Chemical principles the quest for insight 7th edition atkins solutions manual

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FOCUS 6 REACTIONS

6A.2

(a) HC2O4; (b) C6H5NH3+; (c) NH3OH +; (d) HO2; (e) NO2; (f) CrO42conjugate

6A.4

(a)

 H2O(l) + CN(aq)   HCN(l) + OH(aq) acid1 base2 acid2 base1  conjugate  conjugate

 (b) H2O(l) acid1

+

NH2NH2(aq)   NH2NH3+(aq) base2 acid2 

+ OH(aq) base1

conjugate

conjugate

 (c)

H2O(l) acid1

+

CO32 (aq)  HCO3 (aq) base2 acid2 

+

OH(aq) base1

conjugate

conjugate

 (d)

H2O(l) acid1

+

HPO42 (aq)   H2PO4 (aq) + base2 acid2 

OH(aq) base1

conjugate

conjugate

 (e)

H2O(l) acid1

+ NH2CONH2 (aq)  NH2CONH3 (aq) + OH(aq) base2 acid2 base1  conjugate

308

Focus 6 Reactions

6A.6

(a) Brønsted acid: CH3NH3+ Brønsted base: HSO3 (b) Conjugate base to CH3NH3+: CH3NH2 Conjugate acid to HSO3 : H2SO3

6A.8

(a) HClO2 (chlorous acid); conjugate base: ClO2-

(b) HNO3 (nitric acid); conjugate base: NO3-

6A.10 (a) Not a Bronsted acid-base reaction (b) Not a Bronsted acid-base reaction (c) HCO3- + NH3  NH4+ + CO32- acid: NaHCO3; base: NH3 (d) H2S + S2-  2HS-

acid: H2S; base: Na2S

6A.12 (a) As an acid: H2PO3(aq) + H2O(l)  H3O+(aq) + HPO32 (aq) Conjugate acid/base pairs:

H2PO3(aq) / HPO32 (aq) H3O+(aq) / H2O(l)

As a base: H2PO3(aq) + H2O(l)  OH(aq) + H3PO3(aq) Conjugate acid/base pairs:

H3PO3(aq) / H2PO3(aq) H2O(l) / OH(aq)

(b) As an acid: NH3(aq) + H2O(l)  H3O+(aq) + NH2(aq)

Focus 6 Reactions Conjugate acid/base pairs:

NH3(aq) / NH2(aq) H3O+(aq) / H2O(l)

As a base: NH3(aq) + H2O(l)  OH(aq) + NH4+ (aq) Conjugate acid/base pairs:

NH4+ (aq) / NH3(aq) H2O(l) / OH(aq)

6A.14 (a)

Lewis acid (b)

Lewis base; (c)

Lewis base (d)

-

N H

H

Lewis base (e)

Lewis acid

309

310

Focus 6 Reactions

6A.16 (a)

F

+ Lewis acid



Lewis base

product

(b)

Cl

+ Lewis acid 6A.18 (a) acidic;

(b) basic;



Lewis base (c) acidic;

product

(d) acidic

6A.20 In each case use K w  [H 3 O  ][OH  ]  1.0  1014 , then Kw 1.0  1014 [H 3 O ]   [OH  ] [OH  ] 

(a) [H3O+] =

1.0 1014 = 4.8 × 10-12 0.0021

(b) [H3O+] =

1.0 1014 = 2.9 × 10-12 3.4  103

(c) [H3O+] =

1.0 1014 = 1.3 × 10-12 3 7.60 10

6A.22 [H 3 O  ]  [OH  ]  3.9  108 mol  L1

K w  [H 3O  ][OH  ]  (3.9  108 ) 2  1.5  1015

Focus 6 Reactions

311

6A.24 [KNH2]0 = nominal concentration of KNH2  0.40 g KNH 2   1 mol KNH 2  -1 [KNH2]0 =   = 0.016 mol·L   0.450 L   55.13 g KNH 2 

Because KNH2 is a soluble salt, [KNH2]0 = [K+] = [NH 2  ]0 (where

[NH 2  ]0 is the nominal concentration of NH 2  ); thus [K+] and [NH 2  ] = 0.016 mol  L1 .

NH 2  reacts with water: NH 2  (aq)  H 2 O(l)   NH3 (aq)  OH  (aq) Because NH 2  is a strong base, this reaction goes essentially to completion; therefore: [NH2-]0 = [OH-] = 0.016 mol  L1 [H3O+] =

KW 1.0  1014  = 6.3  10-13 mol  L1 [OH - ] 0.016

6B.2 Ca(OH)2 is a strong base. If concentration decreases to 5.2% of its

initial value, (i.e. 0.052  [Ca(OH)2]0), the pOH change (pOH) = log(

0.052×(2[Ca(OH) 2 ]0 ) ) = 1.28 (i.e., pOH will increase 2[Ca(OH) 2 ]0

by 1.28 or pH will decrease by 1.28).

6B.4 (a) The pH of the desired solution = 14 – (-log(0.0175)) = 12.24  25.0 mL  0.0175 M  (b) The actual pH = 14 – (log   = 12.38 18.0 mL  

6B.6

(a) pH   log(0.0356)  1.449 pOH  14.00  1.448  12.55

(b) pH   log(0.0725)  1.140 pOH  14.00  1.140  12.86  Ba 2  2 OH  (c) Ba(OH) 2 

312

Focus 6 Reactions [OH  ]  2  3.46  103 mol  L1  6.92  103 mol  L1 pOH   log(6.92  103 )  2.160 pH  14.00  2.160  11.84

(d)

10.9  103 g  1.94  104 mol KOH 1 56.11 g  mol 1.94  104 mol [OH ]   1.94  102 mol  L1 0.0100 L 

pOH   log(1.94  102 )  1.712 pH  14.00  1.712  12.29

 10.0 mL  1 1 (e) [OH  ]     (5.00 mol  L )  0.0200 mol  L 2500 mL   pOH   log(0.0200)  1.699 pH  14.00  1.699  12.30

 5.0 mL  4 1 5 1 (f) [H 3 O  ]     (3.5  10 mol  L )  7.0  10 mol  L  25.0 mL  pH   log(7.0  105 )  4.15 pOH  14.00  4.15  9.85 6B.8

[H 3 O  ]  10 pH mol  L1 (antilog pH, mol  L1 ) (a) [H3O+] = 10-4.8 = antilog(-4.8) = 1  10-5 mol  L1 (b) [H3O+] = 10-0.7 = antilog(-0.7) = 0.2 mol  L1 (c) [H3O+] = 10-7.4 = antilog(-7.4) = 4  10-8 mol  L1 (d) [H3O+] = 10-8.1 = antilog(-8.1) = 8  10-9 mol  L1

Focus 6 Reactions 6B.10 (a)

[H3O+]

[OH]

( mol·L-1)

( mol·L-1)

(i)

0.50

(ii)

pH

pOH

2.0 × 10-14

0.30

13.70

2.0 × 10-14

0.50

13.70

0.30

(iii)

0.79

7.9 × 10-15

0.10

14.10

(iv)

7.9 × 10-15

0.79

14.10

0.10

(b) (iv) < (ii) < (i) < (iii) 6B.12 (a) pH = 14.12, pOH = -0.12, [OH-] = 100.12 = 1.3 mol·L-1  300.0 mL  . -1  = 39. mol L  10.0 mL 

(b) [OH-] = 1.3 × 

(c) The reaction is K2O(s) + H2O(l)  2 KOH(aq).  1mol K 2 O    2 mol KOH 

The mass of K2O: 39. mol . L-1 × (0.500 L) ×   94.203g K 2 O  2  = 9.2 × 10 g K2O  1mol K 2 O 

×

6C.2

(i) (CH3)2NH (a) (CH 3 ) 2 NH(aq)  H 2 O(l)  (CH 3 ) 2 NH 2  (aq)  OH  (aq) Kb 

[(CH 3 ) 2 NH 2  ][OH  ] [(CH 3 ) 2 NH]

(b) (CH 3 )2 NH 2  (aq)  H 2O(l)  H 3O  (aq)  (CH 3 ) 2 NH(aq)

Ka 

(ii) C14H10N2

[H 3O  ][(CH 3 ) 2 NH] [(CH 3 ) 2 NH 2  ]

313

314

Focus 6 Reactions

(a) C14 H10 N 2 (aq)  H 2 O(l)  C14 H10 N 2 H  (aq)  OH  (aq) [C14 H10 N 2 H  ][OH  ] Kb  [C14 H10 N 2 ] (b) C14 H10 N 2 H  (aq)  H 2O(l)  H 3O  (aq)  C14 H10 N 2 (aq)

Ka 

[H 3O  ][C14 H10 N 2 ] [C14 H10 N 2 H  ]

(iii) C6H5NH2 (a) C6 H 5 NH 2 (aq)  H 2 O(l)  C6 H 5 NH 3 (aq)  OH  (aq) Kb 

[C6 H 5 NH 3 ][OH  ] [C6 H 5 NH 2 ]

(b) C6 H 5 NH 3 (aq)  H 2O(l)  H 3O  (aq)  C6 H 5 NH 2 (aq)

Ka 

[H 3O  ][C6 H 5 NH 2 ] [C6 H 5 NH 3 ]

6C.4 (a) pK b  log K b ; therefore, after taking antilogs, K b  10 pKb .

Kb

pKb

(i) NH3

1.8  105

4.74

(ii) ND3

1.1  105

4.96

(iii) NH2NH2 1.7  106

5.77

(iv) NH2OH 1.1  108

7.96

Base

(b) The larger the Kb, the stronger the base; therefore NH2OH < NH2NH2 < ND3 < NH3 6C.6 The conjugated acid for (C2H5)3N is (C2H5)3NH+

pKa = pKw  pKb = 14.0  2.99 = 11.01 6C.8 Decreasing pKa will correspond to increasing acid strength because pKa

=  log K a . The pKa values (given in parentheses) determine the following ordering:

Focus 6 Reactions

315

(CH 3 )3 NH  (14.00  4.19  9.71)  N 2 H 5  (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.

6C.10 Decreasing pKb will correspond to increasing base strength because pKb

=  log K b . The pKb values (given in parentheses) determine the following ordering: HSO3 (14.00  1.81  12.19)  ClO  (14.00  7.53  6.47)  C10 H14 N 2 (5.98)  (CH 3 )3 N(4.19)

Remember that the pKb for the conjugate base of a weak acid will be given by pKa + pKb = 14.

6C.12 The smaller the value of pKb, the stronger the base; hence, aniline is the

stronger base. 4-Chloroaniline is the stronger acid because it has the electron-withdrawing Cl atom, making it the weaker base; and again we see that aniline is the stronger base.

6C.14 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.

6C.16 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. The higher electronegative atom withdraws electron

316

Focus 6 Reactions

density of the conjugate base and stabilizes it. Therefore, HClO is the stronger acid, with the lower pKa. 6C.18 The smaller the pKa of an acid, the stronger the acid:

pKa(HCN) = 9.31; pKa(NH4+) = pKw  pKb = 14.0  4.75 = 9.25 Hence, NH4+ is a stronger acid.

6C.20 (a) H3PO4 is stronger; it has the more electronegative central atom.

(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 because it has 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 because it has the greater electronegativity of Cl relative to S. (f) HClO is stronger because Cl has a greater electronegativity than I. 6C.22 (a) Methylamine is CH3NH2, and ammonia is NH3. Methylamine can be

thought of as being formed from NH3 by replacement of 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, and 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.

Focus 6 Reactions 6D.2

(a) K a  8.4  104 

317

[H 3O  ][CH 3CH(OH)CO2  ] x2 x2   [CH 3CH(OH)COOH] 0.11  x 0.11

Here we have assumed that x is small enough to be neglected relative to 0.11 mol  L1 . This is a borderline case. We will also solve this exercise without making this approximation and compare the results below. x  [H 3O  ]  9.6  103 mol  L1 pH   log(9.6  103 )  2.02 pOH  14.00  2.02  11.98

Without the approximation, the quadratic equation that must be solved is: 8.4  104 

x2 0.11  x

or x 2  8.4  104 x  9.24  105  0 x

8.4  104  (8.4  104 ) 2  4( 9.24  105 ) 2

x = 0.0092, or 0.010 pH = log(0.0092) = 2.04 pOH = 14.00 – 2.04 = 11.96

% deprotonation = (

0.0092 )  100 = 8.4% 0.11

(b) K a  8.4  104 

x2 3.7  103  x

or x 2  8.4  104 x  3.11  106  0 x = 1.410-3, or 2.210-3

The negative root can be eliminated: x = [H3O+] = 1.4 × 10-3

pH = log (1.4 × 10-3) = 2.85 pOH = 14.00 – 2.85 = 11.15 % deprotonation = (

1.4  103 )  100 = 38% 3.7  103

318

Focus 6 Reactions

(c) 8.4  104 

x2 8.2  105  x

or x 2  8.4  104 x  6.9  108  0 x = 7.5105, or 9.2104

The negative root can be eliminated. x = [H3O+] = 7.5 × 105

pH = log (7.5 × 10-5) = 4.12 pOH = 14.00 – 4.12 = 9.88 7.5  105 % deprotonation = ( )  100 = 91% 8.2  105

6D.4

(a) HNO 2  H 2 O  H 3O   NO 2  [H 3 O  ]  [NO 2  ]  102.63  antilog (  2.63)  2.3  103 mol  L1 Ka 

(2.3  103 ) 2  4.2  104 3 0.015  2.3  10

pK a   log(4.3  104 )  3.38 (b) C4 H 9 NH 2  H 2 O  C4 H 9 NH 3  OH  pOH  14.00  12.04  1.96 [C4 H 9 NH 3 ]  [OH  ]  101.96  antilog (  1.96)  0.011 mol  L1 Kb 

[C 4 H 9 NH 3 ][OH  ] (0.011) 2   1.4  103 [C 4 H 9 NH 2 ] 0.10  0.011

pK b  log(1.4  103 )  2.85

6D.6

(a) K b (pyridine)  1.8  109 C6 H 5 N  H 2 O  C6 H 5 NH   OH  Concentration (mol  L1 )

C6H5N + H2O  C6H5NH+ +

Initial

0.082

—

0

0

Change

x

—

+x

+x

Final

0.082  x

—

+x

+x

OH 

Focus 6 Reactions x2 x2 K b  1.8  10   0.082  x 0.082 9

x  [OH  ]  1.2  105 mol  L1 pOH   log(1.2  105 )  4.92 pH  14.00  4.92  9.08

Percentage ionized =

1.2  105  100%  0.015% 0.082

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

x2 x2 K b  1.0  10   0.0103  x 0.0103 6

x  [OH  ]

x  1.0  104 mol  L1 pOH   log(1.0  104 )  4.00 pH  14.00  4.00  10.00 1.0  104 Percentage protonation   100%  0.97% 0.0103 (c) pK b  14.00  8.52  5.48

K b  3.3  106 Quinine  H 2O  QuinineH   OH 

x2 x2 K b  3.3  10   0.060  x 0.060 6

x  [OH  ]  4.4  104 mol  L1 pOH   log(4.4  104 )  3.36 pH  14.00  3.36  10.64 4.4  104 Percentage protonation   100%  0.73% 0.060 (d) Strychnine  H 2O  StrychnineH   OH 

319

320

Focus 6 Reactions

x  [OH  ] K w 1.00  1014 Kb    1.82  106 9 K a 5.49  10 K b  1.82  106 

x2 x2  0.045  x 0.045

x  [OH  ]  2.9  104 mol  L1 pOH   log(2.9  104 )  3.54 pH  14.00  3.54  10.46 Percentage protonation  6D.8

2.9  104  100%  0.64% 0.045

(a) K a (HClO)  3.0  108 [H 3O  ]  10-pH  105.4  4  106 mol  L1 Let x = nominal concentration of HClO: Concentration (mol  L1 )

HClO

+ H2O  H3O+ + ClO

Nominal

x

—

0

0

Equilibrium

x4×10-6

—

4×10-6

4×10-6

3.0  108 

(4.  106 ) 2 ; x  5  104 mol  L1 6 x  4.  10

(b) K b (pyridine)  1.8  109 pOH  14.00  8.8  5.2 [OH  ]  10-pOH  105.2  6  106 mol  L1 Let x = nominal concentration of C5H5N: Concentration (mol  L1 )

C5H5N + H2O

Nominal

x

Equilibrium

x6 × 10-6

1.8  109 

(6  106 ) 2 ; x  0.02 mol  L1 x  6  106



 C5H6N+

+

OH 

0

0

6 × 10-6

6 × 10-6

Focus 6 Reactions

321

6D.10 (a) H H

C

N

H

H

H

H

methylamine

H

H

C

N

H

H

+

H

conjugate acid

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

Moles of methylamine are found by:

80.0 mL    0.85 g mL1   68.0 g of solution  68.0 g    0.350   23.8 g of methylamine  23.8 g   0.766 mol of methylamine

 31.06 g  mol  1

diluted to 0.3000 L, the resulting solution is 2.55 M in methylamine Concentration (mol  L1 )

CH3NH2 + H2O  CH3NH3+ +

OH 

Initial

2.55

—

0

0

Equilibrium

2.55 − x

—

x

x

CH 3 NH 3   OH   x2  K b  3.6  10  2.55  x CH 3NH 2  4

Solving for x, we find: x  3.01  102 M  OH   pOH   log(3.01  102 )  1.52 pH  14.0  1.52  12.48 6D.12 (a) pH  7, basic: H 2 O(l)  C 2 O 4 2 (aq)  HC2 O 4  (aq)  OH  (aq)

(b) pH  7, neutral: Ca 2 is not an acid and NO3 is not a base. (c) pH  7, acidic: CH 3 NH 3 (aq)  H 2 O(l)  H 3 O  (aq)  CH 3 NH 2 (aq) (d) pH  7, basic: H 2 O(l)  PO 43 (aq)  HPO 4 2 (aq)  OH  (aq) (e) pH  7, acidic: Fe(H 2 O)63 (aq)  H 2 O(l)  H 3 O  (aq)  Fe(H 2 O)5 OH 2 (aq) (f) pH  7, acidic: C5 H 5 NH  (aq)  H 2 O(l)  H 3 O  (aq)  C5 H 5 N(aq)

322

Focus 6 Reactions

6D.14 The rank in increasing solution pH is (b) < (a) < (c) < (d).

Justification: (a) is a strong base, but the concentration is low (pH = 9.00). (b) NO2- is a conjugated base of HNO2, which is a relatively stronger acid (pKa = 3.37). The solution pH = 8.34. (c) is a weak base (Kb = 1.8  10-5) with a pH of 11.28; (d) CN- is a conjugated base of HCN, which is a very weak acid (pKa = 9.31); the resulting CN- solution will be basic, pH = 11.31.

6D.16 (a) Concentration

CH3NH3+ + H2O(l)  CH3NH2 + H3O+

(mol  L1 ) Initial

0.15

—

0

0

Change

x

—

+x

+x

0.15 - x

—

x

x

Equilibrium

(See Table 6C.2 for Kb of CH3NH2, the conjugate base of CH3NH3+.)

Ka 

[CH 3 NH 2 ][H 3O  ] K w 1.00  1014 11     2.8 10 3.6  104 [CH 3 NH 3 ] Kb

2.8  1011 

x2 x2  0.15  x 0.15

x  2.0  106 mol  L1  [H 3O  ] and pH   log(2.0  106 )  5.70 (b) Concentration (mol  L1 ) Fe(H2O)63+(aq) + H2O(l)  H3O+(aq) + Fe(H2O)5OH+(aq) initial

0.063

—

0

0

change

x

—

+x

+x

0.063 - x

—

x

x

equilibrium Ka 

[H 3O  ][Fe(H 2 O)5OH  (aq)] x2 3    3.5 10 [Fe(H 2 O)63 ] 0.063  x

x 2  3.5  103 x  2.2  104  0 x = 0.013, 0.017 The negative root can be discarded.

Focus 6 Reactions

323

[H 3O  ]  0.013, pH   log 0.013  1.89 6D.18 (a) Concentration

(mol  L1 )

SO32

Initial Change

H2O(l) 

HSO3

0.015

—

0

0

x

—

+x

+x

—

x

x

+

Equilibrium 0.015 - x

+ OH 

[See Table 6E.1 for K a (SO32 )  K a2 (H 2SO3 ).]

Kb 

K w 1.00  1014 x2 x2 [HSO3 ][OH  ] 8   8.3  10    K a2 1.2  107 [SO32 ] 0.015  x 0.015

x2 0.015 x  3.5  105 mol  L1  [OH  ]

8.3  108 

pOH   log(3.5  105 )  4.46 pH  14.00  4.46  9.54

(b) Concentration (mol  L1 )

F(aq)

Initial

0.086

—

0

0

Change

x

—

+x

+x

—

x

x

+

Equilibrium 0.086 - x

H2O(l)



HF

+

OH

[See Table 6C.1 for Ka(HF)] Kb 

K w 1.00  1014 x2 x2 [HF][OH  ] 11   2.86  10    Ka 3.5  104 [F  ] 0.086  x 0.086

x2 0.086 x  1.6  106 mol  L1  [OH  ]

2.86  1011 

pOH   log(1.6  106 )  5.80 pH  14.00  pOH  8.20

6D.20 (a) Initial concentration of C6H5NH3+ equals the initial concentration of

C6H5NH3Cl.

324

Focus 6 Reactions 

 1 1   129.583 g mol   0.17 M 0.350 L

 7.8 g  

Given this initial concentration and the Ka for this acid found in Table 6D.1, the percent deprotonation is found as follows: Concentration C6H5NH3+ + H2O(l)

(mol  L1 )

 C6H5NH22+ +

H3O+

Initial

0.17

—

0

0

Change

x

—

+x

+x

Equilibrium

0.17  x

—

x

x

K a  2.3  105 

 [C6 H 5 NH 2+ 2 ][H 3O ] [C6 H 5 NH 3+ ]

2.3  105 

x2 x2  0.17  x 0.17

x  2.0  103 mol  L1  [H 3O  ] Percent dissociation 

2.0  103 mol  L1  100%  1.2% 0.17 mol  L1

6D.22 pH of the nitrate salt of the base is 3.13, [H+] =10-3.13 = 7.4 x 10-4

Hydrolysis reaction: BH+(aq) + H2O(l)  H3O+(aq) + B(aq) [H+] = [H3O+] = [B] = 7.4 x 10-4 M, [BH+] = 0.10-7.4 x 10-4 = 0.10 M Ka =

[ B][ H 3O  ] (7.4  104 ) 2 = = 5.5 x 10-6 [ BH  ] (0.10)

Kb(B) = Kw/Ka = 1.8 x 10-9; The formula of the base is C5H5N 6E.2

The reaction is (after the first, essentially complete ionization) HSeO 4   H 2 O  H 3 O   SeO 4 2 The initial concentrations of HSeO 4  and H 3 O  are both 0.010 mol  L1 because of the complete ionization of H2SeO4 in the first step. The second ionization is incomplete.

Focus 6 Reactions

325

Concentration H3O+

(mol  L1 )

HSeO 4  +

H2O

Initial

0.010

—

0.010

0

Change

x

—

x

x

0.010  x

—

0.010  x

x

Equilibrium K a2  1.2  10

2



+

SeO 4 2

[H 3 O  ][SeO 4 2 ] (0.010  x)( x)   0.010  x [HSeO 4  ]

x 2  0.022 x  1.2  104  0 0.022  (0.022) 2  (4)(1.2  104 )  4.5  103 x 2  [H 3 O ]  0.010  x  (0.010  4.5  103 ) mol  L1  1.5  102 mol  L1 pH   log(1.5  102 )  1.82 6E.4

(a) Second ionization is ignored because K a 2  K a1 .

H 2 S  H 2 O  H 3 O +  HS K a1  1.3  107 

[H 3 O  ][HS ] x2 x2   [H 2 S] 0.10  x 0.10

x  [H 3 O  ]  1.1  104 mol  L1 pH   log(1.1  104 )  3.96 (b) In this situation it may not be justified to ignore the second ionization.

K a1  6.0  104 , K a2  1.5  105 , and K a2 / K a1  40 . This is a marginal case; we can work it both ways, first without ignoring the second ionization. Adopt the following notation: H+ = H3O+ [H+] = [H3O+] = equilibrium concentration of H3O+ H2A = tartaric acid c0  solute molarity  0.15 mol  L1 c0  [H 2 A]  [HA  ]  [A 2 ]

(1)

H  total H present  [H  ]  2[H 2 A]  [HA  ]  2c0 (2)

326

Focus 6 Reactions

The following equilibria occur: H 2 A  H   HA  



HA  H  A

[H  ][HA  ] [H 2 A]

(3)

[H  ][A 2 ] Ka 2  [HA  ]

(4)

K a1 

2

Examination of Eqs. 1 to 4 shows that there are four unknowns, [H  ], [H 2 A], [HA  ], and [A 2 ] , to be determined. However, these four simultaneous equations allow for their determination.

The unknowns [H 2 A], [HA  ], and [A 2 ] may all be expressed in terms of one unknown, [H+].

According to Eq.1, c0  [H 2 A]  [HA  ]  [A 2 ]

From Eq. 3, [H 2 A] 

[H  ][HA  ] K a1

From Eq. 4, [HA  ] 

[H  ][A 2 ] Ka 2

Substituting, [H 2 A] 

[H  ]2 [A 2 ] K a1 K a 2

 [H  ] [H  ]2   Then Eq.1 becomes c0  [A 2 ] 1   Ka 2 K a1 K a2  

Now, subtract 2  Eq.1 from Eq. 2 : 2c0  [H  ]  2[H 2 A]  [HA  ]

Eq. 2

2c0  2[H 2 A]  2[HA  ]  2[A 2 ] [H  ]  [HA  ]  2[A 2 ]  0 [A 2 ], [A 2 ] 

2  Eq.1

[H  ] [H  ] 2 Ka 2

Substitute this into Eq. 5 to give  [H  ]  [H  ]2 [H  ]3   c0  2    [H ]  Ka 2  K a2 K a1 K a 2 

Rearranging into standard cubic form gives

(5)

Focus 6 Reactions

327

 c  [H  ]3 [H  ]2   [H  ] 1  0   2c0  0 K a1 K a 2 Ka 2 Ka 2  

Now, put in the numerical values for c0, Ka1, and Ka2: 1.1  108 [H  ]3  6.7  104 [H  ]2  1.0  104 [H  ]  0.30  0 Solution of this cubic by standard methods gives [H  ]  [H 3 O  ]  9.2  103 mol  L1 pH   log(9.2  103 )  2.04

It is left as an exercise for the reader to show that if the second ionization is ignored: [H  ]  9.5  103 mol  L1 and pH  2.02 The difference is small but perhaps not within experimental error. (c) The second ionization can be ignored because Ka2 << Ka1.

H 2 TeO 4  H 2 O  H 3 O   HTeO 4   K a1  2.1  10

8

[H 3 O  ][HTeO 4  ] x2 x2    [H 2 TeO 4 ] 1.1  103  x 1.1  103

x  [H 3 O  ]  4.8  106 mol  L1 pH  log(4.8  106 )  5.32 6E.6

(a) The pH is given by pH = 12 (pK a1  pK a2 ) . From Table 6E.1, we find: K a1  4.3  107

pK a1  6.37

K a2  5.6  1011

pK a2  10.25

pH  12 (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, pH = 8.31.

6E.8 H2A(aq) + H2O(l)  HA  (aq)

+

H3O+(aq)

(1)

HA  (aq) + H2O(l)  A2  (aq)

+

H3O+(aq)

(2)

The [H+] concentration from the first dissociation is 0.015 M (total

328

Focus 6 Reactions

dissociation). Since the pH of the H2A solution is 1.72, [H+]solution = 1.9 ×10-2 M. The [H+] from the second dissociation will be: (1.9 × 10-2) – 0.015 = 4.0 × 10-3 M; [A2-] = 4.0 × 10-3 M; [HA  ] = 0.015 – 4.0 × 10-3 = 0.011 M

Ka2 =

[H 3O  ][A 2 ] (4.0 103 ) 2 = = 1.5 × 10-3  [HA ] 0.011

6E.10 The equilibrium reactions of interest:

H 2 SO3 (aq)  H 2 O(l)  H 3 O  (aq)  HSO3 (aq)

K a1  1.5  102

HSO3 (aq)  H 2 O(l)  H3 O  (aq)  SO32 (aq)

K a2  1.2  107

Because the second ionization constant is much smaller than the first, we can assume that the first step dominates: Concentration (mol  L1 ) H2SO3(aq) +

H2O(l)  H3O+(aq)

+ HSO3 (aq)

Initial

0.125

—

0

0

Change

x

—

+x

+x

Final

0.125  x

—

+x

+x

K a1 

[H 3 O  ][HSO3 ] [H 2 SO3 ]

1.5  10

2

x2 ( x)( x)   0.125  x 0.125  x

Assume that x << 0.125: x 2  (1.5  102 )(0.125) x  0.043

Because x > 5% of 0.125, the assumption was not valid, and the full expression must be evaluated using the quadratic: x 2  1.5  102 x  (1.5  102 )(0.125)  0

Solving with the quadratic equation gives x  0.036 mol  L1 . x  [H 3 O  ]  [HSO3 ]  0.036 mol  L1 [H 2 SO3 ]  0.125 mol  L1  0.036 mol  L1  0.089 mol  L1

Focus 6 Reactions

329

We can use the other equilibria to determine the remaining concentrations: K a2 

[H 3 O  ][SO32 ] [HSO3 ]

1.2  107 

(0.036)[SO32 ] (0.036)

[SO32 ]  1.2  107 mol  L1 Because 1.2  107  0.036 , the initial assumption that the first dissociation would dominate is valid. To calculate [OH  ] , we use the Kw relationship: K w  [H 3 O  ][OH  ] [OH  ] 

Kw 1.00  1014   2.8  1013 mol  L1  0.036 [H 3 O ]

In summary [H 2 SO3 ]  0.089 mol  L1 , [H 3 O  ]  [HSO3 ]  0.036 mol  L1 , [SO32 ]  1.2  107 mol  L1 , [OH  ]  2.8  1013 mol  L1 6E.12 The equilibrium reactions of interest are now the base forms of the

sulfurous acid equilibrium, so Kb values should be calculated for the following changes: SO32 (aq)  H 2 O(l)  HSO3 (aq)  OH  (aq) K w 1.00  1014 K b1    8.3  108 7 K a2 1.2  10 HSO3 (aq)  H 2 O(l)  H 2 SO3 (aq)  OH  (aq) K b2 

K w 1.00  1014   6.7  1013 2 K a2 1.5  10

Because the second hydrolysis constant is much smaller than the first, we can assume that the first step dominates:

330

Focus 6 Reactions

Concentration (mol  L1 )

SO32 (aq) +

H2O(l) 

HSO3 (aq) + OH  (aq)

Initial

0.170

—

0

0

Change

x

—

+x

+x

Final

0.170x

—

+x

+x

K b1 

[HSO3 ][OH  ] [SO32 ]

8.3  108 

( x )( x ) x2  0.170  x 0.170  x

Assume that x << 0.170: x 2  (8.3  108 )(0.170) x  1.2  104

Because x is less than 1% of 0.170, the assumption was valid. x  [HSO3 ]  [OH  ]  1.2  104 mol  L1 Therefore, [SO32 ]  0.170 mol  L1  1.2  104 mol  L1  0.170 mol  L1.

We can then use the other equilibria to determine the remaining concentrations: K b2 

[H 2SO3 ][OH  ] [HSO3 ]

6.7  1013 

[H 2SO3 ](1.2  104 ) (1.2  104 )

[H 2SO3 ]  6.7  1013 mol  L1 Because 6.7  1013  1.2  104 , the initial assumption that the first hydrolysis would dominate is valid.

To calculate [H3O+], we use the Kw relationship: K w  [H 3O  ][OH  ] [H 3O  ] 

Kw 1.00  1014   8.3  1011 mol  L1  4 [OH ] 1.2  10

In summary, [H 2SO3 ]  6.7  1013 mol  L1 , [OH  ]  [HSO3 ]  1.2  104 mol  L1 , [SO32 ]  0.170 mol  L1 , [H 3O  ]  8.3  1011 mol  L1

Focus 6 Reactions

331

6E.14 The equilibria present in the solution are: H 2S(aq)  H 2 O(l)  H 3O  (aq)  HS (aq)

K a1  1.3  107

HS (aq)  H 2 O(l)  H 3O  (aq)  S2 (aq)

K a2  7.1  1015

The calculation of the desired concentrations follows exactly after the method derived in Eq. (2b) in 6E.4, substituting H2S for H2CO3, HS for HCO3 , and S2 for CO32 . First, calculate the quantity:

f (at pH  9.35 [H 3 O  ]  109.35  4.5  1010 mol  L1 ) :

f  [H 3 O  ]2  [H 3 O  ]K a1  K a1 K a2  (4.5  1010 ) 2  (4.5  1010 )(1.3  107 )  (1.3  107 )(7.1  1015 )  5.9  1017 The fractions of the species present are then given by: [H 3O  ]2 (4.5  1010 ) 2   3.4  103  (H 2S)  17 f 5.9  10

 (HS )   (S2 ) 

[H 3 O  ]K a1 (4.5  1010 )(1.3  107 )   0.99 f 5.9  1017

K a1 K a2 (1.3  107 )(7.1  1015 )   1.6  105 f 5.9  1017

Thus, in a solution at pH 9.35, the dominant species will be HS with a concentration of (0.250 mol  L1 )(0.99)  0.25 mol  L1 . The concentration of H2S will be (3.4  103 )(0.250 mol  L1 )  8.5  104 mol  L1 , and the concentration of S2 will be (1.6  105 )(0.250 mol  L1 )  4.0  106 mol  L1 .

6E.16 For the first ionization of (COOH)2— or H2C2O4— we write:

H 2 C2 O 4  H 2 O  H 3 O   HC2 O 4  , K a1  5.9  102

332

Focus 6 Reactions

Concentration (mol  L1 )

H2C2O4

+

H2O

H3O+ +



HC2 O 4 

Initial

0.12

—

0

0

Change

x

—

+x

+x

0.12  x

—

x

x

Equilibrium

K a1  5.9  102 , K a2  6.5  105 Because Ka2 << Ka1, the second ionization can safely be ignored in the calculation of [H3O+].

K a1  5.9  102 

x2 0.12  x

x 2  0.059 x  0.0071  0 0.059  (0.059) 2  (4)(1)(0.0071)  0.060 mol  L1  [H 3O  ] 2 1.0  1014   1.7  1013 mol  L1 [OH ]  0.060 [H 2 C2 O 4 ]  0.12  0.060  0.060 mol  L1 x

Concentration (mol  L1 )

HC2 O 4  +H2O 

H3O+

Initial

0.060

—

0.060

0

Change

x

—

+x

+x

0.060  x

—

0.060 + x

x

Equilibrium

K a2  6.5  105 

+

C2 O4 2

(0.060  x)( x)  x (because x is small) 0.060  x

or x  [C2 O 4 2 ]  6.5  105 mol  L1 , and [HC2 O 4  ]  0.060  x  0.060  0.000 065  0.060 mol  L1 6E.18 (1) 2 H 2 O(l)  H 3 O  (aq)  OH  (aq)

K w  1.00  1014

(2) NH 3 (aq)  H 2 O(l)  NH 4  (aq)  OH  (aq) or

K b (NH 3 )  1.8  105

Focus 6 Reactions

333

(3) NH 4  (aq)  H 2 O(l)  NH 3 (aq)  H 3 O  (l) K a (NH 4  )  K w / K b (NH 3 )  5.56  1010 (4) CH 3 COOH(aq)  H 2 O(l)  H3 O (l)  CH3 COO  (aq) K a (CH 3 COOH)  1.8  105 or (5) CH 3 COO  (aq)  H 2 O(l)  CH 3 COOH(aq)  OH  (aq) K b (CH 3 COO)   K w / K a (CH 3 COOH)

 5.56  1010 (6) NH 4  (aq)  CH 3 COO  (aq)  NH 3 (aq)  CH 3 COOH(aq) K (NH 4 CH 3 COO) 

[NH 3 ][CH 3 COOH] [NH 4  ][CH 3 COO  ]

This equation is obtained by adding Eqs. 3 and 5 and subtracting Eq. (1): [NH 4  (aq)  H 2 O(l)  NH 3 (aq)  H 3 O  (l)] K w / K b (NH 3 ) [CH 3 COO  (aq)  H 2 O(l)  CH 3 COOH(aq)  OH(aq)] K w / K a (CH 3 COOH) [2 H 2 O(l)  H 3 O  (aq)  OH  (aq)] K w K(NH4CH3COO) = (Ka(NH4+) × Kb(CH3COO-))/Kw = 3.1 × 10-5 Because we are starting with pure ammonium acetate, we will use the last relationship to calculate the concentrations of NH 4  , CH 3COO  , NH3,

and CH3COOH in the solution.

Concentration

(mol  L1 ) NH 4  (aq) + CH 3 COO  (aq)  NH 3 (aq) + CH3COOH(aq) Initial

0.100

0.100

0

0

Change

x

x

+x

+x

0.100  x

+x

+x

Equilibrium 0.100  x

334

Focus 6 Reactions

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.

[NH 3 ][CH 3COOH]  3.1  105 [NH 4  ][CH 3COO  ] x2  3.1  105 2 (0.100  x) x  3.1  105 0.100  x x  (0.100  x) 3.1  105 (1  3.1  105 ) x  (0.100) 3.1  105 x  5.5  104 By using this as the value of the concentrations of NH3 and CH3COOH, we can calculate the concentrations of NH4+, CH 3 COO  , H3O+, and OH  in solution using the above equilibrium relationship. The values calculated are: [NH4+] = [CH3COOH] = 0.099 [H 3 O  ]  [OH  ]  1.00  107 If we substitute these numbers back into each of the equilibrium constant’s 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 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 has the same magnitude.

Focus 6 Reactions 6F.2

335

We can use the relationship derived in the text: [H 3O  ]2  [HA]initial [H 3O  ]  K w  0 [H 3O  ]2  (6.50  108 )[H 3O  ]  (1.00  1014 )  0

Solving using the quadratic equation [H 3O  ]  1.38  107 ; pH  6.860 This value is lower than the value calculated, based on the acid concentration alone (pH   log(6.50  108 )  7.187).

6F.4

We can use the relationship derived in the text: [H 3 O  ]2  [B]initial [H 3 O  ]  K w  0, in which B is any strong base.

[H 3 O  ]2  (8.23  107 )[H 3 O  ]  (1.00  1014 )  0 Solving using the quadratic equation gives: [H 3O  ]  1.20  108 , pH  7.921 This value is slightly higher than is calculated, according to the base concentration alone: (pOH   log(8.23  107 )  6.084, pH  14.00  6.084  7.916)

6F.6 In general, when [H3O+]  10-6 (when the pH lies between 6 and 7), the

autoprotolysis of water must be taken into account. We can use Eq. 3b in 6F.2: x3  K a x 2  (K w  K a  [HA]initial )x  K w  K a  0, where x  [H 3O  ] To solve the [HA]initial, substitute the values of [H3O+] = 1.0 10-6, K w  1.00  1014 , and Ka = 1.4  10-3 (for chloroacetic acid) into this equation and then solve for [HA]initial. [HA]initial = 1.0  10-6 M. That is, when chloroacetic acid concentration is 1.0  10-6 M or lower, autoprotolysis of water must be taken into account.

6F.8 We need to modify Eq. 3b in 6F.2 to solve this problem: x 3  K b x 2  (K w  K b  [A]initial )x  K w  K b  0, where x  [OH  ]

336

Focus 6 Reactions

[A]initial = 0.30 M; (0.15 M Ca(CH3CO2)2 will give off 0.30 M CH3CO2-); Kb = Kw/Ka(CH3CO2H) = 5.6 × 10-10 x = [OH] = 1.3×10-5 M; pOH = 4.89; pH = 9.11 It is clear that the pH does not change, because pOH < 6 and the autoprotolysis of water can still be ignored.

6F.10 (a) In the absence of a significant effect due to the autoprotolysis of

water, the pH values of the 1.89  105 M and 9.64  107 M HClO solutions can be calculated as described earlier.

For 1.89  105 mol  L1 : Concentration (mol  L1 ) HClO(aq)

+

H2O(l)

 H3O+(aq) + ClO  (aq)

Initial

1.89  105

—

0

0

Change

x

—

+x

+x

Final

1.89  105  x

—

+x

+x

Ka 

[H 3 O  ][ClO  ] [HClO]

3.0  108 

x2 ( x)( x)  1.89  105  x 1.89  105  x

Assume x  1.89  105 x 2  (3.0  108 )(1.89  105 ) x  7.5  107

Because x < 5% of 1.89  105 , the assumption was valid. Given this value, the pH is then calculated to be log(7.5  107 )  6.12.

For 9.64  107 mol  L1 :

Focus 6 Reactions

337

Concentration (mol  L1 ) HClO(aq)

+

H2O(l)

 H3O+(aq) + ClO  (aq)

Initial

9.64  107

—

0

0

Change

x

—

+x

+x

Final

9.64  107  x

—

+x

+x

Ka 

[H 3 O  ][ClO  ] [HClO]

3.0  108 

x2 ( x)( x)  9.64  107  x 9.64  107  x

Assume x  9.64  107  x x 2  (3.0  108 )(9.64  107  x) x  1.7  107

x is approximately 10% of 9.64  107 ; the assumption is not reasonable, and so you must calculate the value explicitly for the following expression, using the quadratic equation. x 2  3.0  108 x  (3.0  108 )(9.64  107 )  0 Solving the quadratic equation yields a value of x  1.6  107 , pH = 6.80. (b) To calculate the value, taking into account the autoprotolysis of water, we can use Eq. (3b) in 6F.2: x3  K a x 2  ( K w  K a  [HA]initial ) x  K w  K a  0, where x  [H 3 O  ] .

To solve the expression, you substitute the values of K w  1.00  1014 , the initial concentration of acid, and K a  3.0  108 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+], [OH  ] , [HClO], and [ClO  ] , you will need four equations. As seen in the text, the pertinent equations are: [H 3 O  ][ClO  ] Ka  [HClO]

338

Focus 6 Reactions

K w  [H 3 O  ][OH  ] [H 3 O  ]  [OH  ]  [ClO  ] [HClO]initial  [HClO]  [ClO  ] Using either method should produce the same result.

The values obtained for 1.89  105 mol  L1 are: [H 3 O  ]  7.4  107 mol  L1 , pH  6.13 (compare to 6.12 obtained in (a)) [ClO  ]  7.3  107 mol  L1 [HClO]  1.8  105 mol  L1 [OH  ]  1.3  108 mol  L1 Similarly, for [HClO]initial  9.64  107 : [H 3 O  ]  1.9  107 mol  L1 , pH  6.72 (compare to 6.80 obtained in (a)) [ClO  ]  1.3  107 mol  L1 [HClO]  8.3  107 mol  L1 [OH  ]  5.3  108 mol  L1 For the more concentrated solution, the effect of the autoprotolysis of water is very small. Also the less concentrated solution is more acidic, due to the autoprotolysis of water, than would be predicted if this effect were not operating.

6G.2 (a) When solid sodium dihydrogen phosphate is added to a solution of

phosphoric acid, the equilibrium H 3 PO 4 (aq)  H 2 O(l)  H 3 O  (aq)  H 2 PO 4  (aq) shifts to the left to relieve the stress resulting from the increased H 2 PO 4  .

Consequently, [H3O+] decreases and the pH increases. (b) When HBr, a strong acid, is added to a solution of HCN, the percentage of HCN that is deprotonated decreases because the equilibrium HCN(aq)  H 2 O(l)  H 3 O  (aq)  CN  (aq)

Focus 6 Reactions

339

shifts to the left to relieve the stress imposed by the increased H3O+ supplied by the HBr. (c) When pyridinium chloride (C5H5NHCl) is added to a solution of pyridine, the equilibrium C5 H 5 N(aq)  H 2 O(l)  C5 H 5 NH  (aq)  OH  (aq) shifts to the left to relieve the stress resulting from the increased C5H5NH+. Consequently, the [OH  ] decreases. Because [H 3 O  ][OH  ] is constant, [H3O+] must increase and the pH consequently decreases.

6G.4

 saccharin anion  (a) pH = pKa + log    saccharin  At condition of equal concentration of saccharin and saccharin anions, pH = pKa; pKa = 3.08; Ka = 10-3.08 = 8.3210-4 1 (b) pH = pKa + log   = 3.08 -0.30 = 2.78 2

6G.6

(a) The reaction is HBrO(aq) + H2O(l)  H3O+(aq)

+

BrO(aq)

Concentration

(mol  L1 )

HBrO(aq) + H2O(l)  H3O+(aq) +

BrO(aq)

Initial

0.12

—

0

0.160

Change

x

—

+x

+x

0.12  x

—

x

0.160 + x

Equilibrium

Ka =

[H 3O + ][BrO  ] ( x)(0.160  x) = = 2.0 x 10-9 [HBrO] (0.12  x)

x = [[H3O+]  1.5 x 10-9 mol . L-1

340

Focus 6 Reactions

(b) Concentration (mol  L1 )

CH3NH2(aq) + H2O(l)  CH3NH3+(aq) + OH  (aq)

Initial

0.250

—

0.150

0

Change

x

—

+x

+x

0.250x

—

0.150+x

+x

Equilibrium

Kb 

x(0.150  x) [CH 3 NH 3+ ][OH  ] ; 3.610-4 = (0.250  x) [CH 3 NH 2 ]

Assuming x is small, 0.150 + x  0.150, 0.250  x  0.250 x = [OH] = 6.00  10-4 mol·L-1 [H3O+] =

KW 1.00  1014   1.67  1011 mol·L-1  4 [OH ] 6.00  10

(c) Setup is similar to part (a). [ H 3O  ][ BrO  ] ( x)(0.320  x) = = 2.0 x 10-9 Ka = [ HBrO] (0.160  x)

x = [[H3O+]  1.0 x 10-9 mol . L-1 (d) 0.250 M NaBr will have no effect:

Concentration (mol  L1 )

CH3NH2(aq) + H2O(l)  CH3NH3+(aq) + OH  (aq)

Initial

0.250

—

0

0

Change

x

—

+x

+x

0.250x

—

+x

+x

Equilibrium

Kb 

[CH 3 NH 3+ ][OH  ] ; [CH 3 NH 2 ]

3.610-4 =

x2 (0.250  x)

x2 + 3.6  10-4 x  9.0  10-5 = 0 x = 9.3  10-3 mol·L-1 or 9.7  10-3 mol·L-1 Eliminate the negative answer: x = [OH] = = 9.310-3 mol·L-1

Focus 6 Reactions

[H3O+] =

341

KW 1.00  1014   1.06  1012 mol·L-1  3 [OH ] 9.3  10

6G.8 In each case, the equilibrium involved is:

HPO 4 2 (aq)  H 2 O(l)  H 3O  (aq)  PO 43 (aq) HPO4 2 (aq) and PO4 3 (aq) are conjugate acid and base; therefore, the pH calculation is most easily performed with the HendersonHasselbalch equation:  [PO 43 ]  pH  pK a  log  2   [HPO 4 ]  For H 3 PO 4 , pK a3  12.68

(a)  0.18 mol  L1   12.57; pOH  14.00  12.57  1.43 pH  12.68  log  1   0.23 mol  L 

(b)  0.62 mol  L1   12.82; pOH  14.00  12.82  1.18 pH  12.68  log  1   0.45 mol  L 

(c)  0.16 mol  L1   12.68; pOH  14.00  12.68  1.32 pH  12.68  log  1   0.16 mol  L 

6G.10 (a) (CH 3 ) 2 NH(aq)  H 2 O(l)  (CH 3 ) 2 NH 2  (aq)  OH  (aq)

Total volume = 0.380 L Moles of (CH 3 ) 2 NH  (0.100 L)(0.050 mol  L1 )  5.0  103 mol (CH 3 )2 NH Moles of (CH 3 ) 2 NH 2Cl  (0.280 L)(0.040 mol  L1 )  1.1  102 mol (CH 3 ) 2 NH 2Cl Initial [(CH 3 ) 2 NH] 

5.0  103 mol  1.3  102 mol  L1 0.380 L

342

Focus 6 Reactions

initial [(CH 3 ) 2 NH 2Cl] 

1.1  102 mol  2.9  102 mol  L1 0.380 L

Concentration

(mol  L1 ) H2O(l) + (CH3)2NH(aq)  (CH3)2NH2+(aq) +

OH  (aq)

Initial

—

1.3×10-2

2.9×10-2

0

Change

—

x

+x

+x

-2

x

Equilibrium

-2

1.3×10 - x

—

2.9×10 + x

[(CH 3 )2 NH 2  ][OH  ] (2.9  102  x )( x ) (2.9 102 )( x ) Kb    [(CH 3 ) 2 NH] (1.3  102  x ) (1.3  102 )  5.4  104

x  2.4  104 mol  L1  [OH  ] pOH   log[OH  ]   log(2.4  104 )  3.62 pH  14.00  3.62  10.38

(b) (CH3)2NH(aq) + H2O(l)  (CH3)2 NH2+(aq) + OH  (aq) Total volume 131.0 mL = 0.131 L Moles of (CH 3 ) 2 NH  (0.0450 L)(0.015 mol  L1 )  6.8  104 mol (CH 3 )2 NH Moles of (CH 3 ) 2 NH 2Cl  (0.0860 L)(0.200 mol  L1 )  1.7  102 mol (CH 3 ) 2 NH 2Cl Initial [(CH 3 )2 NH] 

6.8  104 mol  5.2  103 mol  L1 0.131 L

and [(CH 3 ) 2 NH 2Cl] 

1.7  102 mol  0.13 mol  L1 0.131 L

As in part (a), Kb 

[(CH 3 ) 2 NH 2  ][OH  ] (0.13  x )( x ) (0.13)( x )   3 [(CH 3 ) 2 NH] (5.2  10  x ) (5.2  103 )

 5.4  104

Focus 6 Reactions x  2.2  105 mol  L1  [OH  ] pOH   log[OH  ]   log(2.2  105 )  4.66 pH  14.00  4.66  9.34

(c) (CH 3 ) 2 NH(aq)  H 2 O(l)  (CH 3 ) 2 NH 2  (aq)  OH  (aq) Total volume = 140 mL = 0.140 L Moles of (CH 3 ) 2 NH  (0.100 L)(0.080 mol  L1 )  8.0  103 mol (CH 3 )2 NH Moles of (CH 3 ) 2 NH 2Cl  (0.0400 L)(0.075 mol  L1 )  3.0  103 mol (CH 3 )2 NH 2Cl Initial [(CH 3 ) 2 NH 2Cl]  and [(CH 3 ) 2 NH]  Kb 

3.0  103 mol  2.1  102 mol  L1 0.140 L

8.0  103 mol  5.7  102 mol  L1 0.140 L

[(CH 3 ) 2 NH 2  ][OH  ] (2.1  10 2  x )( x ) 2.1  102 ( x )   [(CH 3 ) 2 NH] (5.7  102  x ) 5.7  102

 5.4  104

x  1.5  103 mol  L1  [OH  ] pOH   log[OH  ]   log(1.5  103 )  2.82 pH  14.00  2.82  11.18

6G.12 (a) The initial pH is found by:

 HPO 24  pH  pK a1  log  H 2 PO 4   7.21  log  7.28

0.140 M 0.120 M

(0.0750 L)(0.0100 mol  L1 )  7.50  10 4 mol NaOH (strong base), producing 7.5  104 mol HPO 4 2 from H 2 PO 4 

343

344

Focus 6 Reactions (0.140 mol  L1 )(0.100 L)  1.40  102 mol HPO 4 2 (initially) (0.120 mol  L1 )(0.100 L)  1.20  102 mol H 2 PO 4  (initially)

After adding NaOH: [HPO 4 2 ] 

(1.40  102  7.50  104 ) mol  8.43  102 mol  L1 0.175 L

[H 2 PO 4  ] 

(1.20  102  7.50  104 ) mol  6.43  102 mol  L1 0.175 L

[HPO 4  ] pH  pK a  log [H 2 PO 4 2 ] pH  7.21  log

(8.43  102 mol  L1 )  7.21  0.12  7.33 (6.43  102 mol  L1 )

pH  0.05

(b) (0.0100 L)(0.50 mol  L1 )  5.0  103 mol HNO3 (strong acid), Produing 5.0  103 mol H 2 PO 4  from HPO 4 2 Molarity of HPO 4 2 

(1.40  102  5.0  103 ) mol  8.2  102 mol  L1 0.110 L

Molarity of H 2 PO 4  

(1.20  102  5.0  103 ) mol  1.5  101 mol  L1 0.110 L

pH  pK a  log

[HPO 4  ] [H 2 PO 4 2 ]

(8.2  102 mol  L)  7.21  0.26  6.95 (1.5  101 mol  L) pH  6.95  7.28  0.33 pH  7.21  log

 5.10 g NaBrO   1  1 6G.14 Molarity of NaBrO =     0.429 mol  L  118.90 g/mol   0.100 L 

Focus 6 Reactions

345

Concentration

H2O(l) + HBrO(aq)  H3O+(aq)

(mol  L1 ) Initial

—

0.50

Change

—

Equilibrium

—

Ka = 2.0 × 10 -9 =

[H 3O+ ][BrO- ] x (0.429  x )  [HBrO] 0.50  x

0.50  x

+

BrO  (aq)

0

0.429

+x

+x

x

0.429 + x

x = [H3O+] = 2.3 × 10-9 mol . L-1 pH = -log [H3O+] = 8.64 Change in pH is 8.64 – 4.50 = 4.14 6G.16 HA, acetylsalicylic acid

A  , conjugate base of acetylsalicylic acid HA(aq)  H 2 O(l)  H 3 O  (aq)  A  (aq)

K a  3.2  104

We will use the following formula to evaluate the ratio of [A  ] to [HA] at equilibrium: pH  pK a  log

[A  ] [HA]

The formula is derived in this way:

[H3 O ][A  ] Ka  [HA] K [HA] [H 3 O  ]  a  [A ]  K [HA]  pH  p  a    [A ]  [HA]  pK a  log  [A ]  pK a  log

[A  ] [HA]

Given that pH = 4.13 and Ka = 3.2 x 10-4 (pKa = 3.49)

346

Focus 6 Reactions

4.13 = 3.49 + log

[ A ] [ HA]

[ A ] = 4.37 [ HA]

6G.18 (a) pK a  3.37, pH range  2 to 4

(b) pK a  3.75, pH range  3 to 5 (c) pK a2  10.25, pH range  9 to 11 (d) pK b  4.75, pK a  9.25, pH range  8 to 10 (e) pK b  8.75, pK a  5.25, pH range  4 to 6 6G.20 (a) C6H5COOH and NaC6H5CO2, pKa = 4.19

(b) NH4Cl and NH3, pKb = 4.75, pKa = 9.25 (c) CH3COOH and NaCH3CO2, pKa = 4.75 (d) (C2H5)3 NHCl and (C2H5)3N, pKb = 2.99, pKa = 11.01 6G.22 HPO 4 2 (aq)  H 2 O(l)  H 3 O  (aq)  PO 43 (aq)

K a3 

[H 3 O  ][PO 43 ] [HPO 4 2 ]

pH  pK a 3  log log

[PO 43 ] [HPO 4 2 ]

[PO 43 ]  pH  pK a 3 ; pK a 3  12.68 [HPO 4 2 ]

[PO 43 ]  12.00  12.68  0.68 (a) log [HPO 4 2 ] [PO 43 ]  0.21 [HPO 4 2 ]

Focus 6 Reactions

347

(b) Molarity of PO 4 3  0.21  molarity of HPO4 2  0.21  0.100 mol  L1  2.1  102 mol  L1

Moles of PO 4 3  moles of K 3PO4  2.1  102 mol  L1  1 L  2.1  102 mol 212.27 g K 3PO4 Mass of K 3PO 4  2.1  102 mol   4.46 g K 3PO4 1 mol K 3PO4 (c)

[PO 43 ] 0.100 mol  L1 0.100 mol 0.100 mol K 3 PO 4     0.21 x mol x mol K 2 HPO 4 x mol  L1 [HPO 4 2 ]

x mol K 2 HPO 4 

0.100 mol  0.48 mol K 2 HPO 4 0.21

Mass of K 2 HPO 4  0.48 mol K 2 HPO4 

174.18 g K 2 HPO4 1 mol K 2 HPO4

 84 g K 2 HPO 4 (d) The final volume of the solution is the same for both species, so the ratio can be expressed as a mole ratio.

[PO 43 ] moles of PO 43(x L)(0.150 mol  L-1 )    0.21 [HPO 4 2 ] moles of HPO 4 2(0.0500 L)(0.100 mol  L-1 )

x = 0.0070 L or 7.0 mL

348

Focus 6 Reactions

6H.2 (a) The titration curve is as follows: 12 10

D 8 pH

C 6

B

A

4 2 0 0

1

2

3

4

5

Vol Ca(OH)2

Point A: initial point pH = 2 (pH = -log(0.010) = 2); point B: halfway titration pH = 2.4 (pH = -log[(5)(0.010)/(2 × 6.25)] = 2.4); point C: the stoichiometric point pH = 7.0 (neutral solution); point D: final solution pH = 11.7 (extra base) (b) 2.50 mL; (c) 1.25 mL 3  25.0 mL   10 L   0.412 mol KOH  6H.4 (a) VHCl      2 1 L KOH    1 mL   

 1 mol HCl    1 L HCl      1 mol KOH   0.120 mol HCl   0.0429 L HCl  42.9 mL HCl (b) 2×42.9 mL = 85.8 mL HCl (c) In HCl, [HCl]0  [Cl ] Volume = (0.0250 + 0.0858) L = 0.1108 L  0.150 mol Cl     1 Molarity of Cl  (0.0858 L)    1L    0.1108 L   0.116 mol  L1 

Focus 6 Reactions

349

(d) Number of moles of H 3O  (from acid)  (0.0300 L)(0.120 mol  L1 )  3.60  103 mol Number of moles of OH  (from base)  (0.0150 L)(0.310 mol  L1 )  4.65  103 mol

Excess OH   (4.65  3.60)  103 mol  1.05  103 mol Volume solution = (0.0300 + 0.0150) L = 0.0450 L 1.05  103 mol [OH ]   0.0233 mol  L1 0.0450 L 

pOH   log(0.0233)  1.63 pH  14.00  1.63  12.37

6H.6 The reaction is HCl  KOH   H 2 O  KCl; thus HCl  KOH

(a) Initial : [OH  ]  0.215 mol  L1 pOH   log(0.215)  0.668 pH  14.00  pOH  13.33 (b) After addition of 5.0 mL of 0.116 mol  L1 HCl:  0.215 mol KOH   0.116 mol HCl  (0.0250 L base)    (0.0050 L)   1 L base 1L      4.80  103 mol KOH unreacted 4.80  103 mol OH   0.16 mol  L1 OH  0.0300 L pOH   log(0.16)  0.80 pH  14.00  0.80  13.20

[OH  ] 

(c) After addition of 10.0 mL of 0.116 mol  L1 HCl:  0.215 mol KOH   0.116 mol HCl  (0.0250 L base)    (0.0100 L)   1 L base 1L     3  4.22  10 mol KOH unreacted

350

Focus 6 Reactions 4.22  103 mol OH  [OH ]   0.121 mol  L1 OH  0.0350 L 

pOH   log(0.121)  0.917 pH  14.00  0.917  13.08

(d) At the stoichiometric point: for the reaction of a strong acid with a strong base, the pH at the stoichiometric point is the same as the pH for pure water. Therefore, the pH = 7.0.

Volume HCl at the stoichiometric point:  0.215 mol KOH   1 mol HCl    1 L HCl (0.0250 L KOH)     1 L KOH    1 mol KOH   0.116 mol HCl   0.0463 L HCl

(e) After addition of 5.0 mL of acid beyond the stoichiometric point:  0.116 mol HCl    1 (0.0050 L)    1L    (0.0463  0.0050  0.0250) L   0.0076 mol  L1 H 3 O  pH   log(0.0076)  2.12

(f) After addition of 10 mL of acid beyond the stoichiometric point:  0.116 mol HCl    1 (0.010 L)    1L    (0.0463  0.010  0.0250) L  1   0.014 mol  L H 3 O pH   log(0.014)  1.85

6H.8

Ba(OH) 2  2 HCl   BaCl2  2 H 2 O mass of pure Ba(OH) 2  0.0810 mol HCl   1 mol Ba(OH) 2   (0.0131 L HCl)    1 L HCl    2 mol HCl   171.36 g Ba(OH) 2   200 mL      1 mol Ba(OH)2   25.0 mL   0.727 g Percentage purity 

0.727 g  100%  41.0% 1.773 g