CHAPTER 10

ACIDS AND BASES 10.2 (a) (b) H3H O+

2O (c) 6 5 3C H NH + (d) HS− (e) 34PO −

(f) 4ClO −

10.4 (a) H2

ac

(b) H2

ac

(c) H2

ac

(d) H2

ac

(e) H2

ac

conjugate

O(l) + CN−(aq) HCN(l) + OH−(aq) id1 acid2 base1

O(l) + Nid1

id1

id1

id1

base2

conjugate

conjugate

H2NH2(aq) NH2NH3+(aq) + OH−(aq)

base2 acid2 base1

conjugate

conjugate −(aq)

O(l) + CO3

2− (aq) HCO3− (aq) + OH

base2 acid2 base1

conjugate

conjugate −(aq)

O(l) + HPO4

2− (aq) H2PO4− (aq) + OH

base2 acid2 base1

conjugate

conjugate H−(aq)

O(l) + NH2CONH2 (aq) NH2CONH3

+ (aq) + O

base2 acid2 base1

conjugate

291

10.6 (a) Brønsted acid: NH4+

Brønsted base: 3HSO −

(b) Conjugate base to NH4+ : NH3

Conjugate acid to 3HSO − : H2SO3

10.8 (a) as an acid: H2PO4−(aq) + H2O(l) H3O+(aq) + HPO4

2− (aq)

Conjugate acid/base pairs: H2PO4−(aq) / HPO4

2− (aq)

H3O+(aq) / H2O(l)

as a base: H2PO4−(aq) + H2O(l) OH−(aq) + H3PO4(aq)

Conjugate acid/base pairs: H3PO4(aq) / H2PO4−(aq)

H2O(l) / OH−(aq)

(c) as an acid: HC2O4−(aq) + H2O(l) H3O+(aq) + C2O4

2−(aq)

Conjugate acid/base pairs: HC2O4−(aq) / C2O4

2−(aq)

H3O+(aq) / H2O(l)

as a base: HC2O4−(aq) + H2O(l) OH−(aq) + H2C2O4(aq)

Conjugate acid/base pairs: H2C2O4(aq) / HC2O4−(aq)

H2O(l) / OH−(aq)

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

10.12 (a) 2 OH −(aq) + SiO2(s) SiO32− (aq) + H2O(l)

(b) O

Si

O O

Si

O

O O

Si

O

O O

(c) Lewis Acid: SiO2(s), Lewis Base: OH −(aq)

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

292

14w

31.0 10[H O ]

[OH ] [OH ]

−+

− −

×= =

K

(a) 14

133

1.0 10[H O ] 8.3 100.012

−+ −×

= = ×

(b) 14

103 5

1.0 10[H O ] 1.6 106.2 10

−+ −

−

×= = ×

×

(c) 14

123 3

1.0 10[H O ] 4.3 102.3 10

−+ −

−

×= = ×

×

10.16 (a) 8 13[H O ] [OH ] 3.9 10 mol L+ − −= = × ⋅ −

−×8 2 15w 3

15w w

[H O ][OH ] (3.9 10 ) 1.5 10

p log log(1.5 10 ) 14.82

+ − −

−

= = × =

= − = − × =

K

K K

(b) 8pH log(3.9 10 ) 7.41 pOH−= − × = =

10.18 [KNH2]0 = nominal concentration of KNH2

[KNH2]0 = 12 2

2

0.5 g KNH 1 mol KNH 0.036 mol L0.250 L 55.13 g KNH

−⎛ ⎞⎛ ⎞= ⋅⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

Because KNH2 is a soluble salt, [KNH2]0 = [K+] = (where

is the nominal concentration of

2 0[NH ]−

2 0[NH ]−2NH − ); thus [K+] and =

0.036 .

2[NH ]−

1mol L−⋅

2NH − reacts with water:

2 2 3NH (aq) H O(l) NH (aq) OH (aq)− −+ ⎯⎯→ +

Because is a strong base, this reaction goes essentially to

completion; therefore

2NH −

12 0[NH ] [OH ] 0.036 mol L− −= = ⋅ −

1413 1w

31.0 10[H O ] 2.8 10 mol L

[OH ] 0.036K −

+ −−

×= = = × ⋅ −

293

10.20 pH 13[H O ] 10 mol L (antilog pH, mol L )+ − − −= ⋅ ⋅ 1

−

−⋅

−⋅

−⋅

(a) 5 13[H O ] antilog( 5) 1 10 mol L+ −= − = × ⋅

(b) 2.3 1 3 13[H O ] 10 mol L ; antilog ( 2.3) 5 10 mol L+ − − −= ⋅ − = ×

(c) 7.4 1 8 13[H O ] 10 mol L ; antilog ( 7.4) 4 10 mol L+ − − −= ⋅ − = ×

(d) 10.5 1 11 13[H O ] 10 mol L ; antilog ( 10.5) 3 10 mol L+ − − −= ⋅ − = ×

Acidity increases as pH decreases. The order is thus:

milk of magnesia < blood < urine < lemon juice

10.22 (a) pH log(0.0356) 1.448pOH 14.00 1.448 12.55

= − == − =

(b) pH log(0.0725) 1.12pOH 14.00 1.12 12.86

= − == − =

(c) 22Ba(OH) Ba 2 OH+ −⎯⎯→ +

3 1 3 1

3

[OH ] 2 3.46 10 mol L 6.92 10 mol LpOH log(6.92 10 ) 2.160pH 14.00 2.160 11.84

− − − −

−

= × × ⋅ = × ⋅

= − × == − =

−

(d) 3

41

42 1

2

10.9 10 g 1.94 10 mol KOH56.11 g mol

1.94 10 mol[OH ] 1.94 10 mol L0.0100 L

pOH log(1.94 10 ) 1.712pH 14.00 1.712 12.29

−−

−

−− −

−

×= ×

⋅

×= = ×

= − × == − =

−⋅

(e) 1 110.0 mL[OH ] (5.00 mol L ) 0.0200 mol L2500 mL

pOH log(0.200) 1.70pH 14.00 0.699 12.30

− −⎛ ⎞= × ⋅ =⎜ ⎟

⎝ ⎠= − =

= − =

−⋅

(f) 4 1 53

5

5.0 mL[H O ] (3.5 10 mol L ) 7.0 10 mol L25.0 mL

pH log(7.0 10 ) 4.15pOH 14.00 4.15 9.85

+ − −

−

⎛ ⎞= × × ⋅ = ×⎜ ⎟

⎝ ⎠= − × =

= − =

1− −⋅

294

10.24 Base Kb pKb

(a) NH3 51.8 10−× 4.74

(b) ND3 51.1 10−× 4.96

(c) NH2NH2 61.7 10−× 5.77

(d) NH2OH 81.1 10−× 7.96

(e) NH2OH < NH2NH2 < ND3 < NH3

10.26 (a) 3 2 2 3 2 2

3 2 2b

3 2

(CH ) NH(aq) H O(l) (CH ) NH (aq) OH (aq)

[(CH ) NH ][OH ] [(CH ) NH]

K

+ −

+ −

+ +

=

3 2 2 2 3 3 2

3 3 2a

3 2 2

(CH ) NH (aq) H O(l) H O (aq) (CH ) NH(aq)

[H O ][(CH ) NH] [(CH ) NH ]

K

+ +

+

+

+ +

=

(b) 14 10 2 2 14 10 2

14 10 2b

14 10 2

C H N (aq) H O(l) C H N H (aq) OH (aq)

[C H N H ][OH ] [C H N ]

K

+ −

+ −

+ +

=

14 10 2 2 3 14 10 2

3 14 10 2a

14 10 2

C H N H (aq) H O(l) H O (aq) C H N (aq)

[H O ][C H N ] [C H N H ]

K

+ +

+

+

+ +

=

(c) 6 5 2 2 6 5 3

6 5 3b

6 5 2

C H NH (aq) H O(l) C H NH (aq) OH (aq)

[C H NH ][OH ] [C H NH ]

K

+ −

+ −

+ +

=

6 5 3 2 3 6 5 2

3 6 5 2a

6 5 3

C H NH (aq) H O(l) H O (aq) C H NH (aq)

[H O ][C H NH ] [C H NH ]

K

+ +

+

+

+ +

=

10.28 Decreasing pKa will correspond to increasing acid strength because pKa

= The pKalog .− K a values (given in parentheses) determine the following

ordering:

295

3 3 2 5(CH ) NH (14.00 4.19 9.71) N H (14.00 5.77 8.23)+ +− = < − =

< HCOOH (3.75) < HF (3.45)

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

by

pKa + pKb = 14.

10.30 Decreasing pKb will correspond to increasing base strength because pKb

= blog− K . The pKb values (given in parentheses) determine the following

ordering:

2 4

2 5 3

N H (5.77) BrO (14.00 8.69 5.31) CN (14.00 9.31 4.69) (C H ) N(2.99)

− −< − = < − =<

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

by

pKa + pKb = 14.

10.32 Any base whose conjugate acid lies below water in Table 10.3 will be a

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

than water, and so water will preferentially protonate the base. Based upon

this information, we obtain the following analysis: (a) , strong; 2O −

(b) , weak; (c) Br−4HSO − , weak; (d) 3HCO − , weak; (e) CH3NH2,

weak; (f) H , strong; (g) −3CH − , strong.

10.34 In oxoacids with the same number of oxygen atoms attached to the central

atom, the greater the electronegativity of the central atom, the more the

electrons of the O—H bond are withdrawn, making the bond more polar.

This allows the hydrogen of the OH group to be more readily donated as a

proton to H2O, due to the stronger hydrogen bonds that it forms with the

oxygen of water. Therefore, HClO is the stronger acid, with the lower pKa.

10.36 (a) H3PO4 is stronger; it has the more electronegative central atom.

296

(b) HBrO3 is stronger; there are more O atoms attached to the central

atom in HBrO3, making the H—O bond in HBrO3 more polar than in

HBrO.

(c) We would predict H3PO4 to be a stronger acid due to more oxygens on

the central atom.

(d) H2Te is the stronger acid, because the H—Te bond is weaker than the

H—Se bond.

(e) HCl is the stronger acid. Within a period, the acidities of the binary

acids are controlled by the bond polarity rather than the bond strength, and

HCl has the greater bond polarity, due to the greater electronegativity of

Cl relative to S.

(f) HClO is stronger because Cl has a greater electronegativity than I.

10.38 (a) Methylamine is CH3NH2, ammonia is NH3. Methylamine can be

thought of as being formed from NH3 by replacing one H atom with CH3.

Because CH3 is less electron withdrawing than H, CH3NH2 is a weaker

acid and therefore a stronger base.

(b) Hydroxylamine is HONH2, hydrazine is H2N—NH2. The former can

be thought of as being derived from NH3 by replacement of one H atom

with OH; the latter by replacement of one H atom with NH2. Because the

hydroxyl group is more electron withdrawing than the amino group, NH2,

hydroxylamine is the stronger acid and therefore a weaker base.

10.40 The solution of 0.10 M H2SO4 would have the higher pH (would be the

weaker acid) because the conjugate base, HSO4− ,is less electronegative

than the conjugate base of hydrobromic acid, namely Br−.

10.42 The smaller the value of pKb, the stronger the base; hence, aniline is the

stronger base. 4-Chloroaniline is the stronger acid due to the presence of

the electron-withdrawing Cl atom, making it the weaker base; and again

we see that aniline is the stronger base.

297

10.44 The higher the pKa of an acid, the stronger the corresponding conjugate

base; therefore, the order is

3-hydroxyaniline < aniline < 2-hydroxyaniline < 4-hydroxyaniline

No simple pattern exists, but the position of the —OH group does affect

the basicity.

10.46 (a) 2 2

4 3 3 2a

3

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

[CH CH(OH)COOH] 0.12 0.12

+ −−= × = = ≈

−x xK

x

Here we have assumed that x is small enough to neglect it relative to 0.12

. This is a borderline case. We will also solve this exercise

without making this approximation and compare the results below.

1mol L−⋅

13[H O ] 0.010 mol L

pH log(0.010) 2.00pOH 14.00 2.00 12.00

+ −= = ⋅

= − == − =

x

Without the approximation, the quadratic equation that must be solved is 2

4

2 4 4

4 4 2

3

8.4 100.12

or 8.4 10 1.01 10 0

8.4 10 (8.4 10 ) 4( 1.01 10 )2

0.0096, 0.018.[H O ] 0.0096

−

− −

− −

+

× =−

+ × − × =

− × ± × − − ×=

= −

= =

xx

x x

x

xx

4−

pH log(0.0096) 2.02= − =

(b) 2

4a 38.4 10

1.2 10−

−= × =× −

xKx

or 2 4

3 3

8.4 10 1.0 10 01.5 10 , 1.1 10

− −

− −

+ × − × =

= − × ×

x xx

6

The negative root can be eliminated: 3

3

3

[H O ] 1.1 10

pH log(1.1 10 ) 2.96pOH 14.00 2.96 11.04

+ −

−

= = ×

= − × == − =

x

298

(c) 2

458.4 10

1.2 10−

−× =× −

xx

or 2 4

5 4

8.4 10 1.0 10 01.2 10 , 8.5 10

− −

− −

+ × − × =

= × − ×

x xx

8

The negative root can be eliminated. 5

3

5

[H O ] 1.2 10

pH log(1.2 10 )pH 4.92pOH 14.00 4.92 9.08

+ −

−

= ×

= − ×=

= − =

10.48 (a) 9b (pyridine) 1.8 10−= ×K

6 5 2 6 5C H N H O C H NH OH+ −+ +

Concentration C1(mol L )−⋅ 6H5N + H2O C6H5NH+ + OH−

initial 0.075 — 0 0

change −x — +x +x

final 0.075 − x — +x +x 2 2

9b

5 1

5

1.8 100.075 0.075

[OH ] 1.2 10 mol LpOH log(1.2 10 ) 4.92pH 14.00 4.92 9.00

−

− − −

−

= × = ≈−

= = × ⋅

= − × == − =

x xKx

x

Percentage ionized = 5

21.2 10 100% 1.6 10 %0.075

−−×

× = ×

(b) The setup is similar to that in (a). 2 2

6b 1.0 10 [OH ]

0.0112 0.0112− −= × = ≈ =

−x xK x

x

299

4 1

4

4

1.1 10 mol LpOH log(1.1 10 ) 3.97pH 14.00 3.97 10.03

3.2 10Percentage protonation 100% 2.6%0.0122

− −

−

−

= × ⋅

= − × == − =

×= × =

x

(c) bp 14.00 8.52 5.48= − =K

6b 3.3 10−= ×K

2quinine H O quinineH OH+ −+ +

2 26

b 3.3 100.021 0.021

−= × = ≈−

x xKx

4 1

4

4

[OH ] 2.6 10 mol LpOH log(2.6 10 ) 3.58pH 14.00 3.58 10.42

2.6 10Percentage protonation 100% 1.2%0.021

− − −

−

−

= = × ⋅

= − × == − =

×= × =

x

(d) 2strychnine H O strychnineH OH+ −+ +

146w

b 9a

2 26

b

4 1

4

4

[OH ]1.00 10 1.82 105.49 10

1.82 100.059 0.059

[OH ] 3.3 10 mol LpOH log(3.3 10 ) 3.48pH 14.00 3.48 10.52

3.3 10Percentage protonation 100% 0.56%0.059

−

−−

−

−

− − −

−

−

=

×= = = ×

×

= × = ≈−

= = × ⋅

= − × == − =

×= × =

xK

KK

x xKx

x

300

10.50 (a) HN 2 2 3 2O H O H O NO+ −+ +

2.63 3 13 2

3 24

a 3

4a

[H O ] [NO ] 10 antilog ( 2.63) 2.3 10 mol L

(2.3 10 ) 4.2 100.015 2.3 10

p log(4.3 10 ) 3.38

+ − − −

−−

−

−

−= = = − = × ⋅

×= = ×

− ×

= − × =

K

K

(b) 4 9 2 2 4 9 3C H NH H O C H NH OH+ −+ +

1.96 14 9 3

pOH 14.00 12.04 1.96[C H NH ] [OH ] 10 antilog ( 1.96) 0.011 mol L+ − −

= − =

= = = − = ⋅ −

234 9 3

b4 9 2

3b

[C H NH ][OH ] (0.011) 1.4 10[C H NH ] 0.10 0.011

p log(1.4 10 ) 2.85

+ −−

−

= = =−

= − × =

K

K

×

10.52 (a) 10a (HCN) 4.9 10−= ×K

pH 5.3 63[H O ] 10 10 antilog ( 5.3) 5 10 mol L+ − − − 1−= = = − = × ⋅

Let x = nominal concentration of HCN, then

Concentration HCN + H1(mol L )−⋅ 2O H3O+ + CN −

nominal x — 0 0

equilibrium 65 10−− ×x — 65 10−× 65 10−×

6 2−10

6

2 1 1

(5 10 )4.9 105 10

Solve for : 5 10 mol L 0.05 mol L

−−

− − −

×× =

− ×

= × ⋅ = ⋅

xx x

(b) 9b (pyridine) 1.8 10−= ×K

pOH 5.2 6 1

pOH 14.00 8.8 5.2[OH ] 10 10 antilog ( 5.2) 6 10 mol L− − − −

= − =

= = = − = × ⋅ −

Let x = nominal concentration of C5H5N, then

Concentration C1(mol L )−⋅ 5H5N + H2O C5H6N+ + OH−

nominal x

equilibrium 66 10−− ×x 66 10−× 66 10−×

301

6 29

6

2 1

(6 10 )1.8 106 10

Solve for : 2 10 mol L 0.02 mol L

−−

−

1− − −

×× =

− ×

= × ⋅ = ⋅

xx x

10.54 veronal + H2O H3O+ + veronalate ion

The equilibrium concentrations are 1

5 13

5

5 283

a

[veronal] 0.020 0.0014 0.020 0.020 mol L[H O ] [veronalate ion] 0.0014 0.020 2.8 10 mol L

pH log(2.8 10 ) 4.55[H O ][veronalate ion] (2.8 10 ) 3.9 10

[veronal] 0.020

−

+ − −

−

+ −−

= − × = ⋅

= = × = × ⋅

= − × =

×= = = ×K

10.56 cacodylic acid + H2O H3O+ + cacodylate ion

The equilibrium concentrations are 1

5 13

5

5 23

a

[cacodylic acid] 0.0110 (0.0077 0.0110) 0.0110 0.000 08 0.109 mol L[H O ] [cacodylate ion] 0.0077 0.0110 8.5 10 mol L

pH log(8.5 10 ) 4.07[H O ][cacodylate ion] (8.5 10 )

[cacodylic acid] 0.010

−

+ − −

−

+ −

= − × = − = ⋅

= = × = × ⋅

= − × =

×= =K 76.6 10

9−= ×

C N

H

H

H

H

H

C N+

H

H

H

H

H

H

methylamine conjugate acid

10.58 (a)

(b) First determine the concentration of methylamine following dilution.

Moles of methylamine are found by:

302

( ) ( )( ) ( )

( )( )

1

1

50 mL 0.85 g mL 42.5 g of solution

42.5 g 0.35 14.9 g of methylamine

14.9 g0.479 mol of methylamine

31.06 g mol

diluted to 1.000 L, the resulting solution is 0.479 M in methylamine

−

−

⋅ =

⋅ =

=⋅

Concentration CH1(mol L )−⋅ 3NH2 + H2O CH3NH3+ + OH−

initial 0.479 — 0 0

equilibrium 0.479 − x — x x

[ ]2

3 34b

3 2

2

2

CH NH OH3.6 10

CH NH 0.479solving for x we find:

1.3 10 M OH

pOH log(1.3 10 ) 1.9pH 14.0 1.9 12.1

+ −−

− −

−

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= × = =−

⎡ ⎤= × = ⎣ ⎦= − × =

= − =

xKx

x

10.60 (a) 22 2 4 2 4pH 7, basic: H O(l) C O (aq) HC O (aq) OH (aq)− −> + + −

−

O (aq) Fe(H O) OH (aq)+ ++

(b) 23pH 7, neutral: Ca is not an acid and NO is not a base+ −=

(c) 3 3 2 3 3 2pH 7, acidic: CH NH (aq) H O(l) H O (aq) CH NH (aq)+ +< + +

(d) 3 22 4 4pH 7, basic: H O(l) PO (aq) HPO (aq) OH (aq)− −> + +

(e) 32 6 2pH 7, acidic: Fe(H O) (aq) H O(l)+< +

H 23 2 5

(f) 5 5 2 3 5 5pH 7, acidic: C H NH (aq) H O(l) H O (aq) C H N(aq)+ +< + +

10.62 (a) Concentration

CH1(mol L )−⋅ 3NH3+ + H2O(l) CH3NH2 + H3O+

initial 0.25 — 0 0

change −x — +x +x

equilibrium 0.25 − x — x x

303

(See Table 10.2 for Kb of CH3NH2, the conjugate base of CH3NH3+.)

1411w 3 2

a 4b 3 3

2 211

6 1 63

[CH NH ][H O ]1.00 10 2.8 103.6 10 [CH NH ]

2.8 100.25 0.25

2.6 10 mol L [H O ] and pH log(2.6 10 ) 5.58

+−−

− +

−

− − + −

×= = = × =

×

× = ≈−

= × ⋅ = = − × =

KK

K

x xx

x

3

(b) Concentration

SO1(mol L )−⋅ 23

− + H2O(l) 3HSO − + OH −

initial 0.13 — 0 0

change — +x +x −x

equilibrium 0.13 − x — x x

[See Table 10.9 for a 3 a2 2 3(HSO ) (H SO ).]− =K K

14 2 28w 3

b 7 2a2 3

28

[HSO ][OH ]1.00 10 8.3 101.2 10 [SO ] 0.13 0.13

8.3 100.13

K x xKK x

x

− −−−

− −

−

×= = = × = = ≈

× −

× =4 1

4

1.0 10 mol L [OH ]pOH log(1.0 10 ) 4.00pH 14.00 pOH 10.00

− −

−

= × ⋅ =

= − × == − =

x −

(c) Concentration

Fe(H1(mol L )−⋅ 2O)63+(aq) + H2O(l) H3O+(aq) + Fe(H2O)5OH+(aq)

initial 0.071 — 0 0

change — +x +x −x

equilibrium 0.071 − x — x x

233 2 5

a 32 6

[H O ][Fe(H O) OH (aq)]3.5 10

0.071[Fe(H O) ]

+ +−

+= = × =−

xKx

2 33.5 10−+ ×x x 42.5 10 0−× =

x = 0.014, 0.018−

The negative root can be discarded

3[H O ] 0.014, pH log 0.014 .85+ = = − =

304

10.64 (a) Initial concentration of C6H5NH3+ is:

( ) 1

17.8 g 94.133 g mol

0.24 M0.350 L

−

⎛ ⎞⎜ ⎟⎝ ⎠ =

Given this initial concentration and the Ka for this acid found in Table

10.7, the percent deprotonation is found as follows:

Concentration

C1(mol L )−⋅ 6H5NH3+ + H2O(l) C6H5NH2

2+ + H3O+

initial 0.24 — 0 0

change − — +x +x x

equilibrium 0.24 − x — x x 2+

5 6 5 2 3a +

6 5 3

2 25

3 13

3 1

1

[C H NH ][H O ]2.3 10

[C H NH ]

2.3 100.24 0.24

2.3 10 mol L [H O ]

2.3 10 mol Lpercent dissociation 100% 1.0%0.23 mol L

+−

−

− − +

− −

−

= × =

× = ≈−

= × ⋅ =

× ⋅= ⋅ =

⋅

K

x xx

x

10.66 (a) can act as both an acid and a base. Both actions need to be

considered simultaneously:

3HSO −

23 2 3 3

3 2 2 3

HSO H O H O SO

HSO H O H SO OH

− +

− −

+ +

+ +

−

3−

Summing, 2

3 3 2 3HSO HSO H SO SO− −+ +

The simplest approach is to recognize that there are two conjugate acid-

base pairs and to use the Henderson-Hasselbalch equation twice, once for

each pair.

305

3a1

2 3

23

a23

[HSO ]pH p log

[H SO ]

[SO ]pH p log

[HSO ]

−

−

−

⎛ ⎞= + ⎜ ⎟

⎝ ⎠⎛ ⎞

= + ⎜ ⎟⎝ ⎠

K

K

Summing, 2

3 3a1 a2

2 3 3

[HSO ] [SO ]2pH p p log

[H SO ] [HSO ]

− −

−

⎛ ⎞= + + ×⎜ ⎟

⎝ ⎠K K

Because [ 22 3 3 a1 a2H SO ] [SO ], 2pH p p−= = +K K

Therefore, 1 1a1 a 22 2pH (p p ) (1.81 6.91) 4.36= + = + =K K

Note that it is not necessary to know the concentration of because

it cancels out of the equation. At extremely low concentrations of

however, the approximations upon which the use of this equation are

based are no longer valid.

3HSO ,−

3HSO ,−

(b) Neither silver ion nor nitrate ion is acidic or basic in aqueous solution.

Therefore, the pH is that of neutral water, 7.00.

10.68 (a) 1

10

0.250 mol L KCN 0.0350 L 0.0875 mol L KCN [CN ]0.1000 L

−− −⋅ ×

= ⋅ =

Concentration

H1(mol L )−⋅ 2O(l) + HCN(aq) + CN (aq)− OH (aq)−

initial — 0.0875 0 0

change — −x +x +x

equilibrium — 0.0875 − x x x

145w

b 10a

2 25

3 1

1.00 10 [HCN][OH ]2.0 104.9 10 [CN ]

2.0 100.0875 0.0875

1.3 10 mol L [HCN]

− −−

− −

−

− −

×= = = × =

×

× = ≈−

= × ⋅ =

KK

K

x xx

x

306

(b) 3

3 3

2 13

1.59 g NaHCO 184.01 g NaHCO /mol NaHCO 0.200 L

9.46 10 mol L NaHCO− −

×

= × ⋅

Concentration

H1(mol L )−⋅ 2O(l) + H3HCO (aq)−2CO3(aq) + OH (aq)−

initial — 29.46 10−× 0 0

change — −x +x +x

equilibrium — 29.46 10−× − x x x

1482 3 w

b1 7a13

2 28 5 1

2 2

5

[H CO ][OH ] 1.00 10 2.3 10[HCO ] 4.3 10

2.3 10 , 4.7 10 mol L9.46 10 9.46 10

[OH ]pOH log(4.7 10 ) 4.33, pH 14.00 4.33 9.67

− −−

− −

− −− −

−

−

×= = = = ×

×

× = ≈ = × ⋅× − ×

=

= − × = = − =

KK

K

x x xx

−

10.70 The reaction of interest is:

C6H5CH2(CH3)NH3+(aq) + H2O(l) C6H5CH2(CH3)NH2(aq) +

H3O+(aq)

Ka for this reaction is: 14

11a 4

1 10 1.3 107.8 10

−−

−

×= = ×

×K

The initial concentration of C6H5CH2(CH3)NH3+(aq) is:

( ) ( )-1

-16.48 g 216.12 g mol

0.150 mol L0.200 L

⋅=

Concentration

H1(mol L )−⋅ 2O(l) + C6H5CH2(CH3)NH3+(aq) C6H5CH2(CH3)NH2(aq) + H3O+(aq)

initial — 9.46 210−× 0 0

change — −x +x +x

equilibrium — 9.46 210−× − x x x

307

2 211 6 1

+3

6

1.3 10 , 1.4 10 mol L0.145 0.145

[H O ]

pH log(1.4 10 ) 5.9

x x xx

− −

−

× = ≈ = × ⋅−

=

= − × =

−

O (aq) H O(l) H O (aq) H PO (aq)+ −+ +

−

−

−

−

−

10.72 (a) H P 3 4 2 3 2 4

22 4 2 3 4

2 34 2 3 4

H PO (aq) H O(l) H O (aq) HPO (aq)

HPO (aq) H O(l) H O (aq) PO (aq)

− +

− +

+ +

+ +

(b)

2 4 2 2 3 2 4 2

22 4 2 2 3 2 4 2 2

(CH ) (COOH) (aq) H O(l) H O (aq) (CH ) (COOH)CO (aq)

(CH ) (COOH)CO (aq) H O(l) H O (aq) (CH ) (CO ) (aq)

+ −

− +

+ +

+ +

(c)

2 2 2 2 3 2 2 2

22 2 2 2 3 2 2 2 2

(CH ) (COOH) (aq) H O(l) H O (aq) (CH ) (COOH)CO (aq)

(CH ) (COOH)CO (aq) H O(l) H O (aq) (CH ) (CO ) (aq)

+ −

− +

+ +

+ +

10.74 The reaction is (after the first, essentially complete ionization) 2

4 2 3 4HSeO H O H O SeO− ++ +

The initial concentrations of 14 3HSeO and H O are both 0.010 mol L− + −⋅

due to the complete ionization of H2SeO4 in the first step. The second

ionization is incomplete.

1(mol L )−⋅ 4HSeO − + H2O H3O+ + Se 24

Concentration

O −

initial 0.010 — 0.010 0

change −x — +x +x

equilibrium 0.010 − x — 0.010 + x x

308

22 3 4

a24

2 4

2 43

3 1 23

2

[H O ][SeO ] (0.010 )( )1.2 100.010[HSeO ]

0.022 1.2 10 0

0.022 (0.022) (4)( 1.2 10 )4.5 10

2[H O ] 0.010 (0.010 4.5 10 ) mol L 1.5 10 mol L

pH log(1.5 10 ) 1.82

+ −−

−

−

−−

+ − −

−

+= × = =

−

+ − × =

− + − − ×= = ×

1− −= + = + × ⋅ = × ⋅

= − × =

x xKx

x x

x

x

10.76 (a) Second ionization is ignored because a 2 a1<<K K

+2 2 3H S H O H O HS−+ +

2 27 3

a12

4 13

4

[H O ][HS ]1.3 10

[H S] 0.10 0.10

[H O ] 1.1 10 mol L

pH log(1.1 10 ) 3.96

+ −−

+ − −

−

= × = = ≈−

= = × ⋅

= − × =

x xKx

x

(b) This is a situation where it may not be justified to ignore the second

ionization. . This is a

marginal case; we can work it both ways, first without ignoring the second

ionization. Adopt the following notation:

4 5a1 a2 a2 a16.0 10 , 1.5 10 , and / 40− −= × = × =K K K K

H+ = H3O+

[H+] = [H3O+] = equilibrium concentration of H3O+

H2A = tartaric acid 1

0c solute molarity 0.15 mol L−= = ⋅

−

=

20 2c [H A] [HA ] [A ]−= + + (1)

2 0H total H present [H ] 2[H A] [HA ] 2c+ −= = + + (2)

The following equilibria occur:

2H A H HA+ −+ a12

[H ][HA ][H A]

+ −

=K (3)

2HA H A− + + − 2

a2[H ][A ]

[HA ]

+ −

−=K (4)

309

Examination of Eqs. 1 to 4 shows that there are four unknowns,

, to be determined. However, these four

simultaneous equations allow for their determination.

22[H ], [H A], [HA ], and [A ]+ − −

The unknowns 22[H A], [HA ], and [A ]− − may all be expressed in terms of

one unknown, [H+]. 2

0 2According to Eq.1, c [H A] [HA ] [A ]− −= + +

2a1

2

a 22 2

2a1 a 2

[H ][HA ]From Eq. 3, [H A]

[H ][A ]From Eq. 4, [HA ]

[H ] [A ]Substituting, [H A]

+ −

+ −−

+ −

=

=

=

K

K

K K

22

0a2 a1 a2

[H ] [H ]Then Eq.1 becomes c [A ] 1+ +

− ⎧ ⎫= + +⎨ ⎬

⎩ ⎭K K K (5)

Now, subtract 2 Eq.1 from Eq. 2 :×

0 22c [H ] 2[H A] [HA ]+ −= + + Eq. 2

20 2

2

2c 2[H A] 2[HA ] 2[A ][H ] [HA ] 2[A ] 0

− −

+ − −

= + +− − =

2 Eq.1×

2 2

a2

[H ][A ], [A ][H ] 2

+− −

+=+

K

Substitute this into Eq. 5 to give 2 3

0a2 a2 a1 a2

[H ] [H ] [H ]c 2 [H ]+ +

+⎛ ⎞+ = + +⎜ ⎟

⎝ ⎠K K K

+

K

Rearranging into standard cubic form gives 3 2

00

a1 a 2 a2 a2

c[H ] [H ] [H ] 1 2c 0+ +

+ ⎛ ⎞+ + − −⎜ ⎟

⎝ ⎠K K K K=

=

Now, put in the numerical values for c0, Ka1, and Ka2: 8 3 4 2 41.1 10 [H ] 6.7 10 [H ] 1.0 10 [H ] 0.30 0+ + +× + × − × −

Solution of this cubic by standard methods gives

310

3 13

3

[H ] [H O ] 9.2 10 mol L

pH log(9.2 10 ) 2.04

+ + −

−

= = × ⋅

= − × =

−

It is left as an exercise for the reader to show that, if the second ionization

is ignored, 3 1[H ] 9.5 10 mol L and pH 2.02+ − −= × ⋅ =

The difference is small but, perhaps, not within experimental error.

(c) The second ionization can be ignored because Ka2 << K a1.

2 4 2 3 4

2 28 3 4

a1 3 32 4

6 13

6

H TeO H O H O HTeO

[H O ][HTeO ]2.1 10

[H TeO ] 1.1 10 1.1 10[H O ] 4.8 10 mol L

pH log(4.8 10 ) 5.32

+ −

+ −−

− −

+ − −

−

+ +

= × = = ≈× − ×

= = × ⋅

= − × =

x xKx

x

10.78 (a) The pH is given by pH = 1a1 a22 (p p )+K K . From Table 10.9, we find

7a1 4.3 10−= ×K a1p 6.37=K

11a2 5.6 10−= ×K a2p 10.25=K

12pH (6.37 10.25) 8.31= + =

(b) The nature of the spectator counter ion does not affect the equilibrium

and the pH of a salt solution of a polyprotic acid is independent of the

concentration of the salt, therefore pH = 8.31.

10.80 The pH is given by pH = 1a1 a22 (p p )+K K .

12pH (2.46 7.31) 4.89= + =

10.82 The equilibrium reactions of interest are

2 3 2 3 3H SO (aq) H O(l) H O (aq) HSO (aq)+ −+ + 2a1 1.5 10−= ×K

23 2 3 3HSO (aq) H O(l) H O (aq) SO (aq)− + −+ + 7

a2 1.2 10−= ×K

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

can assume that the first step dominates:

311

Concentration 1(mol L )−⋅ H2SO3(aq) + H2O(l) H3O+(aq) + 3HSO (aq)−

initial 0.125 — 0 0

change — +x +x −x

final 0.125 − x — +x +x

3 3a1

2 32

2

[H O ][HSO ][H SO ]

( )( )1.5 100.125 0.125

+ −

−

=

× = =− −

K

x x xx x

Assume that x << 0.125, then 2 2(1.5 10 )(0.125)

0.043

−= ×=

xx

Because x > 5% of 0.0456, the assumption was not valid, and the full

expression must be evaluated using the quadratic: 2 2 21.5 10 (1.5 10 )(0.125) 0− −+ × − × =x x

Solving with the quadratic equation gives 10.036 mol L−= ⋅x . 1

3 3

1 1 12 3

[H O ] [HSO ] 0.036 mol L

[H SO ] 0.125 mol L 0.036 mol L 0.089 mol L

+ − −

− − −

= = = ⋅

= ⋅ − ⋅ = ⋅

x

We can then use the other equilibria to determine the remaining

concentrations: 2

3 3a2

32

7 3

2 73

[H O ][SO ][HSO ]

(0.036)[SO ]1.2 10

(0.036)[SO ] 1.2 10 mol L

+ −

−

−−

− −

=

× =

1−= × ⋅

K

Because 1.2 , the initial assumption that the first

dissociation would dominate is valid. To calculate [

710 0.036−× <<

OH ]− , we use the Kw

relationship:

312

w 314

13 1w

3

[H O ][OH ]

1.00 10[OH ] 2.8 10 mol L0.036[H O ]

+ −

−− −

+

=

×= = = × ⋅

KK −

2

In summary, 1 1

2 3 3 3 3[H SO ] 0.089 mol L , [H O ] [HSO ] 0.036 mol L , [SO ]− + − −= ⋅ = = ⋅ −

1−

7 1 131.2 10 mol L , [OH ] 2.8 10 mol L− − − −= × ⋅ = × ⋅

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

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

following changes: 2

3 2 314

8wb1 7

a2

3 2 2 314

13wb2 2

a2

SO (aq) H O(l) HSO (aq) OH (aq)

1.00 10 8.3 101.2 10

HSO (aq) H O(l) H SO (aq) OH (aq)

1.00 10 6.7 101.5 10

− − −

−−

−

− −

−−

−

+ +

×= = = ×

×

+ +

×= = = ×

×

KK

K

KK

K

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

can assume that the first step dominates:

Concentration 1(mol L )−⋅ + H2

3SO (aq)−2O(l) + 3HSO (aq)− OH (aq)−

initial 0.125 — 0 0

change — +x +x −x

final 0.125 − x — +x +x

3b1 2

3

28

[HSO ][OH ][SO ]

( )( )8.3 100.125 0.125

− −

−

−

=

× = =− −

K

x x xx x

Assume that x << 0.125, then 2 8

4

(8.3 10 )(0.125)1.0 10

−

−

= ×

= ×

xx

313

Because x < 1% of 0.125, the assumption was valid. 4 1

3

2 1 4 13

[HSO ] [OH ] 1.0 10 mol L

Therefore, [SO ] 0.125 mol L 1.0 10 mol L 0.125 mol L

− − − −

− − − −

= = = × ⋅

= ⋅ − × ⋅ ≅

x1−⋅

We can then use the other equilibria to determine the remaining

concentrations:

2 3b2

34

13 2 34

13 12 3

[H SO ][OH ][HCO ]

[H SO ](1.0 10 )6.7 10

(1.0 10 )[H SO ] 6.7 10 mol L

−

−

−−

−

− −

=

×× =

×

= × ⋅

K

Because , the initial assumption that the first

hydrolysis would dominate is valid.

13 46.7 10 1.0 10−× << × −

To calculate [H3O+], we use the Kw relationship:

w 314

10 1w3 14

[H O ][OH ]

1.00 10[H O ] 1.0 10 mol L[OH ] 1.0 10

+ −

−+ −

− −

=

× −= = = ××

KK

⋅

1

13 12 3

4 1 23 3

1 103

In summary, [H SO ] 6.7 10 mol L , [OH ]

[HSO ] 1.0 10 mol L , [SO ]

0.125 mol L , [H O ] 1.0 10 mol L

− − −

− − − −

− + − −

= × ⋅

= = × ⋅

= ⋅ = × ⋅

10.86 (a) tartaric acid: The two pKa values are 3.22 and 4.82. Because pH = 5.0

lies above both of these values, the major form present will be the doubly

deprotonated ion . (b) hydrosulfuric acid: The two pK2A −a values are

6.89 and 14.15. Because the pH of the solution lies below both of these

values, the dominant form will be the doubly protonated H2A. (c)

phosphoric acid: The three pKa values are 2.12, 7.21, and 12.68. The pH of

the solution lies between the first and second ionization, so the

predominant species should be the singly deprotonated ion 2 4H PO .−

314

10.88 The equilibria present in the solution are:

2 2 3

22 3

H (aq) H O(l) H O (aq) HS (aq)

HS (aq) H O(l) H O (aq) S (aq)

+ −

− + −

+ +

+ +

7a1

15a2

1.3 10

7.1 10

−

−

= ×

= ×

K

K

The calculation of the desired concentrations follows exactly after the

method derived in Eq. 25, substituting H2S for H2CO3, HS− for ,

and for . First, calculate the quantity

:

3HCO −

2S − 23CO −

9.35 10 13(at pH 9.35 [H O ] 10 4.5 10 mol L )+ − − −= = = ×f ⋅

−

23 3 a1 a1 a2

10 2 10 7 7 15

17

[H O ] [H O ]

(4.5 10 ) (4.5 10 )(1.3 10 ) (1.3 10 )(7.1 10 )5.9 10

+ +

− − − −

−

= + +

= × + × × + × ×

= ×

f K K K

The fractions of the species present are then given by 10 2

332 17

[H O ] (4.5 10 )(H S) 3.4 105.9 10

α+ −

−−

×= = = ×

×f

10 73 a1

17

[H O ] (4.5 10 )(1.3 10 )(HS ) 0.995.9 10

α+ − −

−−

× ×= = =

×K

f

17 152 15a1 a2

17

(1.3 10 )(7.1 10 )(S ) 1.6 105.9 10

α− −

− −−

× ×= = = ×

×K K

f

Thus, in a solution at pH 9.35, the dominant species will be with a

concentration of

HS−

1 1(0.250 mol L )(0.99) 0.25 mol L− −⋅ ≅ ⋅ . The

concentration of H2S will be

, and the concentration of

will be

3 1 4(3.4 10 )(0.250 mol L ) 8.5 10 mol L− − −× ⋅ = × 1−⋅

1−⋅2S − 5 1 6(1.6 10 )(0.250 mol L ) 4.0 10 mol L .− − −× ⋅ = ×

10.90 For the first ionization of (COOH)2—or H2C2O4—we write 2

2 2 4 2 3 2 4 a1H C O H O H O HC O , 5.9 10+ − −+ + =K ×

315

Concentration 1(mol L )−⋅ H2C2O4 + H2O H3O+ + 2 4HC O −

initial 0.10 — 0 0

change — +x +x −x

equilibrium 0.10 − x — x x 2 5

a1 a25.9 10 , 6.5 10− −= × = ×K K

Because Ka2 << Ka1, the second ionization can safely be ignored in the

calculation of [H3O+]. 2

2a1

2

21

3

1413 1

12 2 4

5.9 100.10

0.059 0.0059 0

0.059 (0.059) (4)(1)(0.0059)0.053 mol L [H O ]

21.0 10[OH ] 1.9 10 mol L

0.053[H C O ] 0.10 0.053 0.05 mol L

−

− +

−− − −

−

= × =−

+ − =

− + += =

×= = × ⋅

= − = ⋅

xKx

x x

x ⋅ =

1Concentration (mol L )−⋅ 2 4HC O − +H2O H3O+ + C O 2

2 4−

initial 0.053 — 0.053 0

change −x — +x +x

equilibrium 0.053 − x — 0.053 + x x

5a2

2 5 12 4

12 4

(0.053 )( )6.5 10 (because is small)0.053

or [C O ] 6.5 10 mol L , and

[HC O ] 0.053 0.053 0.000 065 0.053 mol L

−

− − −

− −

+= × = ≈

−

= = × ⋅

= − = − = ⋅

x xK xx

x

x

x

10.92 The equilibria involved are:

H2CO3(aq) + H2O(l) HCO3¯(aq) + H3O+ Ka1 = 4.3 × 10⎯7

HCO3⎯(aq) + H2O(l) CO3

2⎯(aq) + H3O+ Ka2 = 5.6 × 10⎯11

Given these two equilibria, two equations can be written:

316

[ ]

[ ]

+3 3

a12 3

2 +3 3

a23

2 32 3 3 3

+ 4.8 53

HCO H O

H CO

CO H O

HCO

also, enough information is provided to construct two other equations:

H CO HCO CO 4.5 10

H O 10 1.6 10

−

−

−

− − −

− −

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦=

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦=⎡ ⎤⎣ ⎦

⎡ ⎤ ⎡ ⎤+ + = ×⎣ ⎦ ⎣ ⎦⎡ ⎤ = = ×⎣ ⎦

K

K

Solving this set of simultaneous equations for the three unknown

concentrations, one finds:

[ ] 32 3

43

2 13

H CO 4.38 10

HCO 1.19 10

CO 4.23 10

−

− −

− −

= ×

⎡ ⎤ = ×⎣ ⎦⎡ ⎤ = ×⎣ ⎦

0

4

=

4

10.94 We can use the relationship derived in the text:

2

3 initial 3 w

2 8 13 3

[H O ] [HA] [H O ] 0

[H O ] (7.49 10 )[H O ] (1.00 10 ) 0

+ +

+ − + −

− − =

− × − × =

K

Solving using the quadratic equation gives

73[H O ] 1.44 10 ; pH 6.842+ −= × =

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

concentration alone 8(pH log(7.49 10 ) 7.126).−= − × =

10.96 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 ] 0,+ ++ − K

2 7 13 3[H O ] (8.23 10 )[H O ] (1.00 10 ) 0+ − + −+ × − × =

Solving using the quadratic equation gives

83[H O ] 1.20 10 , pH 7.922+ −= × =

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

concentration alone

7(pOH log(8.23 10 ) 6.084, pH 14.00 6.084 7.916).−= − × = = − =

317

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

water, the pH values of the 42.50 10−× M and 62.50 10−× M C6H5OH

solutions can be calculated as described earlier.

For 4 12.50 10 mol L :− −× ⋅

Concentration

C1(mol L )−⋅ 6H5OH(aq) + H2O(l) + 3H O (aq)+6 5C H O (aq)−

initial 42.50 10−× — 0 0

change −x — +x +x

final 42.50 10−× − x — +x +x

3 6 5a

6 52

104 4

[H O ][C H O ][C H OH]

( )( )1.3 102.50 10 2.50 10

+ −

−− −

=

× = =× − × −

K

x x xx x

Assume x << 42.50 10−×

2 10

7

(1.3 10 )(2.50 10 )1.8 10

− −

−

= × ×

= ×

xx

4

10 , Because x < 1% of 2.50 4−×

10 mol L :− −

the assumption was valid. Given this

value, the pH is then calculated to be 8log(1.8 10 ) 6.74.−− × =

For 2.50 6 1× ⋅

O (aq)−

Concentration

C1(mol L )−⋅ 6H5OH(aq) + H2O(l) + C H

initial

3H O (aq)+6 5

62.50 10−× — 0 0

change −x — +x +x

final 62.50 10−× − x — +x +x

3 6 5a

6 52

106 6

[H O ][C H O ][C H OH]

( )( )1.3 102.50 10 2.50 10

+ −

−− −

=

× = =× − × −

K

x x xx x

318

Assume x << 62.50 10−×

2 10

8

(1.3 10 )(2.50 10 )1.8 10

− −

−

= × ×

= ×

xx

6

Because x < 1% of 62.50 10 ,−× the assumption is valid and the pH should

be 7.74. However, this number does not make sense because an acid was

added to the water.

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

we can use Eq. 22:

where x = [H3 2a w a initial w a( [HA] )+ − + ⋅ − ⋅ =x K x K K x K K 0, 3O+].

To solve the expression, you substitute the values of

the initial concentration of acid, and

14w 1.00 10 ,−= ×K

10a 1.3 10−= ×K into this equation

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

a graphing calculator. Alternatively, you can use a computer program

designed to solve simultaneous equations. Because the unknowns include

[H3O+], [ [HClO], and [ you will need four equation. As

seen in the text, the pertinent equations are

OH ],− ClO ],−

3 6 5a

6 5

w 3

3 6 5

6 5 initial 6 5 6 5

[H O ][C H O ][C H OH]

[H O ][OH ]

[H O ] [OH ] [C H O ]

[C H OH] [C H OH] [C H O ]

+ −

+ −

+ − −

−

=

=

= +

= +

K

K

Using either method should produce the same result.

The values obtained for 4 12.50 10 mol L− −× ⋅ are

(compare to 6.74 obtained in (a)) 7 13[H O ] 2.1 10 mol L , pH 6.68+ − −= × ⋅ =

7 16 5

4 16 5

8 1

[C H O ] 1.6 10 mol L

[C H OH] 2.5 10 mol L

[OH ] 4.8 10 mol L

− − −

− −

− − −

= × ⋅

≅ × ⋅

= × ⋅

Similarly, for 66 5 initial[C H OH] 2.50 10 :−= ×

(compare to 7.74 obtained in (a)) 7 13[H O ] 1.0 10 mol L , pH 7.00+ − −= × ⋅ =

319

9 16 5

6 16 5

8 1

[C H O ] 3.2 10 mol L

[C H OH] 2.5 10 mol L

[OH ] 9.8 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.

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

water, the pH values of the 51.89 10−× M and 79.64 10−× M HClO

solutions can be calculated as described earlier.

For : 5 11.89 10 mol L− −× ⋅

Concentration

HClO(aq) + H1(mol L )−⋅ 2O(l) H3O+(aq) + Cl O (aq)−

initial 51.89 10−× — 0 0

change — +x +x −x

final 51.89 10−× − x — +x +x

3a

28

5 5

[H O ][ClO ][HClO]

( )( )3.0 101.89 10 1.89 10

+ −

−− −

=

× = =× − × −

K

x x xx x

Assume 51.89 10−<< ×x

2 8

7

(3.0 10 )(1.89 10 )7.5 10

− −

−

= × ×

= ×

xx

5

10 Because x < 5% of 1.89 5−×

10 mol L− −

, the assumption was valid. Given this

value, the pH is then calculated to be 7log(7.5 10 ) 6.12.−− × =

For 9.64 7 1× ⋅ :

320

Concentration

HClO(aq) + H1(mol L )−⋅ 2O(l) H3O+(aq) + Cl O (aq)−

initial 79.64 10−× — 0 0

change — +x +x −x

final 79.64 10−× − x — +x +x

3a

28

7 7

[H O ][ClO ][HClO]

( )( )3.0 109.64 10 9.64 10

+ −

−− −

=

× = =× − × −

K

x x xx x

Assume 79.64 10−<< × −x x

2 8 7

7

(3.0 10 )(9.64 10 )1.7 10

− −

−

= × × −

= ×

x xx

x is approximately 10% of 79.64 10−× ; the assumption is not reasonable,

and so you must calculate the value explicitly for the following

expression, using the quadratic equation.

2 8 8 73.0 10 (3.0 10 )(9.64 10 ) 0− − −+ × − × × =x x

Upon solving the quadratic equation, a value of is obtained,

yielding pH = 6.80.

71.6 10−= ×x

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

we can use Eq. 21:

. 3 2a 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

8a 3.0 10−= ×K into this equation and

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

graphing calculator. Alternatively, you can use a computer program

designed to solve simultaneous equations. Because the unknowns include

[H3O+], [ , [HClO], and [OH ]− ClO ]− , you will need four equations. As

seen in the text, the pertinent equations are

321

3a

[H O ][ClO ][HClO]

+ −

=K

w 3[H O ][OH ]+ −=K

3[H O ] [OH ] [ClO ]+ −= + −

initial[HClO] [HClO] [ClO ]−= +

Using either method should produce the same result.

The values obtained for 5 11.89 10 mol L− −× ⋅ are

7 13[H O ] 7.4 10 mol L , pH 6.13 (compare to 6.12 obtained in (a))+ − −= × ⋅ =

7 1[ClO ] 7.3 10 mol L− −= × ⋅ −

−

−

−

5 1[HClO] 1.8 10 mol L− −≅ × ⋅

8 1[OH ] 1.3 10 mol L− −= × ⋅

7initialSimilarly, for [HClO] 9.64 10 :−= ×

7 13[H O ] 1.9 10 mol L , pH 6.72 (compare to 6.80 obtained in (a))+ − −= × ⋅ =

7 1[ClO ] 1.3 10 mol L− −= × ⋅

7 1[HClO] 8.3 10 mol L− −≅ × ⋅

8 1[OH ] 5.3 10 mol L− −= × ⋅

Note than 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.

10.102 Because the process is only 90% efficient, to remove 50.0 kg of SO2 one

must supply enough CaCO3 to remove 55.6 kg of SO2 (50.0 kg is 90% of

55.6 kg). The moles of to be removed is given by: 2SO

21

55600 g 867 mol SO65.06 g mol− =

( )( )

2 3

13

1 mole of SO is consummed by 1 mole of CaCO . Therefore,

867 mol CaCO 100.09 g mol 86.8 kg− =

322

10.104 (1) 2 32 H O(l) H O (aq) OH (aq)+ −+ 14w 1.00 10−= ×K

(2) 3 2 4NH (aq) H O(l) NH (aq) OH (aq)+ −+ + 5b 3(NH ) 1.8 10−= ×K

or

(3)

4 2 3 3

10a 4 w b 3

NH (aq) H O(l) NH (aq) H O (l)

(NH ) / (NH ) 5.56 10

+ +

+ −

+ +

= =K K K ×

(4) 3 2 3 3

5a 3

CH COOH(aq) H O(l) H O (l) CH COO (aq)

(CH COOH) 1.8 10

+ −

−

+ +

= ×K

or

(5)

3 2 3

b 3 w a 3

10

CH COO (aq) H O(l) CH COOH(aq) OH (aq)

(CH COO) / (CH COOH)

5.56 10

− −

−

−

+ +

=

= ×

K K K

(6)

4 3 3 3

3 34 3

4 3

NH (aq) CH COO (aq) NH (aq) CH COOH(aq)[NH ][CH COOH]

(NH CH COO)[NH ][CH COO ]

+ −

+ −

+ +

=K

This equation is obtained by adding Equations (3) and (5) and subtracting

Equation (1):

4 2 3 3[NH (aq) H O(l) NH (aq) H O (l)]+ ++ + +

w b 3/ (NHK K )

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

w a 3/ (CH COOHK K

2 3[2 H O(l) H O (aq) OH (aq)]+ −− + wK

Because we are starting with pure ammonium acetate, we will use the last

relationship to calculate the concentrations of

4 3 3 3NH , CH COO , NH , and CH COOH in solution.+ −

323

Concentration 1(mol L )−⋅ 4NH (aq)+ + + CH3CH COO (aq) NH (aq)−

3 3COOH(aq)

initial 0.100 0.100 0 0

change −x −x +x +x

equilibrium 0.100 − x 0.100 − x +x +x

For the initial calculation, we will assume that the subsequent

deprotonation of acetic acid by water or the protonation of ammonia by

water will be small compared to the reaction of the ammonium ion with

the acetate ion.

53 3

4 3

[NH ][CH COOH]3.09 10

[NH ][CH COO ]−

−+= ×

25

2

5

5

5 5

4

3.09 10(0.100 )

3.09 100.100

(0.100 ) 3.09 10

(1 3.09 10 ) (0.100) 3.09 10

5.52 10

−

−

−

− −

−

= ×−

= ×−

= − ×

+ × = ×

= ×

xx

xx

x x

x

x

By using this as the value of the concentrations of NH3 and CH3COOH,

we can calculate the concentrations of NH4+, 3CH COO ,− H3O+, and OH−

in solution using the above equilibrium relationship. The values calculated

are

[NH4+] = [CH3COOH] = 0.099

73[H O ] [OH ] 1.00 10+ −= = × −

If we substitute these numbers back into each of the equilibrium constants

expressions, we get very good agreement. This justifies the assumption

that the subsequent hydrolysis reactions of the NH3 and CH3COOH were

small compared to the main reaction.

Alternatively (had this not been the case), we could have solved the

system of equations using a graphing calculator or suitable computer

324

software package by setting up a system of simultaneous equations. The

answer is the same in any case. This problem is simplified by the fact that

the tendency for subsequent hydrolysis of NH3 and CH3COOH have the

same magnitude.

10.106 According to Table 10.3, the amide ion that is formed from the

autoionization of ammonia, 2 3 4NH NH NH ,2+ −+ is a stronger base

than . Therefore, carbonic acid (and other weak acids) is expected to

be stronger in liquid ammonia than in water. Furthermore, carbonic acid is

a stronger acid than NH

OH−

4+; therefore, it will donate a proton to NH3 to

form NH4+. It can then be concluded that carbonic acid will be leveled in

liquid NH3 and will behave as a strong acid.

10.108 (a) The pH is given by pH = ( )1a1 a22 p p+K K . The first and second

dissociations of H3AsO4 are the pertinent values for NaH2AsO4:

( )12pH 2.25 6.77 4.51= + =

(b) For Na2HAsO4 the second and third acid dissociation constants must

be used:

( )12pH 6.77 11.60 9.19= + =

10.110 (a) 3 3 4NH NH NH NH2+ −+ +

(b) acid = 4 2NH , base NH+ −=

(c) 33am amp log log(1 10 ) 33.0−= − = − × =K K

(d) 2am 4 2 4 2

33 17 14

p [NH ][NH ] ( [NH ] [NH ])

[NH ] 1 10 3 10 mol L

+ − + −

+ − − −

= = = =

= × = × ⋅

K x x

(e) 174 2pNH pNH log(3.2 10 ) 17+ − −= = − × =

(f) 4 2 ampNH pNH p 33.0+ −= = ≈K

325

10.112 The osmotic pressure of 0.10 H2SO4 will be slightly higher than that of

0.10 M HCl. When H2SO4 is added to solution it completely deprotonates

into HSO4¯, a small fraction of which deprotonates further to SO4

2-.

Therefore, each mole of H2SO4 gives rise to slightly more than 2 moles of

dissolved ions. HCl, on the other hand, is a strong monoprotic acid giving

exactly two moles of dissolved ions per mole of HCl molecules.

To calculate the moles of dissolved ions in a 0.10 M solution of H2SO4

one assumes that H2SO4 is completely deprotonated, and solves the

equilibrium problem:

Concentration

HSO1(mol L )−⋅ 4¯ (aq) + H2O(l) H3O+(aq) + SO42¯ (aq)

initial 0.10 — 0.10 0

change — +x +x −x

final 0.10 − x — 0.10 + x +x

23 4

a4

2

[H O ][SO ][HSO ]

(0.10 )( )1.2 100.10

−

+ −

−

=

+× =

−

K

x xx

Using the quadratic equation one finds that x = 9.8 × 10−3 and, therefore:

[HSO4-] = 0.0902 M, [H3O+] = 0.1098 M, and [SO4

2¯] = 0.0098 M giving

a total dissolved ion concentration of 0.210 M (compared to a total

dissolved ion concentration of 0.20 for the 0.10 M HCl solution).

Assuming a temperature of 298 K, the osmotic pressure of the 0.10 M

H2SO4 solution is:

( )( )( )1 10.082057 L atm K mol 298 K 0.210 mol L

5.13 atm

− − −Π = = ⋅ ⋅ ⋅

=

RTc 1

While that of the HCl solution is:

( )( )( )1 10.082057 L atm K mol 298 K 0.200 mol L

4.89 atm

− − −Π = = ⋅ ⋅ ⋅ ⋅

=

RTc 1

326

10.114 (where 1 mol refers to the reaction as written) 157 kJ mol−∆ ° = + ⋅H

1

2 1 2

1

2 1

1

2 1

1 21 2

1 1ln

1 12.303 log

1 12.303 log

1 1p p2.303

⎛ ⎞∆ °= − −⎜ ⎟

⎝ ⎠⎛ ⎞∆ °

= − −⎜ ⎟⎝ ⎠⎛ ⎞∆ °

− = ⎜ ⎟⎝ ⎠

⎛ ⎞∆ °− = −⎜ ⎟

⎝ ⎠

K HK R T T

K HK R T T

K HK R T T

HK KR T T

2

2

−

4 :

Let condition 2 be where 25 C,° wp 1=K

1 1 11

1 1 11

57 000 J 1 1p 14298 K2.303 8.314 J K mol

57 000 J 1 1p 14298 K2.303 8.314 J K mol

− −

− −

⎛ ⎞− = −⎜ ⎟⋅ ⋅ ⋅ ⎝ ⎠

⎛ ⎞= + −⎜ ⎟⋅ ⋅ ⋅ ⎝ ⎠

KT

KT

We will not define K1 as the Kw value at some unknown temperature T: 1

w 1 1

3

w

57 000 J mol 1 1p 14298 K2.303 8.314 J K mol

3.0 10 Kp 4.00T

−

− −

⎛ ⎞⋅= + −⎜ ⎟⋅ ⋅ ⋅ ⎝ ⎠

×= +

KT

K

Substituting the value of T = 373 K gives pKw = 12.04. If the solution is

neutral, the pH = pOH = 6.02.

10.116 (a) The structures of alanine, glycine, phenylalanine, and cysteine are

C C

C

O O

N

H

H

HH

H

H

H

H C

C

O O

N

H

H

H

H

alanine glycine

327

C C

C

O O

N

H

H

HH

H

H

H

H C

C

O O

N

H

H

H

H

alanine glycine

All the amino acids have an amine —NH2 function as well as a carboxylic

acid —COOH group.

(b) The pKa value of the —COOH group of alanine is 2.34 and the pKb of

the —NH2 group is 4.31. Often, we find instead of the pKb value, the pKa

value of the conjugate acid of the —NH2 group. The relationship

is used to convert to the pKa bp p p⋅ =K K Kw b value.

(c) To find the equilibrium constant of the reaction of the acid function,

with the base function in the amino acid, we first write the known

equilibrium reactions and corresponding K values:

2 3R COOH(aq) H O(l) H O (aq) R COO (aq)++ + − pKa = 2.34

3a 4.6 10−= ×K

2 2 3R NH (aq) H O(l) RNH (aq) OH (aq)+ −+ + pKb = 4.31

5b 4.9 10−= ×K

If we sum these reactions, we obtain

2 2 3

73 a b

R COOH(aq) R NH (aq) 2 H O(l) H O (aq) OH (aq)

R COO (aq) RNH (aq) 2.3 10

+ −

− + −

+ + +

+ + ⋅ = ×K K

The presence of the H2O, H3O+, and OH− can be eliminated by

subtracting the autoprotolysis reaction of water to give the desired final

equation:

328

2 32 H O(l) H O (aq) OH (aq)+ −+ 14w 1.00 10−= ×K

2 3

1a b w

7

R COOH(aq) R NH (aq) R COO (aq) RNH (aq)

2.3 10

− +

−

+ = +

= ⋅ ⋅

= ×

K K K K

K

(d) A zwitterion is a compound that contains both a positive ion and a

negative ion in the same molecule. Overall, the molecule is neutral, but it

is ionic in that it possesses both positive and negative ions within itself.

This is a common occurrence for the amino acids in which the acid

function reacts to protonate the base site. Given the large value of the

equilibrium constant found in (c), it is more appropriate to write an amino

acid as then as R(NH3R(NH )(COO )+ −2)(COOH).

10.118 (a)

1.0

0.5

0 14 7 0

pH

(b) The major species at pH 7.5 is the doubly deprotonated form.

(c) pH = 5.97

(d) at pH values greater than about 11.4

(e) from approximately pH = 7.0 to 8.4

329