resolução cap 10 atkins

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CHAPTER 10

ACIDS AND BASES

10.1 (a) (b)3 3CH NH + 2 3NH NH + (c) (d)2H CO3 23CO (e) (f)6 5C H O

3 2CH CO

10.3 For all parts (a) (e), H2O and H3O+

form a conjugate acid-base pair in

which H2O is the base and H3O is the acid

(a) +2 4 2 3 4H SO (aq) H O(l) H O (aq) HSO (aq)+ +

2 4 4H SO and HSO (aq) form a conjugate acid-base pair in which

2 4 4H SO is the acid and HSO (aq) is the base.

(b)6 5 3 2 3 6 5 2C H NH (aq) H O(l) H O (aq) C H NH (aq)

+ ++ +

6 5 3 6 5 2C H NH and C H NH (aq)+ form a conjugate acid-base pair in which

6 5 3 6 5 2C H NH is the acid and C H NH (aq) is the base.+

(c) 22 4 2 3 4H PO (aq) H O(l) H O (aq) HPO (aq) ++ +

22 4 4H PO (aq) and HPO (aq)

form a conjugate acid-base pair in which

2

2 4 4H PO (aq) is the acid and HPO (aq) is the base.

(d) 2 3 2HCOOH(aq) H O(l) H O (aq) HCO (aq)+ + +

2HCOOH(aq) and HCO (aq) form a conjugate acid-base pair in which

2HCOOH(aq) is the acid and HCO (aq) is the base.

(e) 2 3 2 3 2 2NH NH (aq) H O(l) H O (aq) NH NH (aq)+ ++ +

2 3 2 2NH NH (aq) and NH NH (aq)+ form a conjugate acid-base pair in

which 2 3 2 2NH NH (aq) is the acid and NH NH (aq) is the base.+

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10.5 (a) Brnsted acid: 3HNO

Brnsted base: 24HPO

(b) conjugate base to 3 3HNO :NO

conjugate acid to 24 2 4HPO :H PO

10.7 (a) as an acid:

HCO

3HCO , 2

3 2 3 3HCO (aq) H O(l) H O (aq) CO (aq) ++ + .

3

4

3-and CO3

2-form a conjugate acid-base pair in which HCO3

-is the

acid and CO32-

is the base.

3HCO , as a base: .

CHO

2 3 2 3H O(l) HCO (aq) H CO (aq) OH (aq) + +

3-and H2CO3 form a conjugate acid-base pair in which HCO3

-is the

base and H2CO3 is the acid. H2O and OH-form a conjugate acid-base pair

in which H2O is the acid and OH-is the base.

(b) , as an acid:

.

form a conjugate acid-base pair in which

H

2

4HPO

2 3

4 2 3 4HPO (aq) H O(l) H O (aq) PO (aq) ++ +

2

4 4HPO (aq) and PO (aq)

2 3

4 4HPO (aq) is the acid and PO (aq) is the base.

2O and H3O+

form a

conjugate acid-base pair in which H2O is the base and H3O+ is the acid.2

4HPO , as a base: .

form a conjugate acid-base pair in which HPO

2

4 2 2 4HPO (aq) H O(l) H PO (aq) OH (aq) + +

2

4 2HPO and H PO

42-

is

the base and H2PO4-is the acid. H2O and OH

-form a conjugate acid-base

pair in which H2O is the acid and OH-is the base.

10.9 (a) basic; (b) acidic; (c) amphoteric; (d) basic

10.11 (a) ;3 2 4 2 2 7SO (g) H SO H S O (l)+

H

S+

O

O

O

O

S+

O

O

O

H

(b)

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(c) sulfuric acid acts as the Lewis base while SO3 acts as a Lewis acid.

10.13 In each case, use then14w 3[H O ][OH ] 1.0 10 ,K+ = =

14

w

3 3

1.0 10[OH ]

[H O ] [H O ]

K + +

= =

(a)14

13 11.0 10[OH ] 5.0 10 mol L0.02

= =

(b)14

9 1

5

1.0 10[OH ] 1.0 10 mol L

1.0 10

= =

(c)

1412 1

3

1.0 10

[OH ] 3.2 10 mol L3.1 10

= =

10.15 (a) 14 2w 3 32.1 10 [H O ][OH ] , where [H O ] [OH ] K x x + += = = = =

14 7 1

3

2.1 10 1.4 10 mol L

pH log[H O ] 6.80

x

+

= =

= =

=

(b)

7 1

3[OH ] [H O ] 1.4 10 mol L

pOH log[OH ] 6.80

+

= =

= =

10.17 Because is a strong base,2Ba(OH)

2

2Ba(OH) (aq) Ba (aq) 2 OH (aq), 100%.+ +

Then

2

2 0 2 0 2 0[Ba(OH) ] [Ba ], [OH ] 2 [Ba(OH) ] , where [Ba(OH) ]+ = =

nominal concentration of 2Ba(OH) .

moles of 32 10.25 g

Ba(OH) 1.5 10 mol171.36 g mol

= =

32 1

2 0

1.5 10 mol[Ba(OH) ] 1.5 10 mol L [Ba ]

0.100 L

2 += = =

2 1 2 1

2 0[OH ] 2 [Ba(OH) ] 2 1.5 10 mol L 2.9 10 mol L = = =

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1413 1w

3 2

1.0 10[H O ] 3.4 10 mol L

[OH ] 2.9 10

K +

= = =

10.19 Because if Taking the antilogs

of both sides gives

3 3pH log[H O ], log[H O ] pH.+ += =

H 1

3[H O ] 10 mol L+ =

(a) 3.3 4 13[H O ] 10 5 10 mol L+ = =

(b) 6.7 1 7 13[H O ] 10 mol L 2 10 mol L

+ = =

(c) 4.4 1 5 13[H O ] 10 mol L 4 10 mol L+ = =

(d) 5.3 1 6 13[H O ] 10 mol L 5 10 mol L

+ = =

10.21 (a) 13 3[HNO ] [H O ] 0.0146 mol L+ = =

pH log(0.0146) 1.84, pOH 14.00 ( 1.84) 12.16= = = =

(b) 13[HCl] [H O ] 0.11 mol L+ = =

pH log(0.11) 0.96, pOH 14.00 0.96 13.04= = = =

(c) M2[OH ] 2 [Ba(OH) ] 2 0.0092

= = 10.018 mol L=

pOH log(0.018) 1.74, pH 14.00 1.74 12.26= = = =

(d)0[KOH] [OH ]

=

1 42.00 mL[OH ] (0.175 mol L ) 7.0 10 mol L

500 mL

= =

1

4pOH log(7.0 10 ) 3.15, pH 14.00 3.15 10.85= = = =

(e) 0[NaOH] [OH ]=

4

1

0.0136 gnumber of moles of NaOH 3.40 10 mol

40.00 g mol

= =

44 1

0

3.40 10 mol[NaOH] 9.71 10 mol L [OH ]

0.350 L

= = =

4pOH log(9.71 10 ) 3.01, pH 14.00 3.01 10.99= = = =

(f) 0 3[HBr] [H O ]+=

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4 1 5

3

75.0 mL[H O ] (3.5 10 mol L ) 5.3 10 mol L

500 mL

+ 1 = =

5pH log(5.3 10 ) 4.28, pOH 14.00 4.28 9.72= = = =

10.23 therefore, after taking antilogs,a1 a1p log ;K = Ka1p

a1 10K

K=

Acid a1pK a1K

(a) 2.123H PO437.6 10

(b) 2.003H PO321.0 10

(c) 2.462H SeO333.5 10

(d) 1.924HSeO

2

1.2 10

(e) The larger , the stronger the acid; thereforea1K

2 3 3 4 3 3 4H SeO H PO H PO HSeO< < <

10.25 (a) 2 2 3 2HClO (aq) H O(l) H O (aq) ClO (aq)+ + +

3 2

a

2

[H O ][ClO ]

[HClO ]K

+

=

2 2 2ClO (aq) H O(l) HClO (aq) OH (aq) + +

2

b

2

[HClO ][OH ]

[ClO ]K

=

(b)2 3HCN(aq) H O(l) H O (aq) CN (aq)

+ + +

3

a

[H O ][CN ]

[HCN]K

+

=

2CN (aq) H O(l) HCN(aq) OH (aq) + +

b

[HCN][OH ]

[CN ]K

=

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(c) 6 5 2 3 6 5C H OH(aq) H O(l) H O (aq) C H O (aq)+ + +

3 6 5

a

6 5

[H O ][C H O ]

[C H OH]K

+

=

6 5 2 6 5C H O (aq) H O(l) C H OH(aq) OH (aq)

+ +

6 5

b

6 5

[C H OH][OH ]

[C H O ]K

=

10.27 Decreasing will correspond to increasing acid strength because

. The values (given in parentheses) determine the

following ordering:

apK

a ap logK K= apK

3 2 2 3(CH ) NH (14.00 3.27 10.73) NH OH (14.00 7.97 6.03)+ + = < =

2 2HNO (3.37) HClO (2.00).< > >

10.39 (a) Nitrous acid is a stronger acid than acetic acid and, therefore, the

acetate ion is a stronger base than the nitrite ion.Since the definition of a

strong acid is one that favors products in the deprotonation reaction, the

presence of acetate ions will shift the nitrous acid deprotonation reaction

toward products by consuming protons making nitrous acid behave like a

strong acid. Carbonic acid is a weaker acid and, therefore, will not behave

like a strong acid in the presence of acetic acid.

(b) Ammonia will act like a strong base because its conjugate acid, the

ammonium ion, is a weaker acid than acetic acid. The presence of acetic

acid will shift the reaction: 3 2 4NH H O NH OH+ + + toward products.

10.41 The larger the the stronger the corresponding acid. 2,4,6-

Trichlorophenol is the stronger acid because the chlorine atoms have a

greater electron-withdrawing power than the hydrogen atoms present in

the unsubstituted phenol.

a ,K

10.43 The larger the of an acid, the stronger the corresponding conjugate

base; hence, the order is aniline < ammonia < methylamine < ethylamine.

Although we should not draw conclusions from such a small data set, we

might suggest the possibility that

apK

(1) arylamines < ammonia < alkylamines

(2) methyl < ethyl < etc.

(Arylamines are amines in which the nitrogen of the amine is attached to a

benzene ring.)

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10.45 (a) Concentration

1

3 2 3(mol L ) CH COOH H O H O CH CO + + 3 2

initial 0 00.29

change x+ x+

equilibrium 0.29 x x x

2 25 3 3 2

a

3

3 1

3

3

[H O ][CH CO ]1.8 10

[CH COOH] 0.29 0.29

[H O ] 2.3 10 mol L

pH log(1.6 10 ) 2.64, pOH 14.00 2.64 11.36

x xK

x

x

+

+

= = =

= =

= = = =

(b) The equilibrium table for (b) is similar to that for (a).

21 3 3 2

a

3

2 1

1 1 2

[H O ][CCl CO ]3.0 10

[CCl COOH] 0.29

or 3.0 10 0.087 0

3.0 10 (3.0 10 ) (4) ( 0.087)0.18, 0.48

2

xK

x

x x

x

+

= = =

+ =

= =

2

The negative root is not possible and can be eliminated.

1

3[H O ] 0.18 mol L

pH log(0.18) 0.74, pOH 14.00 0.74 13.26

x + = =

= = = =

(c) Concentration 1 2 3(mol L ) HCOOH H O H O HCO + + +

initial 0 0.29 0

change x x+ x+


243 2

a

4 3 13

3

[H O ][HCO ]1.8 10

[HCOOH] 0.29 0.29

[H O ] 0.29 1.8 10 7.2 10 mol L

pH log(7.2 10 ) 2.14, pOH 14.00 2.14 11.86

x x xK

x

x

+

+

= = =

= = =

= = = =

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10.47 (a) Concentration1

2 3 4(mol L ) H O NH NH OH + + +

initial 0.057 0 0

change x

x+

x+


254

b

3

5 3 1

3

3

[NH ][OH ]1.8 10

[NH ] 0.057 0.057

[OH ] 0.057 1.8 10 1.0 10 mol L

pOH log(1.0 10 ) 3.00, pH 14.00 3.00 11.00

1.0 10percentage protonation 100% 1.8%

0.057

x x xK

x

x

+

= = =

= = =

= = = =

= =

+

(b) Concentration

1

2 2 3(mol L ) NH OH H O NH OH OH + +

initial 0.162 0 0

change x x+ x+


2 28

b

5 1

5

5

1.1 100.162 0.162

[OH ] 4.2 10 mol L

pOH log(4.2 10 ) 4.38, pH 14.00 4.38 9.62


0.162

x xK

x

x

= =

= =

= = = =

= =

(c) Concentration

1

3 3 2 3 3(mol L ) (CH ) N H O (CH ) NH OH + + +

initial 0.35 0 0

change x x+ x+equilibrium 0.35 x+ x+

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25

3 1

3 1

3

3

6.5 100.35

Assume 0.35

Then 4.8 10 mol L

[OH ] 4.8 10 mol L

pOH log(4.8 10 ) 2.32, pH 14.00 2.32 11.68


0.35

x

x

x

x

=


12/40

Concentration

+1(mol L ) HClO 2 3H O H O ClO+ +

nominal x 0 0

equilibrium

5

2.5 10x

5

2.5 10

5

2.5 10

5 2

8

a 5

(2.5 10 )3.0 10

2.5 10K

x

= =

5 2 5 8

8

2 1 1

(2.5 10 ) (2.5 10 )(3.0 10 )Solve for ;

3.0 10

2.1 10 mol L 0.021 mol L

x x

+ =

= =

(b) pOH 14.00 pH 14.00 10.20 3.80= = =

pOH 3.80 4

2 2

[OH ] 10 10 1.6 10Let nominal concentration of NH NH , thenx

= = = =

Concentration

1(mol L ) 2 2 2 2 3 NH NH H O NH NH OH+ + +

nominal 0 0x

equilibrium 41.6 10x 41.6 10 41.6 10

4 26

b 4

2 1

(1.6 10 )1.7 10

1.6 10Solve for ; 1.5 10 mol L

K

xx x

= =

=

10.53 Concentration

1

6 5 2 3 6 5 2(mol L ) C H COOH H O H O C H CO + + +

initial 0. 0 110 0

change x x+ x+


2H O octylamine octylamineH OH+ + +

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1

3 6 5 2

253 6 5

a

6 5

3

0.024 0.110 mol L [H O ] [C H CO ]

[H O ][C H COO ] (0.024 0.110)6.3 10

[C H COOH] (1 0.024) 0.110

pH log(2.6 10 ) 2.58

x

K

+

+

= = =

= = =

= =

10.55 The change in the concentration of octylamine is

10.067 0.10 0.0067 mol L .x = = Thus the equilibrium table is

Concentration

1(mol L ) 2H O octylamine octylamineH OH+ + +

initial 0.100 0 0

change 0.0067 0.0067+ 0.0067+

equilibrium 0.100 0.0067 0.0067 0.0067

The equilibrium concentrations are

[octylamine] 10.100 0.067 0.10 0.093 mol L= =

1

3 24

b

[OH ] [octylamineH ] 0.0067 mol L

pOH log(0.0067) 2.17, pH 14.00 2.17 11.83

[octylamineH ][OH ] (6.7 10 )4.8 10

[octylamine] 0.093K

+

+

= =

= = = =

= = =

10.57- +

3 2 2 3 2 3CH CH COOH(aq) H O(1) CH CH COO (aq) H O (aq)+ +

Concentration 1(mol L )

initial 0.0147 0 0

change x x+ x+

equilibrium0.0147 x x

+ 2 2

53 2 3a

3 2

[CH CH COO ][H O ] 1.3 10[CH CH COOH] 0.0147 0.0147

x xKx

= = =

44 4.4 104.4 10 , Percent deprotonated 100% 3.0%

0.0147x

= = =

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10.59 (a) 4 2 3 3less than 7, NH (aq) H O(l) H O (aq) NH (aq)+ ++ +

(b) 22 3 3greater than 7, H O(l) CO (aq) HCO (aq) OH (aq)

+ +

+

+

(c) 2greater than 7, H O(l) F (aq) HF(aq) OH (aq) + +

(d) neutral

(e) less than 7,

3 2

2 6 2 3 2 5Al(H O) (aq) H O(l) H O (aq) Al(H O) OH (aq)+ ++ +

(f) less than 7,

2

2 6 2 3 2 5Cu(H O) (aq) H O(l) H O (aq) Cu(H O) OH (aq)+ ++ +

10.61 (a)

1410w

b 5

a

1.00 105.6 101.8 10

KK K

= = =

Concentration

1 3 2 2 3 2(mol L ) CH CO (aq) H O(1) HCH CO (aq) OH (aq) + +

initial 0.63 0 0

change x+ x+


2 2

103 2b

3 2

[HCH CO ][OH ] 5.6 100.63 0.63[CH CO ]

x xKx

= = =

5 51.9 10 [OH ], pOH log(1.9 10 ) 4.72x = = = =

pH 14.00 pOH 14.00 4.72 9.28= = =

(b)14

10w

a 15

b

1.00 105.6 10

1.8 10

KK

K

= = =

1

4 2 3Concentration (mol L ) NH (aq) H O(l) H O (aq) NH (aq) + + + + 3

initial 0.19 0 0

change x x+ x+

equilibrium 0.19 x x

2 2103 3

a

4

[H O ][NH ]5.6 10

[NH Cl] 0.19 0.19

x xK

x

+= = =

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5 1

31.0 10 mol L [H O ]x = = +

+

5pH log(1.0 10 ) 5.00= =

(c) Concentration

1 3 2

2 6 2 3 2 5(mol L ) Al(H O) (aq) H O(l) H O (aq) Al(H O) OH (aq) + +

+ +

initial 0.055 0 0

change x+ x+


2 2 253 2 5

a 3

2 6

[H O ][Al(H O) OH ]1.4 10

0.055 0.055[Al(H O) ]

xK

x

+ +

+= = =

+

4 1

38.8 10 mol L [H O ]x = =

4pH log(8.8 10 ) 3.06= =

(d) Concentration

1

2(mol L ) H O(l) CN (aq) HCN(aq) OH (aq) + +

initial 0.065 0 0

change x x+ x+


14 2 2

5wb 10

a

1.00 10 [HCN][OH ]2.0 10

0.065 0.0654.9 10 [CN ]

K x xK

K x

= = = = =

3 1[OH ] 1.1 10 mol Lx = =

3pOH log(1.1 10 ) 2.96, pH 11.04= = =

10.63 Concentration

1

3 3 2 3 3 2(mol L ) CH NH (aq) H O(l) H O (aq) CH NH (aq) + + + +

initial 0.510 0 0

change x x+ x+


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2 2113 3 2

a

3 3

[H O ][CH NH ]2.8 10

0.510 0.510[CH NH ]

x xK

x

+

+= = =

+

6 1

33.8 10 mol L [H O ]x = =

6pH log(3.8 10 ) 5.4= =

10.65 (a) 250 mL of solution contains , molar

mass=

2 3 25.34 g KC H O

198.14 g mol

2 3 2

2 3 2 2 3 2

2 3 2

1 mol KC H O 1(5.34 g KC H O ) 0218 KC H O

98.14 g KC H O 0.250 LM

=

Concentration1

2 2 3 2 2 3 2(mol L ) H O(l) C H O (aq) HC H O (aq) OH (aq) + +

initial 0.218 0 0

change x x+ x+


14 2 2

5

1.0 10

0.218 0.2181.8 10

x x

x

=

5 1[OH ] 1.1 10 mol L =

10 1

3[H O ] 9.1 10 mol L+ =

10pH log(9.1 10 ) 9.04= =

(b) 100 mL of solution contains molar mass =4

5.75 g NH Br,

197.95 g mol

4

4 4

4

M1mol NH Br 1

(5.75 g NH Br) 0.587 NH Br

97.95 g NH Br 0.100 L

=

Concentration

1

4 2 3 3(mol L ) NH (aq) H O(l) NH (aq) H O (aq) + + + +

initial 0.587 0 0

change x x+ x+

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14 2 2

5

1.0 10

0.587 0.5871.8 10

x x

x

=

5 1

3[H O ] 1.8 10 mol L

+ =

5pH log(1.8 10 ) 4.74= =

10.67 (a)1

13 20.020 mol L NaCH CO 0.150 L 0.0400 mol L0.500 L

=

Concentration

1(mol L ) 2 3 2 3H O(l) CH CO (aq) CH COOH(aq) OH (aq) + +

initial 0.0400 0 0

change x+ x+


1410w 3

b 5

a 3 2

[CH COOH][OH ]1.00 105.6 10

1.8 10 [CH CO ]

KK

K

= = = =

2 2105.6 10

0.0400 0.0400

x x

x

=

6 134.7 10 mol L [CH COOH]x = =

(b) 4 43

4

2.16 g NH Br 1 mol NH Br 1 mL

400 mL 97.95 g NH Br 10 L

1

40.0551(mol NH Br) L=

Concentration

1(mol L ) 4 2 3 3NH (aq) H O(l) H O (aq) NH (aq)+ ++ +

initial 0.0551 0 0

change x+ x+


1410w 3

a 5

b 4

[NH ][H O ]1.00 105.6 10

1.8 10 [NH ]

KK

K

3

+

+

= = = =

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2 2105.6 10

0.0551 0.0551

x

x

=

6 1

35.5 10 mol L [H O ]x = = + and 6pH log(5.5 10 ) 5.26= =

10.69 (a)

C C C

H

H

H

N

H

H

O

O

Na+

(b)1 2

1 12 2

pH = + (p p ) (2.34 9.89) 6.1a aK K = + =

10.71(a) 2 4 2 3 4H SO (aq) H O(l) H O (aq) HSO (aq)

+ + +

2

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

(b)3 4 2 3 2 4H AsO (aq) H O(l) H O (aq) H AsO (aq)

+ + +

2

2 4 2 3 4

2 3

4 2 3 4

H AsO (aq) H O(l) H O (aq) HAsO (aq)

HAsO (aq) H O(l) H O (aq) AsO (aq)

+

+

+ +

+ +

(c)

6 4 2 2 3 6 4 2C H (COOH) (aq) H O(l) H O (aq) C H (COOH)CO (aq)+ + +

2

6 4 2 2 3 6 4 2 2C H (COOH)CO (aq) H O(l) H O (aq) C H (CO ) (aq) ++ +

10.73 The initial concentrations of 14 3HSO and H O are both 0.15 mol L + as a

result of the complete ionization of in the first step. The second

ionization is incomplete.

2H SO4

4

Concentration

1 2

4 2 3(mol L ) HSO H O H O SO

+ + +

initial 0.15 0.15 0

change x x+ x+

equilibrium 0.15 x 0.15 x+ x

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2

2 3 4

a2

4

2 3

2 3

1

1 1

3

[H O ][SO ] (0.15 ) ( )1.2 10

0.15[HSO ]

0.162 1.8 10 0

0.162 (0.162) (4) (1.8 10 )

0.0104 mol L2

[H O ] 0.15 (0.15 0.0104) mol L 0.16 mol L

pH log(0.16) 0.80

x xK

x

x x

x

x

+

+

+= = =

+ =

+ +

= =

= + = + =

= =

10.75 (a) Because the second ionization can be ignored.a2 a1 ,K K


20/40

(c) Because the second ionization can be ignored.a2 a1 ,K K


21/40

Concentration

1

2 3 2 3 3(mol L ) H CO (aq) H O(l) H O (aq) HCO (aq) + + +

initial 0.0456 0 0

change x x+ x+

final 0.0456 x x+ x+

3 3

a1

2 3

27

2 7

4

4 1

3 3

[H O ][HCO ]

[H CO ]

( ) ( )4.3 10

0.0456 0.0456

Assume that 0.0456

Then (4.3 10 ) (0.0456)

1.4 10Because 1% of 0.0456, the assumption was valid.

[H O ] [HCO ] 1.4 10 mol L

K

x x x

x x

x

x

xx

x

+

+

=

= =


22/40

10.83 The equilibrium reactions of interest are now the base forms of the

carbonic acid equilibria, so K values should be calculated for the

following changes:

2

3 2 3

144w

b1 11

a2

CO (aq) H O(l) HCO (aq) OH (aq)1.00 10

1.8 105.6 10

KK

K

+ +

= = =

3 2 2 3

148w

b2 7

a1

HCO (aq) H O(l) H CO (aq) OH (aq)

1.00 102.3 10

4.3 10

KK

K

+ +

= = =

L

Because the second hydrolysis constant is much smaller than the first, we

can assume that the first step dominates:

Concentration

1 2

3 2 3(mol L ) CO (aq) H O(l) HCO (aq) OH (aq) + +

initial 0.0456 0 0

change x x+ x+

final 0.0456 x+ x+

3

b1 2

3

[HCO ][OH ]

[CO ]K

=

24 ( ) ( )1.8 10

0.0456 [0.0456 ]

x x x

x

= =

Assume that 0.0456x

2 4 41.8 10 (1.8 10 ) (0.0456) 0x x + =

Solving using the quadratic equation gives 10.0028 mol L .x =

1

3[HCO ] [OH ] 0.0028 mol Lx = = =

1

Therefore, 2 1 13[CO ] 0.0456 mol L 0.0028 mol L 0.0428 mol L

= =

SM-295


23/40

We can then use the other equilibria to determine the remaining

concentrations:

2 3

b2

3

[H CO ][OH ]

[HCO ]K

=

8 2 3[H CO ](0.0028)2.3 10(0.0028)

=

8 1

2 3[H CO ] 2.3 10 mol L =

Because , the initial assumption that the first

hydrolysis would dominate is valid. To calculate

82.3 10 0.0028


24/40

2

2 3 2 3 3 a1

2 7

3 2 3 3 a2

H SO (aq) H O(l) H O (aq) HSO (aq) 1.5 10

HSO (aq) H O(l) H O (aq) SO (aq) 1.2 10

K

K

+

+

+ + =

+ + =

3

2

The calculation of the desired concentrations follows exactly after the

method derived in Eq. 25, substituting for

First, calculate the

quantity

2H SO

2

2 3 3 3 3 3H CO , HSO for HCO , and SO for CO .

5.5 6 1

3(at pH 5.50 [H O ] 10 3.2 10 mol L ) :f+ = = =

2

3 3 a1 a1 a2

6 2 6 2 2 7

8

[H O ] [H O ]

(3.2 10 ) (3.2 10 ) (1.5 10 ) (1.5 10 ) (1.2 10 )

5.0 10

f K K K + +

= + +

= + +

=

The fractions of the species present are then given by

6 243

2 3 8

6 2

3 a1

3 8

2 72 a1 a2

3 8

[H O ] (3.2 10 )(H SO ) 2.1 10

5.0 10

[H O ] (3.2 10 ) (1.5 10 )(HSO ) 0.96

5.0 10

(1.5 10 ) (1.2 10 )(SO ) 0.036

5.0 10

f

K

f

K K

f

+

+

= = =

= = =

= = =

Thus, in a 10.150 mol L solution at the dominant species will

be with a concentration of

pH 5.50,

3

HSO 1 1(0.150 mol L ) (0.96) 0.14 mol L . =

The concentration of

4 1 5

2 3H SO will be (2.1 10 ) (0.150 mol L ) 3.2 10 mol L1 = and the

concentration of 23SO

will be

1 1(0.036) (0.150 mol L ) 0.0054 mol L . =

10.89 (a) Concentration

13 2 3(mol L ) B(OH) 2 H O H O B(OH) + + + 4

initial 41.0 10 0 0

change x x+ x+

equilibrium 41.0 10 x x x

SM-297


25/40

2 210 3 4

a 4 4

3

7 1

3

7

[H O ][B(OH) ]7.2 10

[B(OH) ] 1.0 10 1.0 10

[H O ] 2.7 10 mol L

pH log(2.7 10 ) 6.57

x xK

x

x

+

+

= = =

= =

= =

Note: this value of 3[H O ]+ is not much different from the value for pure

water, therefore, it is at the lower limit of safely

ignoring the contribution to

71.0 10 mol L ; 1

3[H O ]+ from the autoprotolysis of water. The

exercise should be solved by simultaneously considering both equilibria.

Concentration

1

3 2 3(mol L ) B(OH) 2 H O H O B(OH) + + + 4

equilibrium 41.0 10

x y

Concentration 12 3

(mol L ) 2 H O H O OH + +

equilibrium x z

Because there are now two contributions to 3 3[H O ], [H O ]+ + is no longer

equal to nor is it equal to [ as in pure water. To avoid a

cubic equation, will again be ignored relative to

4[B(OH) ], OH ],

x 4 11.0 10 mol L .

This approximation is justified by the approximate calculation above, and because is very small relative to

concentration of then

aK41.0 10 . Let initiala =

3B(OH) ,

10 a

a

14

w

7.2 10 or

1.0 10

aK xy xyK y

a x a x

K xz

= = =

= =

Electroneutrality requires

wor ; hence, ( ).y z z x y K xz x x y= + = = = Substituting for from above:y

SM-298


26/40

a

w

2

a w

2

w a

14 4 10

w a

7 1

3

7

1 0 10 1 0 10 7 2 10

2 9 10 mol L [H O ]

pH log(2.9 10 ) 6.54

aK x x K

x

x aK K

x K aK

x K aK . . .

x .

+

=

=

= +

= + = +

= =

= =

This value is slightly, but measurably, different from the value 6.57

obtained by ignoring the contribution to 3[H O ]+ from water.

(b) In this case, the second ionization can safely be ignored; a2 a1.K K


27/40

22

a1

2 2 3

2 2 2 3

1

3

1.5 100.10

1.5 10 1.5 10 0

1.5 10 (1.5 10 ) 6.0 10

[H O ] 0.032 mol L2

pH log(0.032) 1.49

xK

x

x x

x

+

= =

+ =

+ +

= = = = =

10.91 The three equilibria involved are:

1

2

3

+ -

- + 3 3 2

3 4 2 4 3

3 4

+ 2

- 2- + 8 3 4

2 4 4 3 -

2 4

+

2- 3- + 13 3

4 4 3

[H O ][H PO ]H PO (aq) H PO (aq) H O (aq), 7.6 10

[H PO ]

[H O ][HPO ]H PO (aq) HPO (aq) H O (aq), 6.2 10

[H PO ]

[H O ][PHPO (aq) PO (aq) H O (aq), 2.1 10

a

a

a

K

K

K

+ = =

+ = =

+ = =

4

-

3-

4

2-

4

O ]

[HPO ]

We also know that the combined concentration of all the phosphate

species is:

[ ] - 2- 3-3 4 2 4 4 4

-pH 2.25 3 -1

H PO +[H PO ] [HPO ] [PO ] 1.5 10

and the hydronium ion concentration is:

[H] 10 10 5.62 10 mol L

2

+ + =

= = =

At this point it is a matter of solving this set of simultaneous equations toobtain the concentrations of the phosphate containing species. We start by

dividing both sides of the equilibrium constant expressions above by the

given hydronium ion concentration to obtain three ratios:

- 2-

5 12 4 4 4

- 2

3 4 2 4 4

[H PO ] [HPO ] [PO ]1.35 , 1.10 10 , and 3.74 10

[H PO ] [H PO ] [HPO ]

= = =3-

1

-

4

3 4

Through rearrangement and substitution of these three ratios, we can

obtain the following expressions:

-

2 4 3 4

2- -5 - -5

4 2 4 3

-5

3 4

3- 11 2- 11 -5

4 4

-16

3 4

[H PO ] 1.35 [H PO ],

[HPO ] 1.1010 [H PO ] 1.1010 1.35 [H PO ]

1.4810 [H PO ], and

[PO ] 3.74 10 [HPO ] 3.74 10 1.4810 [H PO ]

5.5410 [H PO ].

=

= =

=

= =

=

SM-300


28/40

Substituting these expressions back into the sum:

[ ]

[ ] ( ) ( ) (

[ ]

- 2- 3-

3 4 2 4 4 4

-5 -16

3 4 3 4 3 4 3 4

2

-3

3 4

- -3

2 4 3 4

2- -5 -8

4 3 4

3-

4

H PO +[H PO ] [HPO ] [PO ]

H PO + 1.35 [H PO ] 1.4810 [H PO ] 5.5410 [H PO ]

1.5 10

we find:

H PO 6.410 ,

[H PO ] 1.35 [H PO ] 8.610 ,

[HPO ] 1.4810 [H PO ] 9.410 , and

[PO ]

+ + =

= + +

=

=

= =

= =

= -16 -183 45.5410 [H PO ] 3.510 . =

)

=4

=

4

10.93 We can use the relationship derived in the text:

, in which HA is any strong acid.2

3 initial 3 w[H O ] [HA] [H O ] 0K+ +

2 7 1

3 3[H O ] (6.55 10 )[H O ] (1.00 10 ) 0+ + =

Solving using the quadratic equation gives

7

3[H O ] 6.70 10 , pH 6.174.+ = =

This value is slightly lower than the value calculated, based on the acid

concentration alone 7(pH log(6.55 10 ) 6.184).= =

10.95 We can use the relationship derived in the text:

, in which B is any strong base.23 initial 3 w[H O ] [B] [H O ] 0K

+ ++

2 8 1

3 3[H O ] (9.78 10 )[H O ] (1.00 10 ) 0+ + + =

Solving using the quadratic equation gives

8

3[H O ] 6.24 10 , pH 7.205.+ = =

This value is higher than the value calculated, based on the base

concentration alone

8

(pOH log(9.78 10 ) 7.009).

= =

10.97 (a) In the absence of a significant effect due to the autoprotolysis of

water, the pH values of the4

1.00 10

M and 61.00 10 M HBrO

solutions can be calculated as described earlier.

SM-301


29/40

For 4 11.00 10 mol L :

Concentration

1(mol L )2 3HBrO(aq) H O(l) H O (aq) BrO (aq)

+ + +

initial 0 04

1.00 10

change x+ x+

final 41.00 10 x x+ x+

3

a

[H O ][BrO ]

[HBrO]K

+

=

29

4 4

( )( )2.0 10

1.00 10 [1.00 10 ]

x x x

x x

= =

Assume 41.00 10x


30/40

6is 4.5% of 1.00 10 ,x so the assumption is less acceptable. The pH is

calculated to be Because this predicts a basic

solution, it is not reasonable.

8log(4.5 10 ) 7.35. =

(b) To calculate the value taking into account the autoprotolysis of water,we can use equation (22):

3 2

a w a initial w a 3( [HA] ) 0, where [H O ].x K x K K x K K x++ + = =

To solve the expression, you substitute the values of

the initial concentration of acid, and

14

w 1.00 10 ,K=

9

a 2.0 10K= into this equation and

then solve the expression either by trial and error or, preferably, using a

graphing calculator such as the one found on the CD accompanying this

text.

Alternatively, you can use a computer program designed to solve

simultaneous equations. Because the unknowns include

you will need four equations. As

seen in the text, pertinent equations are

3[H O ], [OH ], [HBrO], and [BrO ],+

3

a

w 3

3

initial

[H O ][BrO ]

[HBrO]

[H O ][OH ]

[H O ] [OH ] [BrO ]

[HBrO] [HBrO] [BrO ]

K

K

+

+

+

=

== +

= +

Both methods should produce the same result.

The values obtained are

7 1

3

7 1

5 1

8 1

[H O ] 4.6 10 mol L , pH 6.34 (compare to 6.35 obtained in (a))

[BrO ] 4.4 10 mol L

[HBrO] 1.0 10 mol L

[OH ] 2.2 10 mol L

+

= =

=

=

Similarly, for 6initial[HBrO] 1.00 10 :=

SM-303


31/40

7 1

3

8 1

7 1

8 1


[BrO ] 1.8 10 mol L

[HBrO] 9.8 10 mol L

[OH ] 9.1 10 mol L

+

= =

=

=

Note that for the more concentrated solution, the effect of the

autoprotolysis of water is very small. Notice also that the less concentrated

solution is more acidic, due to the autoprotolysis of water, than would be

predicted if this effect were not operating.

10.99 (a) In the absence of a significant effect due to the autoprotolysis of

water, the pH values of the

solutions can be calculated as described earlier.

5 6M M8.50 10 and 7.37 10 HCN

For 5 18.50 10 mol L :

Concentration

1

2 3(mol L ) HCN(aq) H O(l) H O (aq) CN (aq) + + +

initial 58.50 10 0 0

change x x+ x+

final5

8.50 10 x

x+ x+

[ ]3

a

210

5 5

5

2 10 5

7

[H O ][CN ]

HCN

( ) ( )4.9 10

8.5 10 [8.5 10 ]

Assume 8.5 10

(4.9 10 ) (8.5 10 )

2.0 10

K

x x x

x

x

x

x

+

=

= =


32/40

Concentration

2 3(mol L ) HCN(aq) H O(l) H O (aq) CN (aq) + + +

initial 67.37 10 0 0

change x x+ x+final 67.37 10 x x+ x+

3

a

210

6 6

6

2 10 6

8

6

[H O ][CN ]

[HCN]

( ) ( )4.9 10

7.37 10 [7.37 10 ]

Assume 7.37 10

(4.9 10 )(7.37 10 )

6.0 10

is 1% of 7.37 10 , so the assumption is still reasonable. The pH

is then calculated to be log(6.0

K

x x x

x x

x

x

x

x

+

=

= =


33/40

[ ]3

a

w 3

3

initial

[H O ][CN ]

HCN

[H O ][OH ]

[H O ] [OH ] [CN ]

[HCN] [HCN] [CN ]

K

K

+

+

+

=

=

= +

= +

Both methods should produce the same result.

For 5[HCN] 8.5 10 mol L ,1 = the values obtained are

7 1

3[H O ] 2.3 10 mol L , pH 6.64 (compare to 6.69 obtained in (a))+ = =

7 1

5 1

8 1

6

initial

7 1

3

8 1

6 1

[CN ] 1.8 10 mol L

[HCN] 8.5 10 mol L

[OH ] 4.4 10 mol L

Similarly, for [HCN] 7.37 10 :


[CN ] 3.1 10 mol L

[HCN] 7.3 10 mol L

[

+

=

=

=

= =

=

8 1OH ] 8.6 10 mol L =

Note that for the more concentrated solution, the effect of the

autoprotolysis of water is smaller. Notice also that the less concentrated

solution is more acidic, due to the autoprotolysis of water, than would be

predicted if this effect were not operating.

10.101 (a) Assuming all sulfur is converted to SO2, the amount of SO2 produced

is:

( ) ( )( )

( )( )

3 -1

-1

-1

2

1.00 10 kg 0.025 1000 g kg780 mol of S

32.07 g mol

Therefore, 780 mol 64.07 g mol 50,000 g or 50 kg of SO is produced

=

=

(b) To determine the pH we first must calculate the volume of water in

which this 50 kg of SO2 is dissolved:

( ) ( )2 2

7

3

1000 m 100 cm 1 L2 cm 2.6 Km 5.2 10 L

1 Km 1 m 1000 cmV

= =

SM-306


34/40

The concentration of SO2(aq) is then:

( ) ( )7780 mol 5.2 10 L 1.5 10 M = 5

. We can assume that upon

solution the SO2(aq) is converted to sulfurous acid and the pH is

calculated as described earlier:Concentration

1

2 3 2 3 3(mol L ) H SO (aq) H O(l) H O (aq) HSO (aq) + + +

initial 51.50 10 0 0

change x+ x+

final 51.50 10 x x+ x+

[ ]

3 3

a2 3

2

5

5

3

5

[H O ][HSO ]

H SO

( ) ( )1.55 10

1.5 10

Employing the quadratic formula we find

[H O ] 1.5 10 , giving a pH of

pH log(1.5 10 ) 4.8

K

x x

x

x

+

+

=

=

= =

= =

(c) When dissolved in water, SO3 will form sulfuric acid:

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

The fist deprotonation of sulfuric acid is complete, the concentration of

hydronium ion due to the second deprotonation of sulfuric acid may be

found as described earlier:

Concentration

4 4

1 - 2-

2 3(mol L ) HSO (aq) H O(l) H O (aq) SO (aq) + + +

initial 51.50 10 51.50 10 0

change x+ x+final 51.50 10 51.50 10 x+

SM-307


35/40

2

3 4

a

4

52

5

5

3

5 5 5

[H O ][SO ]

HSO

(1.5 10 ) ( )1.2 10

1.5 10

Employing the quadratic formula we find

[H O ] 1.495 10 ,

and the total hydronium ion concentration to be

1.495 10 1.5 10 3.0 10 .

Theref

K

x x

x

x

+

+

=

+ =

= =

+ = 5ore, pH = log(3.0 10 ) 4.5 =

10.103 (a) 2 7w 3[H O ][OH ] 3.8 10K x+ = = =

7 13

3

[H O ] 1.9 10 mol L

pH log[H O ] 6.72

x+

+= =

= =

(b) There are three data points available:

14

w w25 C ( 1.0 10 ), 40 C (K K = =

14 14

w3.8 10 ), and 37 C (K 2.1 10 ) =

lnT 11/ (K )T wK

0.00335625 C 32.2362

0.00322637 C 31.4943

0.00319540 C 31.1376

The slope is equal to /H R and the intercept equals / .S R

1 1

1 1 1

( 6502 K)(8.314 J K mol ) 54 kJ mol

( 10.43)(8.314 J K mol ) 87 J K mol

H

S

1

1

= =

= =

(c) The equation determined from the graph is for ln In order to write

an equation for the pH dependence of pure water, we must rearrange the

equation.

w .K

First we note the relationship between and the pH of pure water.wK

SM-308


36/40

1/2

3 w

3 w

w

w

w

[H O ]

1pH log[H O ] log

2

6501ln 10.43

65012.303 log 10.43

2823log 4.529

K

K

K

T

KT

KT

+

+

=

= =

=

=

=

w

1pH log

2

1411pH 2.264

K

T

=

= +

10.105 (a) We begin by finding the empirical formula of the compound:

2

2-1

-1

22-1

0.942 g COC : 0.0214 mol CO 0.214 mol C

44.011 g mol

(0.214 mol C)(12.011 g mol ) 0.257 g C

0.0964 g H OH: 0.00535 mol H O 0.0107 mol H

18.0158 g mol

(0.0107 mol H)(1.008

=

=

=

-1

2-1

-1

g mol ) 0.0108 g H

0.246 g H O Na: 0.0107 mol Na22.99 g mol

(0.0107 mol Na)(22.99 g mol ) 0.0246 g Na

O: mass of O 1.200 g 0.257 g 0.0108 g 0.0246 g 0.686 g of O

0.686 g O

16.00

=

=

=

= =

-10.0429 mol O

g mol=

Dividing through by 0.0107 moles, we find the empirical formula to be:

C2HNaO4.

A molar mass of 112.02 gmol-1

indicates that this is also the molecular

formula.

(b)

C C

O

O

O

O

H-

SM-309


37/40

(c) The dissolved substance is sodium oxalate, it is capable of gaining or

loosing a proton and, therefore, amphiprotic. ( )1

122

pH p p 2.7a aK K= + =

10.107 We wish to calculate for the reactionaK

2 3HF(aq) H O(l) H O (aq) F (aq)+ + +

This equation is equivalent to

HF(aq) H (aq) F (aq)+ +

This latter writing of the expression is simpler for the purpose of the

thermodynamic calculations.

The value for this reaction is easily calculated from the free energies

given in the appendix:

G

1 1 1

1 1

/

(18030 J mol ) / [8.314 J K mol ) (298 K)] 4

( 278.79 kJ mol ) ( 296.82 kJ mol ) 18.03 kJ mol

ln

6.9 10

G RT

RT K

K e

K e

1

= =

=

= =

10.109 (a) 2 2 3D O D O D O OD+ + +

(b)2 2 2

15D 0 3 D 0 D 0[D O ][OD ] 1.35 10 , p log 14.870K K

+ = = = =K

(c) 15 8 13[D O ] [OD ] 1.35 10 3.67 10 mol L+ = = =

(d) 8pD log(3.67 10 ) 7.435 pOD= = =

(e)2D 0 2

pD pOD p (D O) 14.870K+ = =

10.111

Concentration

+3 2 3 3CH CH(OH)COOH(aq) + H O(l) H O (aq) + CH CH(OH)COO (aq)

1(mol L )

initial 1.00 0 0

change x x+ x+


SM-310


38/40

+ 243 3

a

3

[H O ][CH CH(OH)COO ]8.4 10

[CH CH(OH)COOH] 1.00-

0.029

0.029Percent deprotonation (100%) 2.9% deprotonated

1.00

xK

x

x

= = =

=

= =

To calculate Tffor this solution we need the concentration in moles of

solute per kilogram of solvent. Assuming a density of 1 gcm-1, 1.00 L of

solution will weigh 1000 g. One liter of solution will contain 1 mole

(90.08 g) of lactic acid. Therefore, one liter of solution will contain 1000 g

90.08 g = 910 g or 0.910 kg of solvent.

The molarity of the solution is:1.029 mol solute

1.130.910 kg solvent

m= .

( )( )-1 -11.86 K Kg mol 1.13 mol Kg 2.1 K The temperature at which the solution will freeze is 271 K

fT = =

10.113 30 C 303 K, 20 C 293 K T T= = = =

a

a

1 1ln

K H H T T

K R T T R TT

= =

This is the vant Hoff equation.4

4 1 1

1.768 10 303 K 293 K ln 0.0017

293 K 303 K 1.765 10 8.314 J K mol

H

= =

10.115 (a) The equilibrium constant for the autoprotolysis of pure deuterium

oxide is given by:

( )( )-1

-1 -115

84800 J mol8.3145J K mol 298 K

1.37 10G

R T K e e

= = =

The concentration of D3O+(aq) at equilibrium is found in the familiar way:

Concentration

1(mol L ) 2 32 D O D O (aq) + OD (aq)+

initial 0 0

change x+ x+

SM-311


39/40

equilibrium x+ x+

15

3

8

3

8

[D O ][OD ] 1.37 10

[D O ] 3.7 10

pD log(3.7 10 ) 7.4

K +

+

= =

=

= =

O and H PO

(b) Given the expression forKin part (b) above, it is apparent that as T

increasesKwill increase resulting in an increase in [D3O+(aq)] and a

decrease in pD.

10.117 (a) and (b) Buffer regions are marked A, B, and C.

1.0

A B C0.5

01470

pH

(c) Region A: H P3 4 2 4

Region B: H P 22 4 4O and HPO

and PO

4

Region C: HPO 2 34 4

(d)

(e) The major species present are similar for both and

and

3H PO

3 4 2 4H AsO : H EO 2

4HEO where E P= or As. For As, there is more

than24HAsO

2 4H AsO ,

with a ratio of approximately 0.63 to 0.37, or

1.7 : 1. For P, the situation is reversed, with more 2 4H PO than HP

in a ratio of about 0.61 to 0.39, or 2.2 : 1.

2

4

1.0

A CB0.5

01470

pH

O

SM-312


40/40

10.119 Given that CO2 will react with water to form carbonic acid, H2CO3, it only

remains to determine the concentration of H3O+

due to the deprotonation

of H2CO3.

( ) ( )

2 -1 -1 4 6

2[CO ] 2.3 10 mol L atm 3.04 10 atm 7.0 10 mol L = = -1

+

2 3 2 3 3H CO (aq) + H O(l) H O (aq) + HCO (aq)

initial 7.0 106 0 0

change x x+ x+

equilibrium -67.0 10 x x

2a 6.37 7

6

63

6

a 10 10 4.27 107.0 10

[H O ] 1.73 10

pH log(1.73 10 ) 5.75

pK xK

x

+

= = = =

= =

= =

resolução cap 10 atkins

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