SKOOG - SOLUCIONÁRIO CAPÍTULO 14.pdf

th

Fundamentals of Analytical Chemistry: 8 ed.

Chapter 14

Chapter 14

14-1

(a) The initial pH of the NH3 solution will be less than that for the solution c ontaining

 NaOH. With the first addition of titrant, the pH of the NH3 solution will decrease rapidly and then level off and become be come nearly constant throughout the middle mid dle part of the titration. In contrast, additions of standard acid to the NaOH solution will cause the pH of the  NaOH solution to decrease gradually and nearly linearly until the equivalence point is approached. The equivalence point pH for the NH3 solution will be well below 7, whereas for the NaOH solution it will be exactly 7. titrant. Thus, the (b) Beyond the equivalence point, the pH is determined b the excess titrant. curves become identical in this region.

14-2

Completeness of the reaction between the analyte and the reagent and the concentrations of the analyte and reagent.

14-3

The limited sensitivity of the eye to small color differences requires that there be a roughly tenfold excess of one or the other form of the indicator to be present p resent in order for the color change to be seen. This change corresponds to a pH range range of ± 1 pH unit about the pK of the indicator.

14-4

Temperature, ionic strength, and the presence of organic solvents and colloidal particles.

14-5

The standard reagents in neutralization titrations are always strong acids or strong bases  because the reactions with this type of reagent are more complete than with those of their weaker counterparts. Sharper end points are are the consequence of this this difference.

th

Fundamentals of Analytical Chemistry: 8 ed. 14-6

The sharper end point will be observed with the solute having the larger  K  b. (a)

For NaOCl,

For hydroxylamine

(b)

(c)

For NH3,

(d)

 K  b

 K  b

 K  b

  

1.00  1014 3.0  10

8

1.00  1014 1.1 10

6

1.00  1014 5.7  10

10

1.00  1014

For sodium phenolate,  K  b



For hydroxyl amine

-9 = 9.110

For methyl amine,

For hydrazine

For NaCN,

14-7

Chapter 14

 K  b 

 K  b

 K  b

 K  b

  

1.00  10

10

11

1.00  1014 1.05  10

8

1.00  1014 6.2  10

 9.1109

Thus, NaOCl

 1.75  105  1.00  104

Thus, sodium phenolate

(part a)

1.00  1014 2.3  10

 3.3  107

10

 4.3  104

Thus, methyl amine

 9.5  107  1.6 103

Thus, NaCN

The sharper end point will be observed with the solute having the larger  K a. a. (a)

(b)

(c)

(d)

For nitrous acid

 K a 

-4 = 7.110

For iodic acid

 K a 

= 1.710

For anilinium

 K a 

-5 = 2.5110

For benzoic acid

 K a 

= 6.2810

For hypochlorous acid

-1

 K a 

-5

Thus, iodic acid

Thus, benzoic acid

-8 = 3.010

For pyruvic acid

 K a 

-3 = 3.210

For salicylic acid

 K a 

= 1.0610

For acetic acid

 K a 

-5 = 1.7510

Thus, pyruvic acid

-3

Thus, salicylic acid

th

Fundamentals of Analytical Chemistry: 8 ed.

14-8



HIn + H2O  p K a = 7.10  K a 

+

Chapter 14

[H 3 O  ][In - ]

-

H3O + In

[HIn]

  K a

(Table 14-1)

-8 = antilog(-7.10) = 7.9410 -

[HIn]/[In ] = 1.43 Substituting these values into the equilibrium expression and rearranging gives + -8 -7 [H3O ] = 7.9410 1.43 = 1.1310 -7

 pH = -log(1.1310 ) = 6.94

14-9

+

InH + H2O

 

[H 3 O  ][In]

+

In + H3O

[InH  ]

For methyl orange, p K a = 3.46  K a 

  K a

(Table 14-1)

-4 = antilog(-3.46) = 3.4710 +

[InH ]/[In] = 1.64 Substituting these values into the equilibrium expression and rearranging gives + -4 -4 [H3O ] = 3.4710 1.64 = 5.6910 -4  pH = -log(5.6910 ) = 3.24

+

14-10 [H3O ] = o

At 0 C, o

At 50 C, o

At 100 C,

 K w

and

1/2

pH = -log( K w)   = -½log K w -15

pH = -½ log(1.1410 ) = 7.47 -14

pH = -½ log(5.4710 ) = 6.63 -13

pH = -½ log(4.910 ) = 6.16

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Fundamentals of Analytical Chemistry: 8 ed. o

-15 p K w  = -log(1.1410 ) = 14.94

14-11 At 0 C, o

-14

At 50 C,

p K w  = -log(5.4710 ) = 13.26

o

-13 p K w  = -log(4.910 ) = 12.31

At 100 C,

14-12  pH + pOH = p K w

14-13

-

and

-2

pOH = -log[OH ] = -log(1.0010 ) = 2.00

(a)

pH = p K w  - pOH = 14.94 - 2.00 = 12.94

(b)

pH = 13.26 - 2.00 = 11.26

(c)

pH = 12.31 - 2.00 10.31

14.0 g HCl 1.054 g soln 100 g soln



mL soln

+

[H3O ] = 4.047 M

14-14

Chapter 14



1 mmol HCl 0.03646 g HCl

and

pH = -log4.047 = -0.607

9.00 g NaOH 1.098 g soln 100 g soln



mL soln

-

[OH ] = 2.471 M

= 4.047 M

and

1 mmol NaOH



0.04000 g NaOH

= 2.471 M

pH = 14.00 - (-log2.471) = 14.393 -

14-15 The solution is so dilute that we must take into account the contribution of water to [OH ] +

which is equal to [H3O ]. Thus, -

-8

+

-8

[OH ] = 2.0010  + [H3O ] = 2.0010  + - 2

-8

-

-14

[OH ]  –  2.0010 [OH ] –  1.0010 -

[OH ] = 1.10510

1.00  10 14 -

[OH ]

= 0

-7

 pOH = -log 1.10510

-7

= 6.957

and

pH = 14.00 –   6.957 = 7.04

14-16 The solution is so dilute that we must take into account the contribution of water to +

-

[H3O ] which is equal to [OH ]. Thus,

th

Fundamentals of Analytical Chemistry: 8 ed.

Chapter 14 1.00  10 

14

+

-8

-

-8

[H3O ] = 2.0010  + [OH ] = 2.0010  + + 2

-8

+

[H3O ]  –  2.0010 [H3O ] –  1.0010 + -7 [H3O ] = 1.10510

cHCl 

pH = -log 1.10510   = 6.96

0.05832 g Mg(OH) 2 / mmol

= 1.749 mmol Mg(OH)2 taken

= (75.00.0600 –  1.7492)/75.0 = 0.01366 M +

[H3O ] = 0.01366 (b) 

= 0 -7

and

0.102 g Mg(OH) 2

14-17 In each part,

(a)

-14

[H 3 O  ]

and

pH = -log(0.01366) = 1.87

15.00.0600 = 0.900 mmol HCl added. Solid Mg(OH)2 remains and 2+

[Mg ] = 0.900 mmol HCl  K sp 

1 mmol Mg 2



2 mmol HCl

1 15.0 mL soln

  = 0.0300 M

-12 2+ - 2 = 7.110   = [Mg ][OH ] -

-12

1/2

-5

[OH ] = (7.110 /0.0300)   = 1.5410 -5

 pH = 14.00 - (-log(1.5410 )) = 9.19 (c) 

2+ 30.000.0600 = 1.80 mmol HCl added, which forms 0.90 mmol Mg . 2+

-2

[Mg ] = 0.90/30.0 = 3.0010

-12 1/2 -5 [OH ] = (7.110 /0.0300)   = 1.5410 -5  pH = 14.00 - (-log(1.5410 )) = 9.19

(d)

2+

[Mg ] = 0.0600 M -12 1/2 -5 [OH ] = (7.110 /0.0600)   = 1.0910 -5

 pH = 14.00 - (-log(1.0910 )) = 9.04

th

Fundamentals of Analytical Chemistry: 8 ed.

Chapter 14

14-18 In each part, (20.0 mL HCl  0.200 mmol HCl/mL) = 4.00 mmol HCl is taken (a)

cHCl

4.00 mmol HCl

+

= [H3O ] =

20.0  25.0mL soln

  = 0.0889 M

 pH = -log 0.0899 = 1.05 (b)

Same as in part (a); pH = 1.05

(c)

cHCl 

-2 = (4.00 –  25.0  0.132)/(20.0 + 25.0) = 1.55610  M

+ -2 [H3O ] = 1.55610 M

(d)

and

-2 As in part (c), cHCl  = 1.55610

-2

pH = -log 1.55610   = 1.81 and pH = 1.81

+

(The presence of NH4  will not alter the pH significantly.) (e)

c NaOH 

-2

= (25.0  0.232 –   4.00)/(45.0) = 4.0010  M

 pOH = -log 4.0010

14-19 (a) (b)

+

[H3O ] = 0.0500

-2

= 1.398

and

and

pH = 14.00 –   1.398 = 12.60

pH = -log(0.0500) = 1.30

2

2

  = ½ {(0.0500)(+1)  + (0.0500)(-1) } = 0.0500

 H O   =

0.85

(Table 10-2)

3

a

H3O 

  = 0.860.0500 = 0.0425

 pH = -log(0.043) = 1.37

14-20 (a)

[OH ] = 20.0167 = 0.0334 M

 pH = 14 –   (-log(0.0334)) = 12.52 (b)

2

2

  = ½ {(0.0167)(+2)  + (0.0334)(-1) } = 0.050

 OH   = a

OH

0.81

(Table 10-2)

  = 0.810.0334 = 0.0271

th

Fundamentals of Analytical Chemistry: 8 ed. a

a

OH

  aH O   =

H3O 

Chapter 14

-14 1.0010

3

-14 -13   = 1.0010 /0.0271 = 3.6910

-13  pH = -log(3.6910 ) = 12.43

14-21 HOCl + H2O +



-

[H3O ] = [OCl ] + 2

+

-

H3O   + OCl

and

=

 K a

[HOCl] =

+

[H 3 O  ][OCl - ] [HOCl]

-8   = 3.010

+

cHOCl –  [H3O

]

-8

[H3O ] /(cHOCl  –  [H3O ]) = 3.010

+ 2

-8

+

-8

rearranging gives the quadratic: 0 = [H3O ]  + 310 [H3O ] - cHOCl3.010 +

cHOCl

[H3O ]

pH

(a)

0.100

-5 5.47610  

4.26

(b)

0.0100

-5 1.73110  

4.76

(c) 

1.0010  

-4

-

14-22 OCl + H2O

-6

1.71710  

  HOCl

-

[HOCl] = [OH ]

5.76

-

+ OH

 K  b =

-

and

[OCl ] =

 K w  K a



[HOCl][OH - ] -

[OCl ]



1.00  10 14 3.0  10

8

 3.33  10 7

-

c NaOCl –  [OH ]

- 2 -7 [OH ] /(c NaOCl -[OH ]) = 3.3310 - 2

-7

-

rearranging gives the quadratic: 0 = [OH ]  + 3.3310 [OH ] - c NaOCl3.3310 -

c NaOCl

[OH ]

pOH

pH

(a)

0.100

-4 1.82310

3.74

10.26

(b)

0.0100

-5 5.75410

4.24

9.76

(c) 

-4 1.0010  

-6 5.60610

5.25

8.75

-7

th

Fundamentals of Analytical Chemistry: 8 ed.



14-23  NH3 + H2O +

+

-

NH4  + OH

-

[NH4 ] = [OH ] - 2

and

Chapter 14

1.00  1014

 K  b =

[NH3] =

-

[OH ] /( c NH3  -[OH ]) = 1.7510

5.7  10

10

 1.75  105

-

 –  [OH ] 3

c NH

-5

- 2

-5

-

rearranging gives the quadratic: 0 = [OH ]  + 1.7510 [OH ] -

c NH

[OH ]

pOH

pH

(a)

0.100

-3 1.31410

2.88

11.12

(b)

0.0100

-4 4.09710

3.39

10.62

(c) 

-4 1.0010  

-5 3.39910

4.47

9.53

+

 K a 

3

+

14-24  NH4 + H2O



H3O + NH3

+

[H3O ] = [NH3]

+

and

+ 2

[NH4 ] =

+

[H3O ] /( c NH  –  [H3O ]) = 5.710

c NH

3

-5

1.7510

-10 = 5.710

c

+

 NH4

 –  [H3O ]

-10

4

+ 2

-10

+

rearranging gives the quadratic: 0 = [H3O ]  + 5.710 [H3O ] c

+



 NH4

[H3O ]

pH 5.12

(a)

0.100

-6 7.55010  

(b)

0.0100

2.38710  

5.62

(c) 

-4 1.0010  

-7 1.38510  

6.62

14-25 C5H11 N + H2O +

-6



+

-

C5H11 NH  + OH -

[C5H11 NH ] = [OH ]

and

 K  b =

[C5H11 N] =

1.00  1014 7.5  10 cC

5 H11 N

12

c

 NH4

 1.333 103 -

 –  [OH ]

-10

5.710

th

Fundamentals of Analytical Chemistry: 8 ed.

Chapter 14

- 2 -3 [OH ] /( cC5H11 N  -[OH ]) = 1.33310 - 2

-3

-

rearranging gives the quadratic: 0 = [OH ]  + 1.33310 [OH ] -

cC

[OH ]

pOH

pH

(a)

0.100

-2 1.09010

1.96

12.04

(b)

0.0100

-3 3.04510

2.52

11.48

(c) 

-4 1.0010  

-5 9.34510

4.03

9.97

+

 K a 

5 H11 N

14-26 HIO3 + H2O +



H3O + IO3

-

[H3O ] = [IO3 ]

and

+ 2

-

[HIO3] =

+

[H3O ] /( cHIO3  –  [H3O ]) = 1.710

+

-1

cHIO

[H3O ]

pH

(a)

0.100

-2 7.06410  

1.15

(b)

0.0100

-3 9.47210  

2.02

(c) 

-4 1.0010  

-5 9.99410  

4.00

14-27 (a)

cHA

= 43.0 g HA 

HA + H2O



+

-

[H3O ] = [A ]

1 mmol HA 0.090079 g HA +

-

H3O + A and

1.33310

 –  [H3O ] 3

cHIO

+ 2

+

5 H11 N

-1 = 1.710

-1

+

rearranging gives the quadratic: 0 = [H3O ]  + 1.710 [H3O ] -

3

-3

cC



1 500 mL soln

 K a 

cHIO

3

1.710

-1

= 0.9547 M HA

-4 = 1.3810 +

[HA] = 0.9547 –  [H3O ]

+ 2 + -4 [H3O ] /(0.9547 –  [H3O ]) = 1.3810 +

rearranging and solving the quadratic gives: [H3O ] = 0.0114 and pH = 1.94 (b)

cHA  

= 0.954725.0/250.0 = 0.09547 M HA

th

Fundamentals of Analytical Chemistry: 8 ed.

Chapter 14 +

Proceeding as in part (a) we obtain: (c)

cHA  

-4 = 0.0954710.0/1000.0 = 9.54710  M HA +

Proceeding as in part (a) we obtain:

14-28 (a)

cHA

-3

[H3O ] = 3.5610  and pH = 2.45

= 1.05 g HA 

HA + H2O



+

-

1 mmol HA

1



0.22911 g HA 100 mL soln +

-

H3O + A

[H3O ] = [A ]

-4

[H3O ] = 3.0010  and pH = 3.52

 K a 

= 0.04583 M HA

= 0.43

+ [HA] = 0.04583 –  [H3O ]

and

+ 2 + [H3O ] /(0.04583 –  [H3O ]) = 0.43 +

rearranging and solving the quadratic gives: [H3O ] = 0.0418 and pH = 1.38 (b)

cHA  

= 0.0458310.0/100.0 = 0.004583 M HA +

Proceeding as in part (a) we obtain: (c)

cHA  

-5 = 0.00458310.0/1000.0 = 4.58310  M HA +

Proceeding as in part (a) we obtain:

HA + H2O cHA  



+

-

H3O + A

-5

[H3O ] = 4.58310  and pH = 4.34

amount HA taken = 20.00 mL 

14-29 Throughout 14-29: (a)

-3

[H3O ] = 4.53510  and pH = 2.34

 K a 

0.200 mmol mL

  = 4.00 mmol

-4 = 1.8010

-2 = 4.00/45.0 = 8.8910 +

-

[H3O ] = [A ]

and

+

[HA] = 0.0889 –  [H3O ]

+ 2 + -4 [H3O ] /(0.0889 –  [H3O ]) = 1.8010 +

-3

rearranging and solving the quadratic gives: [H3O ] = 3.9110  and pH = 2.41 (b)

amount NaOH added = 25.0  0.160 = 4.00 mmol therefore, we have a solution of NaA

th

Fundamentals of Analytical Chemistry: 8 ed. -

A + H2O c

A-



Chapter 14

-

OH   + HA

 K  b 

-14 -4 -11 = 1.0010 /(1.8010 ) = 5.5610

-2   = 4.00/45.0 = 8.8910

-

[OH ] = [HA]

-

and

-

[A ] = 0.0889 –  [OH ]

- 2 -11 [OH ] /(0.0889 –  [OH ]) = 5.5610 -

-6

rearranging and solving the quadratic gives: [OH ] = 2.2210  and pH = 8.35 (c)

amount NaOH added = 25.0  0.200 = 5.00 mmol therefore, we have an excess of NaOH and the pH is determined by its concentration -

[OH ] = (5.00 - 4.00)/45.0 = 2.2210

-2

 pH = 14 –   pOH = 12.35 (d)

amount NaA added = 25.0  0.200 = 5.00 mmol [HA] = 4.00/45.0 = 0.0889 -

[A ] = 5.00/45.00 = 0.1111 +

-4

[H3O ]0.1111/0.0889 = 1.8010 + -4 [H3O ] = 1.44010

and

pH = 3.84

14-30 Throughout 14-30 the amount of NH3 taken is 4.00 mmol (a)

NH3 + H2O c NH

3



-

OH + NH4

+

1.00  10 14

 K  b =

5.7  10

 1.75  105

10

-2   = 4.00/60.0 = 6.6710

+

-

[NH4 ] = [OH ]

and

-

[NH3] = 0.0667 –  [OH ]

- 2 -5 [OH ] /(0.0667 –  [OH ]) = 1.7510 -

-3

rearranging and solving the quadratic gives: [OH ] = 1.0710  and pH = 11.03 (b)

amount HCl added = 20.0  0.200 = 4.00 mmol

th

Fundamentals of Analytical Chemistry: 8 ed.

Chapter 14

therefore, we have a solution of NH4Cl +

 NH4 + H2O c



 NH4



+

H3O + NH3

  = 4.00/60.0 = 6.6710 +

[H3O ] = [NH3]

and

 K a 

-10 = 5.710

-2

+ + [NH4 ] = 0.0667 –  [H3O ]

+ 2 + -10 [H3O ] /(0.0667 –  [H3O ]) = 5.710 +

-6

rearranging and solving the quadratic gives: [H3O ] = 6.1610  and pH = 5.21 (c)

amount HCl added = 20.0  0.250 = 5.00 mmol therefore, we have an excess of HCl and the pH is determined by its concentration +

-2

[H3O ] = (5.00 - 4.00)/60.0 = 1.6710  pH = 1.78 (d)

amount NH4Cl added = 20.0  0.200 = 4.00 mmol +

[NH3] = 4.00/60.0 = 0.0667

[NH4 ] = 4.00/60.0 = 0.0667

+ -10 [H3O ]0.0.0667/0.0667 = 5.7010 + -10 [H3O ] = 5.7010

(e)

and

pH = 9.24

amount HCl added = 20.0  0.100 = 2.00 mmol +

[NH3] = (4.00-2.00)/60.0 = 0.0333

[NH4 ] = 2.00/60.0 = 0.0333

+ -10 [H3O ]0.0.0333/0.0333 = 5.7010 + -10 [H3O ] = 5.7010

14-31 (a)

+

NH4 + H2O



[NH3] = 0.0300

and

pH = 9.24

+

-5

H3O + NH3 

and

5.7010

=

[H 3 O  ][NH 3 ]

+

[NH4 ] = 0.0500

+ -10 -10 [H3O ] = 5.7010   0.0500/0.0300 = 9.5010

[ NH 4 ]

th

Fundamentals of Analytical Chemistry: 8 ed.

Chapter 14

-14 -10 -5 [OH ] = 1.0010 /9.5010   = 1.0510 -10

 pH = -log (9.5010 ) = 9.022 (b)

2

2

  = ½ {(0.0500)(+1)  + (0.0500)(-1) } = 0.0500

 NH

From Table 10-2

= 0.80

 NH

and

a

 NH [ NH 4 ] 5.70  105  0.80  0.0500  =    NH [ NH3 ] 1.00  0.0300  K a

H3 O

 = 1.0

3

4



4

-10 7.6010

3

-10

 pH = -log (7.6010 ) = 9.12

14-32 In each part of this problem a buffer mixture of a weak acid, HA, and its conjugate base, +

-

 NaA, is formed. In each case we will assume initially that [H3O ] and [OH ] are much -

smaller than the molar concentration of the acid an d conjugate so that [A ]  c NaA and [HA]  cHA. These assumptions then lead to the following relationship: +

[H3O ] = (a)

cHA

c NaA

 K a cHA / c NaA

= 9.20 g HA 

1 mol HA

1



90.08 g HA 1.00 L soln

= 11.15 g HA 

1 mol NaA



  = 0.1021 M

1

112.06 g NaA 1.00 L soln

  = 0.0995 M

+ -4 -4 [H3O ] = 1.3810 0.1021/0.0995 = 1.41610 +

-

 Note that [H3O ] (and [OH ])