Exercícios Resolvidos - Cap. 10 (Ímpares) - Ácidos e Bases - -Princípios de Química - Atkins

CHAPTER 10

ACIDS AND BASES 10.1 (a) (b) 3 3CH NH +

2 3NH NH + (c) (d) 2H CO32

3CO −

(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

22 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.+

SM-274

10.5 (a) Brønsted acid: 3HNO

Brønsted 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 ,− 23 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

24HPO −

2 34 2 3 4HPO (aq) H O(l) H O (aq) PO (aq)− ++ +

24 4HPO (aq) and PO (aq)−

2 34 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

24 2 2 4HPO (aq) H O(l) H PO (aq) OH (aq)− −+ +

24 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

OH

(b)

SM-275

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

10.13 In each case, use then 14w 3[H O ][OH ] 1.0 10 ,K + − −= = ×

14w

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 15

1.0 10[OH ] 1.0 10 mol L1.0 10

−− −

−

×= = ×

×−⋅

(c) 14

12 13

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)

22Ba(OH) (aq) Ba (aq) 2 OH (aq), 100%.+ −⎯⎯→ +

Then

22 0 2 0 2 0[Ba(OH) ] [Ba ], [OH ] 2 [Ba(OH) ] , where [Ba(OH) ]+ −= = ×

nominal concentration of 2Ba(OH) .

moles of 32 1

0.25 gBa(OH) 1.5 10 mol171.36 g mol

−−= = ×

⋅

32 1

2 01.5 10 mol[Ba(OH) ] 1.5 10 mol L [Ba ]

0.100 L

−2− − +×

= = × ⋅ =

2 1 2 12 0[OH ] 2 [Ba(OH) ] 2 1.5 10 mol L 2.9 10 mol L− − −= × = × × ⋅ = × ⋅− −

SM-276

1413 1w

3 2

1.0 10[H O ] 3.4 10 mol L[OH ] 2.9 10

K −+ −

− −

× −= = = ××

⋅

10.19 Because if σε δ Taking the antilogs

of both sides gives

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

pH 13[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) M 2[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 L500 mL

− −⎛ ⎞= × ⋅ = ×⎜ ⎟

⎝ ⎠1− −⋅

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

(e) 0[NaOH] [OH ]−=

41

0.0136 gnumber of moles of NaOH 3.40 10 mol40.00 g mol

−−= = ×

⋅

44 1

03.40 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 ]+=

SM-277

4 1 53

75.0 mL[H O ] (3.5 10 mol L ) 5.3 10 mol L500 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 = − K a1pa1 10 KK −=

Acid a1pK a1K

(a) 2.12 3H PO437.6 10−×

(b) 2.00 3H PO321.0 10−×

(c) 2.46 2H SeO333.5 10−×

(d) 1.92 4HSeO 21.2 10−×

(e) The larger , the stronger the acid; therefore a1K

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 2a

2

[H O ][ClO ][HClO ]

K+ −

=

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

2b

2

[HClO ][OH ][ClO ]

K−

−=

(b)

2 3HCN(aq) H O(l) H O (aq) CN (aq)+ −+ +

3a

[H O ][CN ][HCN]

K+ −

=

2CN (aq) H O(l) HCN(aq) OH (aq)− −+ +

b[HCN][OH ]

[CN ]K

−

−=

SM-278

(c)

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

3 6 5a

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 5b

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).< <

Remember that the for the conjugate acid of a weak base will be

given by

apK

a bp p 1K K+ = 4.

10.29 Decreasing bpK will correspond to increasing base strength because

b bp log .K K= − The bpK values (given in parentheses) determine the

following ordering:

3

5 5 3

(14.00 3.45 10.55) CH COO (14.00 4.75 9.25)C H N (8.75) NH (4.75).

F − −− = < − =< <<

Remember that the bpK for the conjugate base of a weak acid will be

given by a bp p 1K K+ = 4.

10.31 Any acid whose conjugate base lies above water in Table 10.3 will be a

strong acid; that is, the conjugate base of the acid will be a weaker base

than water, and so water will accept the H+ preferentially. Based upon

this information, we obtain the following analysis: (a) , strong;

(b) weak; (c)

3HClO

2H S, 4HSO − , weak (Note: even though is a 2H SO4

SM-279

strong acid, is a weak acid. Its conjugate base is ); (d)

weak acid; (e)

4HSO − 24SO −

3 3CH NH ,+3HCO ,− weak; (f) strong; (g)

weak.

3HNO ,

4CH ,

10.33 For oxoacids, the greater the number of highly electronegative O atoms

attached to the central atom, the stronger the acid. This effect is related to

the increased oxidation number of the central atom as the number of O

atoms increases. Therefore, is the stronger acid, with the lower 3HIO ap .K

10.35 (a) HCl is the stronger acid, because its bond strength is much weaker

than the bond in HF, and bond strength is the dominant factor in

determining the strength of binary acids.

(b) is stronger; there is one more O atom attached to the Cl atom

in than in HClO. The additional O in helps to pull the

electron of the H atom out of the H—O bond. The oxidation state of Cl is

higher in than in HClO.

2HClO

2HClO 2HClO

2HClO

(c) is stronger; Cl has a greater electronegativity than Br, making

the H—O bond more polar than in

2HClO

2HClO 2HBrO .

(d) is stronger; Cl has a greater electronegativity than P. 4HClO

(e) is stronger. The explanation is the same as that for part (b).

has one more O atom.

3HNO

3HNO

(f) is stronger; C has greater electronegativity than Ge. See part

(c).

2H CO3

10.37 (a) The — group that is bonded to the carboxyl group, —COOH, in

trichloroacetic acid, is more electron withdrawing than the — group

in acetic acid. Thus, trichloroacetic acid is the stronger acid.

3CCl

3CH

SM-280

(b) The — group in acetic acid has electron-donating properties,

which means that it is less electron withdrawing than the —H attached to

the carboxyl group in formic acid, HCOOH. Thus, formic acid is a slightly

stronger acid than acetic acid. However, it is not nearly as strong as

trichloroacetic acid. The order is

3CH

3 3CCl COOH HCOOH CH COOH.>> >

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 it’s 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.)

SM-281

10.45 (a) Concentration

13 2 3(mol L ) CH COOH H O H O CH CO− −⋅ + + 3 2

initial — 0 0 0.29

change x− — x+ x+

equilibrium 0.29 x− — x x2 2

5 3 3 2a

3

3 13

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 xKx

x

+ −−

+ − −

−

= × = = ≈−

= = × ⋅

= − × = = − =

(b) The equilibrium table for (b) is similar to that for (a). 2

1 3 3 2a

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

xKx

x x

x

+ −−

−

− −

= × = =−

+ × − =

− × ± × − −= = −

2

The negative root is not possible and can be eliminated. 1

3[H O ] 0.18 mol LpH log(0.18) 0.74, pOH 14.00 0.74 13.26x + −= = ⋅

= − = = − =

(c) Concentration 12 3(mol L ) HCOOH H O H O HCO− + −⋅ + +

initial — 0 0.29 0


equilibrium 0.29 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 xKx

x

+ −−

+ − −

−

⋅= = ≈ = ×

−

= = × × = × ⋅

= − × = = − =

−

SM-282


12 3 4(mol L ) H O NH NH OH− +⋅ + + −

initial — 0.057 0 0

change — x− x+ x+

equilibrium — 0.057 x− x x2

54b

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 11.001.0 10percentage protonation 100% 1.8%

0.057

x x xKx

x

+ −−

− − −

−

−

⋅= = ≈ = ×

−

= = × × = × ⋅

= − × = = − =

×= × =

−

−+

(b) Concentration

12 2 3(mol L ) NH OH H O NH OH OH− +⋅ +

initial 0.162 — 0 0



8b

5 1

5

5

1.1 100.162 0.162

[OH ] 4.2 10 mol LpOH log(4.2 10 ) 4.38, pH 14.00 4.38 9.62

4.2 10percentage protonation 100% 0.026%0.162

x xKx

x

−

− − −

−

−

= × = ≈−

= = × ⋅

= − × = = − =

×= × =

(c) Concentration

13 3 2 3 3(mol L ) (CH ) N H O (CH ) NH OH− + −⋅ + +

initial — 0.35 0 0


equilibrium 0.35 x− — x+ x+

SM-283

25

3 1

3 1

3

3

6.5 100.35

Assume 0.35Then 4.8 10 mol L[OH ] 4.8 10 mol LpOH log(4.8 10 ) 2.32, pH 14.00 2.32 11.68


xx

xx

−

− −

− − −

−

−

× =−

<<

= × ⋅

= × ⋅

= − × = = − =

×= × =

(d) 6b a bp 14.00 p 14.00 8.21 5.79, 1.6 10K K K −= − = − = = ×

22 2

6b

4 1

4

4

codeine H O codeineH OH

1.6 100.073 0.073

[OH ] 1.1 10 mol LpOH log(1.1 10 ) 3.96, pH 14.00 3.96 10.04


x xKx

x

+ −

−

− − −

−

−

+ +

= × = ≈−

= = × ⋅

= − × = = − =

×= × =

L

10.49 (a) 2 2 3 2HClO H O H O ClO+ −+ → +

pH 1.23 2

23 2

a2

a

[H O ] [ClO ] 10 10 0.06 mol L

[H O ][ClO ] (0.06) 0.09 (1 sf )[HClO ] 0.10 0.06

p log(0.09) 1.0

K

K

+ − − −

+ −

1−= = = = ⋅

= = =−

= − =

(b) +3 7 2 2 3 7 3C H NH H O C H NH +OH−+ L

2.14 3 13 7 3

3 243 7 3

b 33 7 2

4b

pOH 14.00 11.86 2.14[C H NH ] [OH ] 10 7.2 10 mol L

[C H NH ][OH ] (7.2 10 ) 5.6 10[C H NH ] 0.10 7.2 10

p log(5.6 10 ) 3.25

K

K

+ − − − −

+ − −−

−

−

= − =

= = = × ⋅

×= = =

− ×

= − × =

×

10.51 (a) pH 4.60 53pH 4.60, [H O ] 10 10 2.5 10 mol L 1+ − − −= = = = × −⋅

Let nominal concentration of HClO, thenx =

SM-284

Concentration

+ 1(mol L )−⋅ HClO 2 3H O H O ClO+ −+

nominal — x 0 0

equilibrium 52.5 10x −− × — 52.5 10−× 52.5 10−×

5 2

8a 5

(2.5 10 )3.0 102.5 10

Kx

−−

−

×= × =

− ×

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 3NH NH H O NH NH OH+ −+ +

nominal — 0 0 x

equilibrium 41.6 10x −− × — 41.6 10−× 41.6 10−×

4 26

b 4

2 1

(1.6 10 )1.7 101.6 10

Solve for ; 1.5 10 mol L

Kx

x 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


equilibrium 0.110 x− —

x x

2H O octylamine octylamineH OH+ −+ +

SM-285

13 6 5 2

253 6 5

a6 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 LpOH log(0.0067) 2.17, pH 14.00 2.17 11.83

[octylamineH ][OH ] (6.7 10 ) 4.8 10[octylamine] 0.093

K

− + −

+ − −−

= = ⋅= − = = − =

×= = = ×

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


equilibrium0.0147 x− — x x+ 2 2

53 2 3a

3 2

[CH CH COO ][H O ]1.3 10

[CH CH COOH] 0.0147 0.0147x xK

x

−−= = × =

−≈

44 4.4 104.4 10 , Percent deprotonated 100% 3.0%

0.0147x

−− ×

= × = × =

SM-286

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) 14

10wb 5

a

1.00 10 5.6 101.8 10

KK

K

−−

−

×= = = ×

×

Concentration

13 2 2 3 2(mol L ) CH CO (aq) H O(1) HCH CO (aq) OH (aq)− − −⋅ + +

initial 0.63 — 0 0



2 2

103 2b

3 2

[HCH CO ][OH ]5.6 10

0.63 0.63[CH CO ]x xK

x

−−

−= = × =−

≈

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

pH 14.00 pOH 14.00 4.72 9.28= − = − =

(b) 14

10wa 15

b

1.00 10 5.6 101.8 10

KK

K

−−

−

×= = = ×

×

14 2 3Concentration (mol L ) NH (aq) H O(l) H O (aq) NH (aq)− + +⋅ + + 3

initial 0.19 — 0 0



103 3a

4

[H O ][NH ]5.6 10

[NH Cl] 0.19 0.19x xK

x

+−= = × =

−≈

SM-287

5 131.0 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


equilibrium 0.055 x− — x x2 2 2

53 2 5a 3

2 6

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

0.055 0.055[Al(H O) ]x xK

x

+ +−

+= = × =−

≈

+

−

4 138.8 10 mol L [H O ]x − −= × ⋅ =

4pH log(8.8 10 ) 3.06−= − × =

(d) Concentration

12(mol L ) H O(l) CN (aq) HCN(aq) OH (aq)− −⋅ + +

initial — 0.065 0 0


equilibrium — 0.065 x− x x14 2 2

5wb 10

a

1.00 10 [HCN][OH ]2.0 100.065 0.0654.9 10 [CN ]

K x xKK 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



SM-288

2 2113 3 2

a3 3

[H O ][CH NH ]2.8 10

0.510 0.510[CH NH ]x xK

x

+−

+= = × =−

≈

+

6 133.8 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 22 3 2 2 3 2

2 3 2

1 mol KC H O 1(5.34 g KC H O ) 0218 KC H O98.14 g KC H O 0.250 L

M⎛ ⎞⎛ ⎞=⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

Concentration 1

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


equilibrium — 0.218 x− x x14 2 2

5

1.0 100.218 0.2181.8 10

x xx

−

−

×= ≈

−×

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 =

45.75 g NH Br,

197.95 g mol−⋅

44 4

4

M1 mol 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


SM-289

equilibrium 0.587 x− — x x14 2 2

5

1.0 100.587 0.5871.8 10

x xx

−

−

×= ≈

−×

5 13[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 L0.0400 mol L

0.500 L

−−⋅ ×

= ⋅

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

initial — 0.0400 0 0


equilibrium — 0.0400 x− x x14

10w 3b 5

a 3 2

[CH COOH][OH ]1.00 10 5.6 101.8 10 [CH CO ]

KK

K

−−−

− −

×= = = × =

×

2 2105.6 10

0.0400 0.0400x x

x−× = ≈

−

6 134.7 10 mol L [CH COOH]x − −= × ⋅ =

(b) 4 43

4

2.16 g NH Br 1 mol NH Br1 mL400 mL 97.95 g NH Br10 L−

⎛ ⎞⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠

140.0551 (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


equilibrium 0.0551 x− — x x14

10w 3a 5

b 4

[NH ][H O ]1.00 10 5.6 101.8 10 [NH ]

KK

K3

+−−

− +

×= = = × =

×

SM-290

2 2105.6 10

0.0551 0.0551x x

x−× = ≈

−

6 135.5 10 mol L [H O ]x − −= × ⋅ = + and 6pH log(5.5 10 ) 5.26−= − × =

10.69 (a)

C C CH

HH

NH

HO

O

Na+

(b) 1 2

1 12 2pH = + (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)+ −+ +

24 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)+ −+ +

22 4 2 3 4

2 34 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)+ −+ +

26 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


equilibrium 0.15 x− — 0.15 x+ x

SM-291

22 3 4

a24

2 3

2 31

1 13

[H O ][SO ] (0.15 ) ( )1.2 100.15[HSO ]

0.162 1.8 10 0

0.162 (0.162) (4) (1.8 10 )0.0104 mol L

2[H O ] 0.15 (0.15 0.0104) mol L 0.16 mol LpH log(0.16) 0.80

x xKx

x x

x

x

+ −−

−

−

−−

+ −

+= × = =

−

+ − × =

− + + ×= =

−

⋅

= + = + ⋅ = ⋅

= − =

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

Concentration 1

2 3 2 3 3(mol L ) H CO H O H O HCO− +⋅ + + −

initial — 0 0.010 0



73 3a1

2 35 1

35

[H O ][HCO ]4.3 10

[H CO ] 0.010 0.010

[H O ] 6.6 10 mol L

pH log(6.6 10 ) 4.18

x xKx

x

+ −−

+ − −

−

= = ≈ =−

= = × ⋅

= − × =

×

2−

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

Concentration

12 2 3(mol L ) (COOH) H O H O (COOH)CO− +⋅ + +

initial 0.10 — 0 0


equilibrium 0.10 x− — x x2

2 3 2a1

22 2 3

2 2 2 31

[H O ][(COOH)CO ]5.9 10

[(COOH) ] 0.10

5.9 10 5.9 10 0

5.9 10 (5.9 10 ) (4)(5.9 10 )0.053 mol L

2pH log(0.053) 1.28

xKx

x x

x

+ −−

− −

− − −−

= × = =−

+ × − × =

− × + × + ×= =

= − =

⋅

SM-292

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

Concentration 12 2 3(mol L ) H S H O H O HS− + −⋅ + +


7 3a1

24 1

3

4

[H O ][HS ]1.3 10

[H S] 0.20 0.20

[H O ] 1.6 10 mol L

pH log(1.6 10 ) 3.80

x xKx

x

+ −−

+ − −

−

= × = = ≈−

= = × ⋅

= − × =

10.77 (a) The pH is given by 1a1 a22pH (p p ).K K= + From Table 10.9, we find

2a1 a1

7a2 a2

12

1.5 10 p 1.82

1.2 10 p 6.92pH (1.82 6.92) 4.37

K K

K K

−

−

= × =

= × =

= + =

(b) The pH of a salt solution of a polyprotic acid is independent of the

concentration of the salt, therefore pH 4.37.=

10.79 (a) The pH is given by 1a1 a22pH (p p ).K K= + For the monosodium salt,

the pertinent values are a1 a2p and pK K :

12pH (3.14 5.95) 4.55= + =

(b) For the disodium salt, the pertinent values are a2 a3p and pK K :

12pH (5.95 6.39) 6.17= + =

10.81 The equilibrium reactions of interest are 7

2 3 2 3 3 a1

2 13 2 3 3 a2

H CO (aq) H O(l) H O (aq) HCO (aq) 4.3 10

HCO (aq) H O(l) H O (aq) CO (aq) 5.6 10

K

K 1

+ − −

− + −

+ + =

+ + = −

×

×

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

can assume that the first step dominates:

SM-293

Concentration

12 3 2 3 3(mol L ) H CO (aq) H O(l) H O (aq) HCO (aq)− +⋅ + + −

initial 0.0456 — 0 0


final 0.0456 x− — x+ x+

3 3a1

2 3

27

2 7

4

4 13 3

[H O ][HCO ][H CO ]

( ) ( )4.3 100.0456 0.0456

Assume that 0.0456Then (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 xx x

xx

xx

x

+ −

−

−

−

+ − − −

=

× = =− −

<<

= ×

= ×<

= = = × ⋅

This means that the concentration of 12 3H CO is 0.0456 mol L−⋅ −

10.00014 mol L 0.0455 mol L .−⋅ = ⋅ 1− We can then use the other equilibria

to determine the remaining concentrations: 2

3 3a2

34 2

11 34

2 113

[H O ][CO ][HCO ]

(1.4 10 )[CO ]5.6 10

(1.4 10 )[CO ] 5.6 10 mol L

K+ −

−

1

− −−

−

− −

=

×× =

×

= × ⋅ −

−

Because the initial assumption that the first

ionization would dominate is valid.

11 45.6 10 1.4 10 ,−× << ×

To calculate [ we use the relationship: OH ],−wK

w 314

11 1w4

31

2 3 3 34 1

2 11 1 113

[H O ][OH ]

1.00 10[OH ] 7.1 10 mol L[H O ] 1.4 10

In summary, [H CO ] 0.0455 mol L , [H O ] [HCO ]

1.4 10 mol L ,[CO ] 5.6 10 mol L , [OH ] 7.1 10 mol L .

KK

+ −

−− −

+ −

1

−

− + −

− −

− − − − −

=

×= = = × ⋅

×

= ⋅ =

= × ⋅

= × ⋅ = × ⋅ −

SM-294

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

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

following changes: 2

3 2 314

4wb1 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 7a1

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

1.00 10 2.3 104.3 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


final 0.0456 x− — x+ x+

3b1 2

3

[HCO ][OH ][CO ]

K− −

−=

24 ( ) ( )1.8 10

0.0456 [0.0456 ]x x x

x x−× = =

− −

Assume that 0.0456x <<

Then 2 4(1.8 10 ) (0.0456)x −= ×

32.9 10x −= ×

Because of 0.0456, the assumption was not valid and the full

expression should be solved using the quadratic equation:

5%x >

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

We can then use the other equilibria to determine the remaining

concentrations:

2 3b2

3

[H CO ][OH ][HCO ]

K−

−=

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

(0.0028)−× =

8 12 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−× <<

3[H O ]+ , we use the

relationship:

wK

w 3[H O ][OH ]K + −=

1412 1w

31.00 10[H O ] 3.6 10 mol L

0.0028[OH ]K −

+ −−

× −= = = × ⋅

In summary, 8 12 3 3[H CO ] 2.3 10 mol L , [OH ] [HCO ]− − − −= × ⋅ = =

1−⋅

,

1 2 1 123 30.0028 mol L , [CO ] 0.0428 mol L , [H O ] 3.6 10 mol L− − − + −⋅ = ⋅ = ×

10.85 (a) phosphorous acid: The two values are 2.00 and 6.59. Because

lies between the dominant form will be the

singly deprotonated HA

apK

pH 6.30= a1 a2p and pK K

− ion.

(b) oxalic acid: The two values are 1.23 and 4.19. Because

lies above the species present in largest concentration

will be the doubly deprotonated

apK

pH 6.30= a 2pK ,

2A − ion.

(c) hydrosulfuric acid: The two values are 6.89 and 14.15. Because

lies below both values, the species present in highest

concentrations will be the fully protonated form.

apK

pH 6.30= apK

2H A

10.87 The equilibria present in solution are

SM-296

22 3 2 3 3 a1

2 73 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

22 3 3 3 3 3H CO , HSO for HCO , and SO for CO .− − −

5.5 6 13(at pH 5.50 [H O ] 10 3.2 10 mol L ) :f + − − −= = = × ⋅

23 3 a1 a1 a 2

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 2

432 3 8

6 23 a1

3 8

2 72 a1 a2

3 8

[H O ] (3.2 10 )(H SO ) 2.1 105.0 10

[H O ] (3.2 10 ) (1.5 10 )(HSO ) 0.965.0 10

(1.5 10 ) (1.2 10 )(SO ) 0.0365.0 10

fK

fK K

f

α

α

α

+ −−

−

+ − −−

−

− −−

−

×= = = ×

×

× ×= = =

×

× ×= = =

×

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

be with a concentration of

pH 5.50,

3HSO − 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 L 1− − −× ⋅ = × −⋅ and the

concentration of 23SO − will be

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


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

−

initial 41.0 10−× — 0 0


equilibrium — 41.0 10 x−× − x x

SM-297

2 210 3 4

a 4 43

7 13

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 xKx

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−× − — 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

aK 41.0 10 . Let initiala−× =

3B(OH) ,

10 aa

14w

7.2 10 or

1.0 10

aKxy xyK ya x a x

K xz

−

−

= × = ≈ =−

= × =

Electroneutrality requires

wor ; hence, ( ).x y z z x y K xz x x y= + = − = = −

Substituting for from above: y

SM-298

aw

2a w

2w a

14 4 10w a

7 13

7

1 0 10 1 0 10 7 2 10

2 9 10 mol L [H O ]

pH log(2.9 10 ) 6.54

aKx x K

xx 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; a 2 a1.K K<<

Concentration 1(mol L )−⋅ 3 4 2 3 2 4H PO H O H O H PO+ −+ +

initial 0.015 — 0 0


equilibrium 0.015 – x — x x 2

3a1

2 3 4

3 3 2

3

3 1

3

7.6 100.015

7.6 10 1.14 10 0

7.6 10 (7.6 10 ) 4.56 10[H O ]

27.5 10 mol L

pH log(7.5 10 ) 2.12

xKx

x x

x

−

− −

4− − −+

− −

−

= × =−

+ × − × =

− × + × + ×= =

= × ⋅

= − × =

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

Concentration

12 3 2 3 3(mol L ) H SO H O H O HSO− +⋅ + + −

initial 0.1 — 0 0



SM-299

22

a1

2 2 3

2 2 2 31

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 L

2pH log(0.032) 1.49

xKx

x x

x

−

− −

− − −+ −

= × =−

+ × − × =

− × + × + ×= = = ⋅

= − =

10.91 The three equilibria involved are:

1

2

3

+ -- + 3 3 2

3 4 2 4 33 4

+ 2- 2- + 8 3 4

2 4 4 3 -2 4+

2- 3- + 13 34 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 10and 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 to

obtain 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 42- -5 - -54 2 4 3

-53 4

3- 11 2- 11 -54 4

-163 4

[H PO ] 1.35 [H PO ],

[HPO ] 1.10×10 [H PO ] 1.10×10 1.35 [H PO ]

1.48×10 [H PO ], and

[PO ] 3.74 10 [HPO ] 3.74 10 1.48×10 [H PO ]

5.54×10 [H PO ].

− −

= ⋅

= ⋅ = ⋅ ⋅

= ⋅

= × ⋅ = × ⋅ ⋅

= ⋅

SM-300

Substituting these expressions back into the sum:

[ ][ ] ( ) ( ) (

[ ]

- 2- 3-3 4 2 4 4 4

-5 -163 4 3 4 3 4 3 4

2

-33 4

- -32 4 3 4

2- -5 -84 3 4

3-4

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

H PO + 1.35 [H PO ] 1.48×10 [H PO ] 5.54×10 [H PO ]

1.5 10we find:H PO 6.4×10 ,

[H PO ] 1.35 [H PO ] 8.6×10 ,

[HPO ] 1.48×10 [H PO ] 9.4×10 , and

[PO ]

−

+ + =

= ⋅ + ⋅ + ⋅

= ×

=

= ⋅ =

= ⋅ =

= -16 -183 45.54×10 [H PO ] 3.5×10 .⋅ =

)

=

4

=

4

10.93 We can use the relationship derived in the text:

, in which HA is any strong acid. 23 initial 3 w[H O ] [HA] [H O ] 0K+ +− −

2 7 13 3[H O ] (6.55 10 )[H O ] (1.00 10 ) 0+ − + −− × − × =

Solving using the quadratic equation gives

73[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 13 3[H O ] (9.78 10 )[H O ] (1.00 10 ) 0+ − + −+ × − × =

Solving using the quadratic equation gives

83[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 the 4

1.00 10−

× M and 61.00 10−× M HBrO

solutions can be calculated as described earlier.

SM-301

For 4 11.00 10 mol L :− −× ⋅

Concentration

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

initial — 0 0 41.00 10−×


final 41.00 10 x−× − — x+ x+

3a

[H O ][BrO ][HBrO]

K+ −

=

29

4 4

( )( )2.0 101.00 10 [1.00 10 ]

x x xx x

−− −× = =

× − × −

Assume 41.00 10x −<< × 2 9(2.0 10 ) (1.00 10 )x − −= × × 4

74.5 10x −= ×

Because 41% of 1.00 10x −< × , the assumption was valid. Given this

value, the pH is then calculated to be 7log(4.5 10 ) 6.35.−− × =

For 6 11.00 10 mol L :− −× ⋅

Concentration 1(mol L )−⋅ HBrO(aq) + 2 3H O(l) H O (aq) BrO (aq)+ −+

initial 61.00 10−× — 0 0


final 61.00 10 x−× − — x+ x+

3a

[H O ][BrO ][HBrO]

K+ −

=

29

6 6

( )( )2.0 101.00 10 [1.00 10 ]

x x xx x

−− −× = =

× − × −

Assume 61.00 10x −<< ×

2 9

8

(2.0 10 ) (1.00 10 )4.5 10

xx

− −

−

= × ×

= ×

6

SM-302

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

14w 1.00 10 ,K −= ×

9a 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 ],+ − −

3a

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

7 13

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


final 58.50 10 x−× − — x+ x+

[ ]3

a

210

5 5

5

2 10 5

7

[H O ][CN ]HCN

( ) ( )4.9 108.5 10 [8.5 10 ]

Assume 8.5 10(4.9 10 ) (8.5 10 )

2.0 10

K

x x xx x

xxx

+ −

−− −

−

− −

−

=

× = =× − × −

<< ×

= × ×

= ×

5Because 1% of 8.50 10 ,x −< × the assumption was valid. Given this

value, the pH is then calculated to be 7log(2.0 10 ) 6.69.−− × =

6 1For 7.37 10 mol L :− −× ⋅

SM-304

Concentration

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

initial 67.37 10−× — 0 0


final 67.37 10 x−× − — x+ x+

3a

210

6 6

6

2 10 6

8

6

[H O ][CN ][HCN]

( ) ( )4.9 107.37 10 [7.37 10 ]

Assume 7.37 10(4.9 10 ) (7.37 10 )

6.0 10is 1% of 7.37 10 , so the assumption is still reasonable. The pH

is then calculated to be log(6.0

K

x x xx x

xxxx

+ −

−− −

−

− −

−

−

=

× = =× − × −

<< ×

= × ×

= ×

< ×

− × 810 ) 7.22. This answer isnot reasonable because we know HCN is an acid.

− =

(b) To calculate the value, taking into account the autoprotolysis of water,

we can use equation (21): 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

141.00 10 ,wK −= ×

10a 4.9 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, the pertinent equations are

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

SM-305

[ ]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

6initial

7 13

8 1

6 1

[CN ] 1.8 10 mol L[HCN] 8.5 10 mol L[OH ] 4.4 10 mol LSimilarly, 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

-12

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

73

1000 m 100 cm 1 L2 cm 2.6 Km 5.2 10 L1 Km 1 m 1000 cm

V ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

×

SM-306

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

12 3 2 3 3(mol L ) H SO (aq) H O(l) H O (aq) HSO (aq)− +⋅ + +

initial 51.50 10−× — 0 0


final 51.50 10 x−× − — x+ x+

[ ]3 3

a2 3

25

53

5

[H O ][HSO ]H SO

( ) ( )1.55 101.5 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 xx

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


final 51.50 10 x−× − — 51.50 10 x−× − x+

SM-307

23 4

a4

52

5

53

5 5 5

[H O ][SO ]HSO

(1.5 10 ) ( )1.2 101.5 10

Employing the quadratic formula we find[H O ] 1.495 10 ,

and the total hydronium ion concentration to be1.495 10 1.5 10 3.0 10 .Theref

K

x xx

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:

14w w25 C ( 1.0 10 ), 40 C (K K−° = × ° =

14 14w3.8 10 ), and 37 C (K 2.1 10 )− −× ° = ×

ln T 11/ (K )−T wK

0.003356 25 C° 32.2362−

0.003226 37 C° 31.4943−

0.003195 40 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

HS

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

1/23 w

3 w

w

w

w

[H O ]1pH log[H O ] log2

6501ln 10.43

65012.303 log 10.43

2823log 4.529

K

K

KT

KT

KT

+

+

=

= − = −

= − −

= − −

= − −

w1pH log2

1411pH 2.264

K

T

= −

= +

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

22-1

-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 H0.246 g H O

Na: 0.0107 mol Na 22.99 g mol

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

⋅ =

=⋅

⋅ == − − − =

-1 0.0429 mol Og mol

=⋅

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

C2HNaO4.

A molar mass of 112.02 g⋅mol-1 indicates that this is also the molecular

formula.

(b)

C C

O

O

O

O

H -

SM-309

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

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

122pH p p 2.7a aK K= + =

10.107 We wish to calculate for the reaction aK

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 molln

6.9 10

G RT

RT KK 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 2pD 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



SM-310

+ 243 3

a3

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

[CH CH(OH)COOH] 1.00-0.029

0.029Percent deprotonation (100%) 2.9% deprotonated1.00

xKx

x

−−= = =

=

= =

×

To calculate Tf for this solution we need the concentration in moles of

solute per kilogram of solvent. Assuming a density of 1 g⋅cm-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 1.13 0.910 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 KfT∆ = ⋅ ⋅ ⋅ =

10.113 30 C 303 K, 20 C 293 KT T= ° = = ° =�

a

a

1 1lnK H H T TK R T T R TT

′⎛ ⎞ ∆ ° ∆ ° −⎛ ⎞ ⎛= − =⎜ ⎟ ⎜ ⎟ ⎜′⎝ ⎠ ⎝⎝ ⎠

�

�

⎞⎟⎠

This is the van’t Hoff equation. 4

4 1 1

1.768 10 303 K 293 Kln 0.0017293 K 303 K1.765 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 -1 1584800 J mol

8.3145J K mol 298 K 1.37 10GR TK 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

equilibrium — x+ x+ 15

38

38

[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 for K in part (b) above, it is apparent that as T

increases K will 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

0

14 7 0

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 − 24HEO − where E P= or As. For As, there is more

than 24HAsO −

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.

24

1.0

A CB0.5

0 14 7 0

pH

O −

SM-312

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 62[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 × 10−6 — 0 0


equilibrium -67.0 10 x× − — 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

− − −−

+ −

−

⎛ ⎞= = = × = ⎜ ⎟

×⎝ ⎠= = ×

= − × =

SM-313

Exercícios Resolvidos - Cap. 10 (Ímpares) - Ácidos e Bases - -Princípios de Química - Atkins

Documents

Transcript of Exercícios Resolvidos - Cap. 10 (Ímpares) - Ácidos e Bases - -Princípios de Química - Atkins