; here, we see much better agreement.) (b) S = Sproducts - Sreactants

= [1 mol JKmol

13.4 + 0.5 molJ

Kmol130.7 ]

[1 mol JKmol

29.12 + 1 mol J

Kmol0 ] = 49.6

JK

G = H TS = 277 kJ 298.15 K J49.6K

1kJ

1000J 292 103 J

Eox = o 3292 10 J J= 3.03 = 3.03 V

C C1 mol e 96,485mol e

GnF

As mentioned in part (a) of this question, we have calculated the oxidation potential for the half reaction Li(s) Li+(aq) + e-. The reduction potential is the reverse half-reaction and has a potential of -3.03 V. This is in excellent agreement with -3.040 V given in Appendix D.

79. (D) (a)

0. 438 mol NaClL

58.443 g NaCl1 mol NaCl

= 26.6 g NaCl

20. 0512 mol MgClL

22

95.211 g MgCl1 mol MgCl

= 4.87 g MgCl2

18 tbs NaHCO3 33

10 g NaHCO1 tbs NaHCO

= 180 g NaHCO3

10 tbs NaCl 10 g NaCl1 tbs NaCl

= 100 g NaCl

8 tsp MgSO4 7 H2O 4 24 2

1 tbs 10 g MgSO 7 H O3 tsp 1 tbs MgSO 7 H O

= 27 g MgSO4 7 H2O

(b) The pH of the lake will be determined by the amphiprotic bicarbonate ion, (HCO3),

which hydrolyzes in water (1a

K = 4.4 107 or 1apK = 6.36 and 2aK = 4.7 1011 or

2apK = 10.33). We saw earlier (Chapter 18, Exercise 100) that the pH of a

solution of alanine, a diprotic species is independent of concentration (as long as it is

31.5 g of salt per liter of seawater

307 g of salt per gal. of lake water (3.7854 L/gal)


1092

relatively concentrated). The pH = 1 2a a(p p )2

K K = (6.36 10.33)2 = 8.35

This is not as basic as the actual pH of the lake. Addition of borax (sodium salt of boric acid) would aid in increasing the pH of the solution (since borax contains an anion that is the conjugate base of a weak acid). The lake may be more basic due to the presence of other basic anions, namely carbonate ion (CO32).

(c) Tufa are mostly made of calcium carbonate (CaCO3(s)). Since they form near

underwater springs, one must assume that the springs have a high concentration of calcium ion (Ca2+(aq)). We then couple this with the fact that the lake has a high salinity (high carbonate and bicarbonate ion content). We can thus conclude that two major reactions are most likely responsible for the formation of a tufa (see below):

Ca2+(aq) + CO32(aq) CaCO3(s) Ca2+(aq) + 2 HCO3(aq) Ca(HCO3)2(q) CaCO3(s) + H2O(l) + CO2(g)

SELF-ASSESSMENT EXERCISES 80. (E) (a) A dimer is a chemical or biological entity consisting of two structurally similar

subunits called monomers, which are joined by bonds, which can be strong or weak. (b) An adduct is a product of a direct addition of two or more distinct molecules, resulting in a single reaction product containing all atoms of all components, with formation of two chemical bonds and a net reduction in bond multiplicity in at least one of the reactants. (c) Calcination is a thermal treatment process applied to ores and other solid materials in order to bring about a thermal decomposition, phase transition, or removal of a volatile fraction. (d) An amphoteric oxide is an oxide that can act both as an acid and as a base. An example is aluminum oxide Al2O3. (e) A three-center two-electron bond is an electron deficient chemical bond where three atoms share two electrons. The combination of three atomic orbitals form three molecular orbitals: one bonding, one non-bonding, and one anti-bonding. The two electrons go into the bonding orbital, resulting in a net bonding effect and constituting a chemical bond among all three atoms.

81. (E) (a) A diagonal relationship exists between certain pairs of diagonally adjacent

elements in the second and third periods of the periodic table. These pairs (Li and Mg, Be and Al, B and Si, etc.) exhibit similar properties. For example boron and silicon are both semiconductors. (b) One way to prepare deionized water is by passing it through an ion exchange column. The latter is composed of insoluble chemicals which remove both positive and negative ions from water. (c) Thermite is a pyrotechnic composition of a metal powder and a metal oxide, which produces an aluminothermic reaction known as a thermite reaction.


1093

(d) The inert pair effect is the tendency of the outermost s electrons to remain un-ionized or unshared in post-transition compounds.

82. (E) (a) A peroxide is a compound containing an oxygen-oxygen single bond. Superoxide

is an anion with the chemical formula O2- .

(b) Calcium oxide (CaO) is commonly known as burnt lime, lime or quicklime. Calcium hydroxide (Ca(OH)2) on the other hand is commonly known as slaked lime, hydrated lime, or slack lime. (c) Soap is an anionic surfactant used in conjunction with water for washing and cleaning. Soap consists of sodium or potassium salts of fatty acids and is obtained by reacting common oils or fats with a strong alkaline solution in a process known as saponification. Detergents, on the other hand, are prepared synthetically. (d) A silicate is a compound containing an ion in which one or more central silicon atoms are surrounded by electronegative ligands. Silicones are compounds containing silicon-oxygen and silicon-carbon bonds that have the capability to act as bonding intermediates between glass and organic compounds, and to form polymers with useful properties such as impermeability to water, flexibility and resistance to chemical attack. (e) The sol-gel process is a chemical solution deposition technique widely used in the fields of materials science and ceramic engineering. Typical precursors are metal alkoxides and metal chlorides, which undergo various forms of hydrolysis and polycondensation reactions.

83. (E) (c) MgO 84. (E) (f) PbO2 85. (M) (a) BCl3NH3(g) (an adduct); (b) KO2(s); (c) Li2O(s); (d) Ba(OH)2(aq) and H2O2(aq);

H2O2(aq) slowly disproportionates into H2O(l) and O2(g).

86. (M) The thermite reaction is evidence that aluminum will readily extract oxygen from

Fe2O3. Aluminum can be used for making products that last and for structural purposes,

because aluminum develops a coating of Al2O3 that protects the metal beneath it.

87. (M) (b) Ca(s) and CaH2(s) 88. (M) (a) Li2CO3(s)

Li2O(s)+CO2 (g) (b) CaCO3(s)+2HCl(aq) CaCl2 (aq)+H2O(l)+CO2 (g) (c) 2Al(s)+2NaOH(aq)+6H2O(l) 2Na[Al(OH)4 ](aq)+2H2 (g) (d) BaO(s)+H2O(l) Ba(OH)2 (aq,limited solubility) (e) 2Na2O2 (s)+2CO2 (g) 2Na2CO3(s)+O2 (g)


1094

89. (M) (a) MgCO3(s)+2HCl(aq) MgCl2 (aq)+H2O(l)+CO2 (g) (b) 2Na(s)+2H2O(l) 2NaOH(aq)+H2 (g) followed by the reaction in Exercise 82(c) (c) 2NaCl(s)+H2SO4 (concd.,aq)

Na2SO4 (s)+2HCl(g) 90. (M) (a) K2CO3(aq)+Ba(OH)2 (aq) BaCO3(s)+2KOH(aq) (b) Mg(HCO3)2 (aq)

MgCO3(s)+H2O(l)+CO2 (g) (c) SnO(s)+C(s) Sn(l)+CO(g) (d) CaF2 (s)+H2SO4 (concd. aq)

CaSO4 (s)+2HF(g) (e) NaHCO3(s)+HCl(aq) NaCl(aq)+H2O(l)+CO2 (g) (f) PbO2 (s)+4HBr(aq) PbBr2 (s)+Br2 (l)+2H2O(l) (g) SiF4 (g)+4Na(l) Si(s)+4NaF(s) 91. (M) CaSO4 2H2O(s)+(NH4 )2CO3(aq) CaCO3(s)+(NH4 )2SO4 (aq)+2H2O(l) . The

reaction proceeds to the right because Ksp for CaCO3 /JPEG2000ColorACSImageDict > /JPEG2000ColorImageDict > /AntiAliasGrayImages false /CropGrayImages true /GrayImageMinResolution 300 /GrayImageMinResolutionPolicy /OK /DownsampleGrayImages true /GrayImageDownsampleType /Bicubic /GrayImageResolution 300 /GrayImageDepth -1 /GrayImageMinDownsampleDepth 2 /GrayImageDownsampleThreshold 1.50000 /EncodeGrayImages true /GrayImageFilter /DCTEncode /AutoFilterGrayImages true /GrayImageAutoFilterStrategy /JPEG /GrayACSImageDict > /GrayImageDict > /JPEG2000GrayACSImageDict > /JPEG2000GrayImageDict > /AntiAliasMonoImages false /CropMonoImages true /MonoImageMinResolution 1200 /MonoImageMinResolutionPolicy /OK /DownsampleMonoImages true /MonoImageDownsampleType /Bicubic /MonoImageResolution 1200 /MonoImageDepth -1 /MonoImageDownsampleThreshold 1.50000 /EncodeMonoImages true /MonoImageFilter /CCITTFaxEncode /MonoImageDict > /AllowPSXObjects false /CheckCompliance [ /None ] /PDFX1aCheck false /PDFX3Check false /PDFXCompliantPDFOnly false /PDFXNoTrimBoxError true /PDFXTrimBoxToMediaBoxOffset [ 0.00000 0.00000 0.00000 0.00000 ] /PDFXSetBleedBoxToMediaBox true /PDFXBleedBoxToTrimBoxOffset [ 0.00000 0.00000 0.00000 0.00000 ] /PDFXOutputIntentProfile (None) /PDFXOutputConditionIdentifier () /PDFXOutputCondition () /PDFXRegistryName () /PDFXTrapped /False

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Petrucci Chap 21 Answers