SKOOG - SOLUCIONÁRIO CAPÍTULO 10.pdf

th

Fundamentals of Analytical Chemistry: 8 ed.

Chapter 10

Chapter 10

10-1

solution. The (a) Activity, aA, is the effective concentration of a chemical species A in solution. activity coefficient , γA, is the numerical factor necessary to convert the molar

concentration of the chemical species A to activity as shown below: aA=γA[A]

refers to an ideal system within which each (b) The thermodynamic equilibrium constant refers chemical species is unaffected by any others. A concentration equilibrium constant takes takes into account the influence influence exerted by solute species upon one another. The thermodynamic equilibrium constant is numerically constant and ind ependent of ionic strength; the concentration equilibrium constant depends upon molar concentrations of reactants and products as well as other chemical species that may not participate in the equilibrium.

10-2

Activity coefficients have the following properties: 1. The activity coefficient depends on the solution ionic strength. 2. In very dilute solutions, the activity a ctivity coefficient approaches unity. 3. For a given ionic strength, the activity coefficient becomes smaller as the charge of the chemical species increases. 4. At any ionic strength, the activity coefficients are approximately equal for che mical species having the same charge state.

10-3

(a) MgCl 2

+ 2 NaOH →← Mg(OH ) 2 ( s ) + 2 NaCl 2+

+

Replacing divalent-Mg with Na causes the activity coefficient to increase (i.e.,

th


Chapter 10

approach 1.0). Thus, the ionic strength decreases. (b) HCl + NaOH

→ ←

H 2 O + NaCl

The activity coefficient remains relatively constant when the NaOH (strong base) is added to HCl (strong acid). There is no change in the charge states states of the ions present present in the solution equilibria. The ionic strength strength is unchanged. (c) HOAc + NaOH

→ ←

H 2 O + NaOAc

The activity coefficient will decrease when the NaOH (strong base) is added to the acetic +

-

acid (weak acid) generating water, Na and OAc (conjugate base). Thus, the ionic strength increases.

10-4

(a) The ionic strength will increase when FeCl3 is added to HCl. (b) 3 NaOH + FeCl 3

→ ←

Fe(OH ) 3 ( s ) + 3 NaCl -

-

Addition of FeCl3 to NaOH replaces a univalent OH with univalent Cl ; thus, the ionic strength is unchanged. (c) 3AgNO3

+ FeCl 3 → Fe( NO 3 ) 3 + 3AgCl( s ) ← +

3+

Addition of FeCl3 replaces the univalent Ag with the trivalent Fe ; thus, the ionic strength increases.

10-5

2+

+

The initial slope of the change of activity coefficient for Ca is steeper than that for K

because activity coefficients for multiply-charged ions deviate from ideality (unity) more than activity coefficients for singly-charged ions.

10-6

The chemical species NH3 is not charged; therefore, the activity coefficient is unity.

th


10-7

10-8

(a)

µ=

1

(b)

µ=

1

(c)

µ=

1

(d)

µ=

1

2

2

2

2

Chapter 10

([0.040](2)

2

([0.40](1)

2

+ [0.20](2) 2 ) = 0.60

([0.10](3)

2

+ [0.30](1) 2 + [0.20](2) 2 + [0.40](1) 2 ) = 1.2

([0.060](3)

− log γ A =

2

+ [0.040]( 2) 2 ) = 0.16

+ [0.180](1) 2 + [0.030](2) 2 + [0.060](1) 2 ) = 0.45

µ 1 + 3.3α A µ 0.51Z A

2

Z

µ

µ

α

-log γ

γ

(a)

3

0.075

0.274

0.9

0.693

0.20

(b)

2

0.012

0.110

0.45

0.193

0.64

(c)

4

0.080

0.283

1.1

1.139

0.073

(d)

4

0.060

0.245

1.1

1.058

0.087

10-9 (a)

µ = 0.05 γ Fe + = 0.24 µ = 0.10 γ Fe + = 0.18 3

γ Fe + (µ = 0.075) = 0.18 + 3

3

0.025 0.05

(0.24 − 0.18) = 0.21

(b)

µ = 0.01 γ Pb + = 0.665 µ = 0.05 γ Pb + = 0.45 2

γ Pb + (µ = 0.012) = 0.45 + 2

2

0.038 0.05

(0.665 − 0.45) = 0.61

th


Chapter 10

(c)

µ = 0.05 γ Ce + = 0.10 µ = 0.10 γ Ce + = 0.063 4

4

γ Ce + (µ = 0.080) = 0.063 + 4

0.02 0.05

(0.10 − 0.063) = 0.078

(d)

µ = 0.05 γ Sn + = 0.10 µ = 0.10 γ Sn + = 0.063 4

4

γ Sn + (µ = 0.060) = 0.063 + 4

0.04 0.05

(0.10 − 0.063) = 0.093

10-10 (a)

µ = 5.0 × 10 −2 γ Ag + = 0.80 γ SCN − = 0.81 K sp

'

=

K sp

γ Ag + γ SCN −

=

1.1 × 10 −12

= 1.7 × 10 −12

(0.80)(0.81)

(b)

µ = 5.0 × 10 −2 γ Pb + = 0.46 γ I− = 0.80 2

K sp

'

=

K sp

=

γ Pb + γ I− 2

7.9 × 10 −9 (0.45)(0.80)

2

= 2.7 × 10 −8

(c)

µ = 5.0 × 10 −2 γ La + = 0.24 γ IO − = 0.82 3

3

K sp

'

=

K sp

=

γ La + γ IO − 3

1.0 × 10

−11

(0.24)( 0.82)

3

= 7.6 × 10 −11

3

(d)

µ = 5.0 × 10 −2 γ Mg + = 0.52 γ NH + = 0.80 γ PO

3−

2

4

K sp

'

=

K sp

γ Mg + γ NH + γ PO

= 3−

2

4

4

3.0 × 10 −13 (0.52)(0.80)(0.16)

= 0.16

4

= 4.5 × 10 −12

th


Zn 10-11 Zn(OH ) 2 ( s ) → ←

2+

+ 2OH −

K sp

Chapter 10

= 3.0 × 10 −16

(a)

1

µ=

(0.0100(1)

2

+ 0.0100(1) 2 ) = 0.0100

2 From Table 10 − 2,

γ Zn + = 0.675

γ OH − = 0.90

2

K 'sp

= a Zn + aOH − 2 = γ Zn + [Zn 2+ ] × γ OH − 2 [OH − ] 2

2+

2

−

[Zn ][OH ]

2

=

3.0 × 10 −16

γ Zn + γ OH −

2

=

2

Solubility = S = [Zn 2+ ] = S(2S) 2

1 2

3.0 × 10 −16 (0.675)(0.90)

2

= 5.49 × 10 −16

[OH − ]

= 5.49 × 10 −16 −16

1

⎛ 5.49 × 10 ⎞ 3 ⎟⎟ = 5.2 × 10 −6 M 4 ⎝ ⎠

S = ⎜⎜ (b)

1

µ=

((2 × 0.0167)(1)

2


γ Zn + = 0.48

γ OH − = 0.81

2

K ' sp

+ 0.0167(2) 2 ) = 0.050

= a Zn + aOH − 2 = γ Zn + [ Zn 2+ ] × γ OH − 2 [OH − ] 2

2+

2

−

[Zn ][OH ]

2

=

3.0 × 10 −16

γ Zn + γ OH −

2

2

Solubility = S = [Zn 2+ ] = S(2S)

2

1 2

=

3.0 × 10 −16 (0.48)(0.81)

[OH − ]

= 9.53 × 10 −16 −16

1

⎛ 9.53 × 10 ⎞ 3 ⎟⎟ = 6.2 × 10 −6 M 4 ⎝ ⎠

S = ⎜⎜

2

= 9.53 × 10 −16

th


Chapter 10

(c)

≡

0.25 M KOH

0.250 mole

0.0250 M ZnCl 2

+

[ K ] = [OH

−

L

≡

[Cl ] =

L 1000 mL

0.0250 mole L

5.0 × 10 −3 mole

×

100.0 mL

( 5.0 × 10 − ]=

−

×

×

1000 mL L

L 1000 mL

100.0 mL

2 × (2.0 × 10 −3 mole) 100.0 mL

×

× 80.0 mL = 2.0 × 10 −3 mole ZnCl 2

= 5.0 × 10 −2 M

− ( 2 × ( 2.0 × 10 −3 )) )

3

× 20.0 mL = 5.0 × 10 −3 mole KOH

×

1000 mL L

1000 mL L

= 1.0 × 10 −2 M

= 4.0 × 10 −2 M

[Zn 2+ ] = 0 1

µ=

(0.05(1)

2

+ 0.010(1) 2 + 0.04(1) 2 ) = 0.05


γ Zn + = 0.48

γ OH − = 0.81

2

2+

−

[Zn ][OH ]

2

=

3.0 × 10 −16

γ Zn + γ OH −

=

2

2

Solubility = S = [Zn 2+ ]

= a Zn + aOH − 2 = γ Zn + [Zn 2+ ] × γ OH − 2 [OH − ]

K ' sp

2

2

3.0 × 10 −16 (0.48)(0.81)

S(1.0 × 10 −2 ) 2

2

= 9.53 × 10 −16

= 9.53 × 10 −16

⎛ 9.53 × 10 −16 ⎞ −12 ⎟ S = ⎜⎜ M −2 2 ⎟ = 9.53 × 10 × ( 1 . 0 10 ) ⎝ ⎠ (d)

0.100 M KOH

≡

0.0250 M ZnCl 2 +

[ K ] =

0.100 mole L

≡

×

L 1000 mL

0.0250 mole

5.0 × 10 −3 mole 100.0 mL

L

×

×

1000 mL L

× 20.0 mL = 2.0 × 10 −3 mole KOH

L 1000 mL

× 80.0 mL = 2.0 × 10 −3 mole ZnCl 2

= 2.0 × 10 −2 M

[OH − ] = 0 −

[Cl ] =

2 × (2.0 × 10 −3 mole) 100.0 mL

×

1000 mL L

= 4.0 × 10 −2 M

th


[Zn 2+ ] = 1

µ=

2.0 × 10

(0.02(1)

2

−3

−

1

Chapter 10

−3

(2.0 × 10 )

2 100.0 mL

×

1000 mL L

= 1.0 × 10 −2 M

+ 0.040(1) 2 + 0.01( 2) 2 ) = 0.05


γ Zn + = 0.48

γ OH − = 0.81

2

K 'sp

= a Zn + aOH − 2 = γ Zn + [Zn 2+ ] × γ OH − 2 [OH − ] 2

2+

2

−

[Zn ][OH ]

2

3.0 × 10 −16

=

γ Zn + γ OH −

2

=

2

1

Solubility = S = −2

2 −

3.0 × 10 −16 (0.48)(0.81)

2

= 9.53 × 10 −16

[OH − ]

(1.0 × 10 )[OH ]2

= 9.53 × 10 −16 −16

1

⎛ 9.53 × 10 ⎞ 2 = 3.09 × 10 −7 M −2 ⎟ ⎟ ⎝ (1.0 × 10 ) ⎠

[OH − ] = ⎜⎜

10-12

S=

1

µ=

1

2

(3.09 × 10 − M ) = 1.5 × 10 − 7

7

2

+ (2 × (0.0333))(1) 2 ) = 0.100

(0.0333(2) 2

M

(a)

AgSCN( s ) → Ag + ←

+ SCN −

γ Ag+ = 0.75 γ SCN − = 0.76

(1)

+

1.1 × 10 −12

−

[Ag ][SCN ] =

γ Ag + γ SCN −

=

K ' sp

1.1 × 10 −12 (0.75)(0.76)

Solubility = S = [Ag + ] = [SCN − ] S2

= 1.93 × 10 −12

S = (1.93 × 10

−12

1

)

2

= 1.4 × 10 −6 M 1

( 2)

S = (1.1 × 10

−12

)2

= a Ag+ aSCN − = γ Ag+ [Ag + ] × γ SCN − [SCN − ]

= 1.0 × 10 −6 M

= 1.93 × 10 −12

th


Chapter 10

(b) 2+ PbI 2 ( s ) → Pb ←

+ 2I −

γ Pb + = 0.36 γ I− = 0.75

(1)

2+

−

[ Pb ][ I ]

=

2

7.9 × 10 −9

γ Pb + γ I −

2

=

2

1

Solubility = S = [Pb 2+ ] = S(2S)

2

= 3.90 × 10

2

= a Pb + a I − 2 = γ Pb + [ Pb 2+ ] × ( γ I− [I − ]) 2 2

7.9 × 10 −9 (0.36)(0.75)

2

2

= 3.90 × 10 −8

[I − ]

−8

1

−8

⎛ 3.90 × 10 ⎞ 3 ⎟⎟ = 2.1 × 10 −3 M 4 ⎝ ⎠ 1 7.9 × 10 −9 3 S=( ) = 1.3 × 10 −3 M

S = ⎜⎜ ( 2)

K ' sp

2

4

(c)

BaSO 4 ( s ) → Ba ← (1)

2+

+ SO 4 2−

γ Ba + = 0.38 γ SO 2−

[ Ba 2+ ][SO 4 ] =

= 0.35

2−

2

4

1.1 × 10

−10

γ Ba + γ SO

2−

2

=

K ' sp

1.1 × 10 2−

= 8.3 × 10 −10

S = (8.3 × 10

−10

1

)

2

= 2.9 × 10 −5 M 1

( 2)

S = (1.1 × 10

−10

)2

= 1.0 × 10 −5 M

2−

2

−10

(0.38)(0.35)

4

Solubility = S = [Ba 2+ ] = [SO 4 ] S2

= a Ba + aSO

4

= γ Ba + [ Ba 2 + ] × γ SO

= 8.3 × 10 −10

2−

2−

2

4

[SO 4 ]

th


Chapter 10

(d) 2+ Cd 2 Fe(CN ) 6 ( s ) → 2 Cd ←

+ Fe(CN ) 6 4−

γ Cd + = 0.38 γ Fe ( CN ) − = 0.020

(1)

2

4

6

K ' sp

= aCd + 2 a Fe( CN )

4−

2

6

= ( γ Cd + [Cd 2+ ]) 2 × γ Fe ( CN ) 3.2 × 10

4−

2+ 2

[Cd ] [ Fe(CN ) 6 ] =

−17

γ Cd + γ Fe ( CN )

1 2

=

2

4−

2

Solubility = S =

4−

4−

2

6

[ Fe(CN ) 6 ]

3.2 × 10 −17 2

(0.38) (0.020)

= 1.1 × 10 −13

6

4−

[Cd 2+ ] = [ Fe(CN ) 6 ]

( 2S) S = 1.1 × 10 −13 2

1

−13

⎛ 1.1 × 10 ⎞ 3 ⎟⎟ = 1.4 × 10 −5 M S = ⎜⎜ ⎝ 4 ⎠ 1 3.2 × 10 −17 3 ( 2) S = ( ) = 2.0 × 10 −6 M 4

10-13

µ=

1

(0.0167(2) 2

2

+ (2 × (0.0167))(1) 2 ) = 0.050

(a)

AgIO 3 ( s ) → Ag + ←

+ IO 3 −

γ Ag+ = 0.80 γ IO − = 0.82

(1)

3

+

3.1 × 10 −8

−

[Ag ][ IO3 ] =

γ Ag+ γ IO −

=

K ' sp

3.1 × 10 −8 (0.80)(0.82)

3

−

Solubility = S = [Ag + ] = [IO 3 ] S2

= 4.7 × 10 −8

S = (4.7 × 10

−8

1

)

2

= 2.2 × 10 −4 M 1

( 2)

−8

S = (3.1 × 10 ) 2

= a Ag+ a IO − = γ Ag+ [Ag + ] × γ IO − [ IO 3 − ]

= 1.8 × 10 −4 M

3

= 4.7 × 10 −8

3

th


Chapter 10

(b) 2+ Mg(OH ) 2 ( s ) → Mg ←

+ 2OH −

γ Mg + = 0.52 γ OH − = 0.81

(1)

2+

−

[ Mg ][OH ]

7.1 × 10 −12

=

2

γ Mg + γ OH −

2

=

2

1

Solubility = S = [Mg 2+ ] = S(2S) 2

2

= a Mg + aOH − 2 = γ Mg + [ Mg 2+ ] × ( γ OH − [OH − ]) 2 2

7.1 × 10 −12 (0.52)(0.81)

2

2

= 2.1 × 10 −11

[OH − ]

= 2.1 × 10 −11 −11

1

⎛ 2.1 × 10 ⎞ 3 ⎟⎟ = 1.7 × 10 −4 M ⎝ 4 ⎠ 1 7.1 × 10 −12 3 S=( ) = 1.2 × 10 −4 M

S = ⎜⎜ ( 2)

K ' sp

2

4

(c)

BaSO 4 ( s ) → Ba 2+ ← (1)

+ SO 4 2−

γ Ba + = 0.46 γ SO

2−

2

2+

2−

[ Ba ][SO 4 ] =

= 0.44

4

1.1 × 10 −10

γ Ba + γ SO

2−

2

=

K 'sp

= a Ba + aSO

1.1 × 10 −10 (0.46)(0.44)

4

2−

Solubility = S = [SO 4 ] (0.0167 M ) × S = 5.4 × 10 −10

⎛ 5.4 × 10 −10 ⎞ ⎟⎟ = 3.3 × 10 −8 M S = ⎜⎜ ⎝ 0.0167 ⎠ ⎛ 1.1 × 10 −10 ⎞ ⎟⎟ = 6.6 × 10 −9 M ( 2) S = ⎜⎜ ⎝ 0.0167 ⎠

2−

2

4

= γ Ba + [ Ba 2+ ] × γ SO

= 5.4 × 10 −10

2−

2−

2

4

[SO 4 ]

th


Chapter 10

(d)

La ( IO 3 ) 3 ( s ) → La 3+ ←

+ 3IO 3 −

γ La + = 0.24 γ IO − = 0.82

(1)

K ' sp

3

3

−

[ La 3+ ][IO 3 ]3

1.0 × 10

=

−11

γ La + γ IO −

=

3

3

= a La + a IO − 3 = γ La + [ La 3+ ] × ( γ IO − [ IO 3 − ]) 3

1.0 × 10

3

3

3

−11

(0.24)(0.82)

3

3

= 7.6 × 10 −11

3

3+

Solubility = S = [La ] = S(3S) 3

1 3

−

[IO 3 ]

= 7.6 × 10 −11 1

−11

⎛ 7.6 × 10 ⎞ 4 ⎟⎟ = 1.3 × 10 −3 M S = ⎜⎜ ⎝ 27 ⎠ 1

−11

⎛ 1.0 × 10 ⎞ 4 ⎟⎟ = 7.8 × 10 −4 M S = ⎜⎜ ⎝ 27 ⎠

( 2)

10-14

µ=

1

(0.05000(1) 2

+ 0.05000(1) 2 ) = 0.050

2

Cu (a) CuCl( s ) → ←

+

+ Cl − +

If we assume that Cu has an effective diameter of 0.3 like similarly charged cations, then

γ Cu+ = 0.80 γ Cl− = 0.80

(1) +

−

[Cu ][Cl ] =

1.9 × 10 −7

γ Cu + γ Cl − 2

=

K ' sp

= aCu+ aCl − = γ Cu+ [Cu + ] × γ Cl− [Cl − ]

1.9 × 10 −7 (0.80)(0.80)

= 2.97 × 10 −7

Solubility = S = [Cu + ] = [Cl − ] S

2

= 2.97 × 10 −7

S = (2.97 × 10

−7

1

)

2

S = (1.9 × 10

= 5.4 × 10 −4 M 1

)

= 4.4 × 10 −4 M ( 4.4 × 10 −4 − 5.4 × 10 −4 ) relative error = × 100% = − 19% 5.4 × 10 − 4

( 2)

−7

2

th


Chapter 10

(b)

Fe(OH ) 2 ( s ) → Fe 2+ ←

+ 2OH −

γ Fe + = 0.48 γ OH − = 0.81

(1)

K ' sp

2

2+

−

[ Fe ][OH ]

2

4.1 × 10 −15

=

γ Fe + γ OH −

=

2

2

1

Solubility = S = [Fe 2+ ] = S(2S)

2

= 1.3 × 10

2

= a Fe + aOH − 2 = γ Fe + [ Fe 2+ ] × ( γ OH − [OH − ]) 2 2

4.1 × 10 −15 (0.48)(0.81)

2

2

= 1.3 × 10 −14

[OH − ]

−14

1

−14

⎛ 1.3 × 10 ⎞ 3 ⎟⎟ = 1.6 × 10 −5 M ⎝ 4 ⎠

S = ⎜⎜

1

−15

⎛ 4.1 × 10 ⎞ 3 ⎟⎟ = 1.1 × 10 −5 M ⎝ 4 ⎠ 1.1 × 10 −5 − 1.6 × 10 −5 × 100% = − 31% relative error = 1.6 × 10 −5 S = ⎜⎜

( 2)

(c) 3+ Fe(OH ) 3 ( s ) → Fe ←

+ 3OH −

γ Fe + = 0.24 γ OH − = 0.81

(1)

K ' sp

3

3+

−

[ Fe ][OH ]

3

=

2 × 10 −39

γ Fe + γ OH −

3

3

Solubility = S = [Fe 3+ ] = S(3S) 3

1 3

=

= a Fe + aOH − 3 = γ Fe + [ Fe 3+ ] × ( γ OH − [OH − ]) 3 3

2 × 10 −39 (0.24)(0.81)

3

3

= 2 × 10 −38

[OH − ]

= 2 × 10 −38 −38

1

⎛ 2 × 10 ⎞ 4 ⎟⎟ = 1.55 × 10 −10 M ⎝ 27 ⎠

S = ⎜⎜

−39

1

⎛ 2 × 10 ⎞ 4 ⎟⎟ = 9.3 × 10 −11 M ⎝ 27 ⎠ 9.3 × 10 −11 − 1.55 × 10 −10 × 100% = − 40% relative error = 1.55 × 10 −10

( 2)

S = ⎜⎜

th


Chapter 10

(d)

La ( IO 3 ) 3 ( s ) → La 3+ ← (1)

+ 3IO 3 −

γ La + = 0.24 γ IO − = 0.82

K ' sp

3

3

−

[ La 3+ ][IO 3 ]3

=

1.0 × 10

−11

γ La + γ IO −

=

3

3

= a La + a IO − 3 = γ La + [ La 3+ ] × ( γ IO − [ IO 3 − ]) 3

1.0 × 10

3

3

3

−11

(0.24)(0.82)

3

3

= 7.6 × 10 −11

3

3+

Solubility = S = [La ] = S(3S) 3

1 3

−

[IO 3 ]

= 7.6 × 10 −11 −11

1

⎛ 7.6 × 10 ⎞ 4 ⎟⎟ = 1.3 × 10 −3 M S = ⎜⎜ ⎝ 27 ⎠ −11

1

⎛ 1.1 × 10 ⎞ 4 ⎟⎟ = 8.0 × 10 −4 M ( 2) S = ⎜⎜ ⎝ 27 ⎠ 8.0 × 10 −4 − 1.3 × 10 −3 relative error = × 100% = − 38% 1.3 × 10 −3 3Ag (e) Ag 3 ( AsO 4 )( s ) → ←

+

+ AsO 4 3−

3-

If we assume that AsO4 has an effective diameter of 0.4 like similarly charged anions, then (1)

γ Ag+ = 0.80 γ AsO − = 0.16 3

4

3−

[Ag + ]3 [AsO 4 ] =

6 × 10

− 23

γ Ag+ γ AsO

=

3

4−

Solubility = S = [AsO 3 ] =

3−

K ' sp

3

4

3

− 23

(0.80) (0.16)

[Ag + ]

3−

3−

3

6 × 10

4

1

= a Ag+ 3a AsO − = ( γ Ag+ [Ag + ]) 3 × γ AsO = 7.3 × 10 −22

4

[AsO 4 ]

th


Chapter 10

(3S) 3 S = 2.9 × 10 −22 −22

1

⎛ 7.3 × 10 ⎞ 4 ⎟⎟ = 2.3 × 10 −6 M ⎝ 27 ⎠

S = ⎜⎜

− 23

1

⎛ 6 × 10 ⎞ 4 ⎟⎟ = 1.2 × 10 −6 M ⎝ 27 ⎠ 1.2 × 10 −6 − 2.3 × 10 −6 relative error = × 100% = − 46% −6 2.3 × 10 S = ⎜⎜

( 2)

10-15 (a)

µ=

1

(0.200(1) 2

2

(0.51)(1) 2 0.20

− log γ H O+ =

1 + (3.3)(0.9) 0.20

3

=

= 0.0979 γ H O+ = 0.798 3

(0.51)(1) 2 0.20

− log γ OAc− = K ' a

+ 0.200(1) 2 ) = 0.200

1 + (3.3)(0.425) 0.20

γ H O+ [ H 3O + ]γ OAc − [OAc − ] 3

[ HOAc]

[ H 3O + ] =

(1.01 × 10 −5 )(0.10)

[ H 3O + ] =

(1.75 × 10 −5 )(0.10)

= 0.140 γ OAc− = 0.724

= γ H O+ γ OAc− K a = (0.798)(0.724)(1.75 × 10 −5 ) = 1.01 × 10 −5 3

= 1.00 × 10 −3 M

= 1.32 × 10 −3 M 1.32 × 10 −3 − 1.0 × 10 −3 × 100% = 32% relative error = 1.0 × 10 −3

(b)

µ=

1

(0.200(1) 2

2

− log γ NH + = 4

− log γ OH − =

+ 0.200(1) 2 ) = 0.200

(0.51)(1) 2 0.20 1 + (3.3)(0.25) 0.20 (0.51)(1) 2 0.20

1 + (3.3)(0.35) 0.20

= 0.167 γ H O+ = 0.681 3

= 0.150 γ OAc− = 0.708

th


Chapter 10

γ NH + [ NH 4 + ]γ OH − [OH − ]

⎛ 1.0 × 10 −14 ⎞ −6 ⎟ K ' b = = γ NH + γ OH − K b = (0.681)(0.708)⎜⎜ −10 ⎟ = 9.32 × 10 [ NH 3 ] ⎝ 5.70 × 10 ⎠ 1.0 × 10 −14 − −6 −4 + −11 [OH ] = (9.32 × 10 )(0.050) = 6.83 × 10 M [ H 3O ] = M −4 = 1.46 × 10 6.83 × 10 1.0 × 10 −14 − −5 −4 + = 1.07 × 10 −11 M [OH ] = (1.75 × 10 )(0.050) = 9.36 × 10 M [ H 3O ] = −4 9.36 × 10 −11 −11 1.07 × 10 − 1.46 × 10 × 100% = 27% relative error = 1.46 × 10 −11 4

4

(c)

µ=

1

(0.0600(1) 2

2

− log γ H O+ = 3

(0.51)(1) 2 0.060 1 + (3.3)(0.9) 0.060

− log γ ClCH COO− = 2

K ' a

+ 0.0600(1) 2 ) = 0.060

=

= 0.07232 γ H O+ = 0.847 3

(0.51)(1) 2 0.060 1 + (3.3)(0.425) 0.060

= 0.0930 γ ClCH COO− = 0.807

γ H O+ [ H 3O + ]γ OAc− [ClCH 2 COO − ] 3

[ClCH 2 COOH ]

2

= γ H O+ γ ClCH COO− K a 3

= (0.847)(0.807)(1.36 × 10 −3 ) = 9.30 × 10 −4 [ H 3O + ] = +

[ H 3O ] = relative error =

(9.3 × 10 − 4 )(0.01) −3

= 3.05 × 10 −3 M

(1.36 × 10 )(0.01) 3.69 × 10 −3

= 3.69 × 10 −3 M

− 3.05 × 10 −3 × 100% = 21% −3 3.05 × 10

2

th


Chapter 10

10-16

A

B

C

D

(c o n c )

p s (conc)

E

F

(act)

p s (act)

G

H

I

1 Prob lem 10-16(a) 10-16(a) 2 3 Ksp(AgIO3) 4

3.10E-08

5 6

c

(Ba(NO3)2)

p c (Ba(NO3)2)

s

s

+

(Ag )

-

(IO3 )

7

0.0001

4.00

1.76E-04

3.75

1.81E-04

3.74

4.76E-04

0.975

0.975

8

0.0002

3.70

1.76E-04

3.75

1.82E-04

3.74

7.76E-04

0.969

0.969

9

0.0004

3.40

1.76E-04

3.75

1.84E-04

3.74

1.38E-03

0.959

0.959

1.86E-04

3.73

2.58E-03

0.944

0.946

10

0.0008

3.10

1.76E-04

3.75

11

0.001

3.00

1.76E-04

3.75

1.87E-04

3.73

3.18E-03

0.939

0.941

12

0.002

2.70

1.76E-04

3.75

1.92E-04

3.72

6.18E-03

0.917

0.920

13

0.004

2.40

1.76E-04

3.75

1.98E-04

3.70

1.22E-02

0.888

0.894

14

0.008

2.10

1.76E-04

3.75

2.06E-04

3.69

2.42E-02

0.851

0.861

2.09E-04

3.68

3.02E-02

0.837

0.849

15

0.01

2.00

1.76E-04

3.75

16

0.02

1.70

1.76E-04

3.75

2.21E-04

3.66

6.02E-02

0.787

0.807

17

0.04

1.40

1.76E-04

3.75

2.37E-04

3.63

1.20E-01

0.729

0.760

18

0.08

1.10

1.76E-04

3.75

2.56E-04

3.59

2.40E-01

0.664

0.711

19

0.1

1.00

1.76E-04

3.75

2.64E-04

3.58

3.00E-01

0.642

0.695

1.76E-04

3.75

2.89E-04

3.54

6.00E-01

0.574

0.647

3.18E-04

3.50

1. 1.20E+00

0.509

0.602

20

0.2

0.70

21

0.4

0.40

1.76E-04

3.75

22

0.8

0.10

1.76E-04

3.75

3.50E-04

3.46

2. 2.40E+00

0.450

0.564

23

1

0.00

1.76E-04

3.75

3.60E-04

3.44

3. 3.00E+00

0.433

0.553

24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 Spr eadsheet Document ation 40 B7=-LOG(A7)

F7=-LOG(E7) F7=-LOG(E7)

41 C7=SQRT(A$4)

G7=0.5((A7)*(2)*(2)+(2*A 7)*(1)*(1)+(C7)*(1)*(1)+(C7)*(-1)*(-1))

42 D7=-LOG(C7)

H7=10^ (-0.51*(1)*(1)*SQRT(G7)/(1+(3.3*0.25*SQRT(G7)))

43 E7=SQRT(A$4/(H7*I7))

I7=10^ (-0.51*(-1)*(-1)*SQRT(G7)/(1+(3.3*0.425*SQRT(G7)))

th


A

B

Chapter 10

C

D

(c o n c )

p s (conc)

E

F

(act)

p s (act)

G

H

I

1 Prob lem 10-16(b) 10-16(b) 2 3 Ksp(Mg(OH)2) 4

7.10E-12

5 6

c

(Ba(NO3)2)

p c (Ba(NO3)2)

s

s

2+

(Mg )

-

(OH )

7

0.0001

4.00

1.21E-04

3.92

1.30E-04

3.88

1.13E-03

0.865

0.963

8

0.0002

3.70

1.21E-04

3.92

1.32E-04

3.88

1.53E-03

0.847

0.957

9

0.0004

3.40

1.21E-04

3.92

1.34E-04

3.87

2.33E-03

0.818

0.948

1.38E-04

3.86

3.93E-03

0.777

0.934

10

0.0008

3.10

1.21E-04

3.92

11

0.001

3.00

1. 1.21E-04

3.92

1.39E-04

3.86

4.73E-03

0.761

0.928

12

0.002

2.70

1. 1.21E-04

3.92

1.45E-04

3.84

8.73E-03

0.703

0.906

13

0.004

2.40

1. 1.21E-04

3.92

1.54E-04

3.81

1.67E-02

0.636

0.876

14

0.008

2.10

1. 1.21E-04

3.92

1.65E-04

3.78

3.27E-02

0.563

0.839

1.69E-04

3.77

4.07E-02

0.539

0.825

15

0.01

2.00

1.21E-04

3.92

16

0.02

1.70

1.21E-04

3.92

1.85E-04

3.73

8.07E-02

0.466

0.778

17

0.04

1.40

1.21E-04

3.92

2.04E-04

3.69

1.61E-01

0.401

0.725

18

0.08

1.10

1.21E-04

3.92

2.26E-04

3.65

3.21E-01

0.344

0.669

19

0.1

1.00

1.21E-04

3.92

2.34E-04

3.63

4.01E-01

0.329

0.651

1.21E-04

3.92

2.59E-04

3.59

8.01E-01

0.286

0.596

2.86E-04

3.54

1.60E+00

0.254

0.547

20

0.2

0.70

21

0.4

0.40

1.21E-04

3.92

22

0.8

0.10

1.21E-04

3.92

3.12E-04

3.51

3.20E+00

0.230

0.504

23

1

0.00

1.21E-04

3.92

3.20E-04

3.49

4.00E+00

0.224

0.492

24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 Spreadsheet Documentation 40 B7=-LOG(A7)


41 C7=(A$4/(4))^ (1/3)

G7=0.5((A7)*(2)*(2)+(2*A 7)*(-1)*(-1)+(C7)*(2)*(2)+(2*C7)*(-1)*(-1))

42 D7=-LOG(C7)

H7=10^ (-0.51*(2)*(1)*SQRT(G7)/(1+(3.3*0.80*SQRT(G7)))

43 E7=(A$4/(4*H7*I7*I7))^ (1/3)

I7=10^ (-0.51*(-1)*(-1)*SQRT(G7)/(1+(3.3*0.35*SQRT(G7)))

th

Fundamentals of Analytical Chemistry: 8 ed. A 1

B

Chapter 10

C

D

(c o n c )

p s (conc)

E

F

(act)

p s (act)

G

H

I

Prob lem 10-16(c) 10-16(c)

2 3

Ksp(BaSO4)

4

1.10E-10

5 6

c

(Ba(NO3)2)

p c (Ba(NO3)2)

s

s

2+

(Ba )

2-

(SO4 )

7

0.0001

4.00

1.10E-06

5.96

1.32E-06

5.88

4.09E-04

0.912

0.912

8

0.0002

3.70

5.50E-07

6.26

7.10E-07

6.15

8.04E-04

0.880

0.879

9

0.0004

3.40

2.75E-07

6.56

3.92E-07

6.41

1.60E-03

0.838

0.836

10

0.0008

3.10

1.38E-07

6.86

2.25E-07

6.65

3.20E-03

0.784

0.781

1.89E-07

6.72

4.00E-03

0.764

0.760

11

0.001

3.00

1.10E-07

6.96

12

0.002

2.70

5.50E-08

7.26

1.15E-07

6.94

8.00E-03

0.693

0.687

13

0.004

2.40

2.75E-08

7.56

7.48E-08

7.13

1.60E-02

0.612

0.601

14

0.008

2.10

1.38E-08

7.86

5.19E-08

7.28

3.20E-02

0.523

0.507

15

0.01

2.00

1.10E-08

7.96

4.69E-08

7.33

4.00E-02

0.493

0.476

3.58E-08

7.45

8.00E-02

0.404

0.380

16

0.02

1.70

5.50E-09

8.26

17

0.04

1.40

2.75E-09

8.56

2.92E-08

7.54

1.60E-01

0.322

0.292

18

0.08

1.10

1.38E-09

8.86

2.49E-08

7.60

3.20E-01

0.253

0.218

19

0.1

1.00

1.10E-09

8.96

2.38E-08

7.62

4.00E-01

0.234

0.198

20

0.2

0.70

5.50E-10

9.26

2.06E-08

7.69

8.00E-01

0.183

0.146

1.74E-08

7.76

1. 1.60E+00

0.146

0.108

21

0.4

0.40

2.75E-10

9.56

22

0.8

0.10

1.38E-10

9.86

1.40E-08

7.85

3. 3.20E+00

0.119

0.082

23

1

0.00

1.10E-10

9.96

1.29E-08

7.89

4. 4.00E+00

0.113

0.076

24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39

Spreadsheet Document ation

40

B7=-L OG(A7)

F7=-LOG(E7)

41

C7=(A$4/(A 7))

G7=0.5((A7)*(2)*(2)+(2*A 7)*(1)*(1))

42

D7=-LOG(C7)

H7=10^ (-0.51*(2)*(2)*SQRT(G7)/(1+(3.3*0.50*SQRT(G7)))

43

E7=(A$4/(A 7*H7*I7))

I7=10^ (-0.51*(-2)*(-2)*SQRT(G7)/(1+(3.3*0.40*SQRT(G7)))

th

Fundamentals of Analytical Chemistry: 8 ed. A 1

B

Chapter 10

C

D

(c o n c )

p s (conc)

E

F

(act)

p s (act)

G

H

I

Prob lem 10-16(d) 10-16(d)

2 3

Ksp(La(IO3)3

4

1.00E-11

5 6

c

(Ba(NO3)2)

p c (Ba(NO3)2)

s

s

3+

(La )

-

(IO3 )

7

0.0001

4.00

7.80E-04

3.11

9.63E-04

3.02

4.98E-03

0.540

0.927

8

0.0002

3.70

7.80E-04

3.11

9.69E-04

3.01

5.28E-03

0.532

0.925

9

0.0004

3.40

7.80E-04

3.11

9.78E-04

3.01

5.88E-03

0.517

0.921

10

0.0008

3.10

7.80E-04

3.11

9.96E-04

3.00

7.08E-03

0.491

0.915

1.00E-03

3.00

7.68E-03

0.480

0.912

11

0.001

3.00

7. 7.80E-04

3.11

12

0.002

2.70

7. 7.80E-04

3.11

1.04E-03

2.98

1.07E-02

0.434

0.899

13

0.004

2.40

7. 7.80E-04

3.11

1.10E-03

2.96

1.67E-02

0.373

0.878

14

0.008

2.10

7. 7.80E-04

3.11

1.19E-03

2.93

2.87E-02

0.304

0.850

15

0.01

2.00

7.80E-04

3.11

1.22E-03

2.91

3.47E-02

0.282

0.839

1.35E-03

2.87

6.47E-02

0.216

0.800

16

0.02

1.70

7.80E-04

3.11

17

0.04

1.40

7.80E-04

3.11

1.52E-03

2.82

1.25E-01

0.162

0.754

18

0.08

1.10

7.80E-04

3.11

1.72E-03

2.76

2.45E-01

0.120

0.704

19

0.1

1.00

7.80E-04

3.11

1.80E-03

2.75

3.05E-01

0.110

0.687

20

0.2

0.70

7.80E-04

3.11

2.03E-03

2.69

6.05E-01

0.083

0.637

2.29E-03

2.64

1.20E+00

0.066

0.591

21

0.4

0.40

7.80E-04

3.11

22

0.8

0.10

7.80E-04

3.11

2.54E-03

2.60

2.40E+00

0.054

0.550

23

1

0.00

7.80E-04

3.11

2.61E-03

2.58

3.00E+00

0.051

0.538

24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 Spreadsheet Document ation 40 B7=-LOG(A7)


41 C7=(A$4/(A 7))

G7=0.5((A7)*(2)*(2)+(2*A 7)*(-1)*(-1)+(C7)*(3)*(3)+(3*C7)*(-1)*(-1))

42 D7=-LOG(C7)

H7=10^ (-0.51*(3)*(3)*SQRT(G7)/(1+(3.3*0.90*SQRT(G7)))

43 E7=(A$4/(27*H7*I7^ 3))^ (1/4)

I7=10^ (-0.51*(-1)*(-1)*SQRT(G7)/(1+(3.3*0.40*SQRT(G7)))

th


Chapter 10

10-17

1 2 3

A B Prob lem 10-17

C

D

E

F

4 Z 0.001 0.005 0.01 0.05 x 0.9 1 0.967 0.934 0.913 0.854 5 6 0.6 1 0.966 0.930 0.907 0.834 7 0.425 1 0.965 0.927 0.902 0.819 8 0.35 1 0.965 0.926 0.900 0.812 9 0.3 1 0.965 0.925 0.899 0.807 0.25 1 0.964 0.925 0.897 0.801 10 11 0.8 2 0.872 0.756 0.690 0.517 0.6 2 0.870 0.747 0.676 0.483 12 0.5 2 0.868 0.743 0.668 0.464 13 14 0.45 2 0.868 0.740 0.664 0.455 0.4 2 0.867 0.738 0.660 0.444 15 0.9 3 0.737 0.539 0.443 0.242 16 0.4 3 0.726 0.505 0.393 0.161 17 1.1 4 0.587 0.347 0.252 0.098 18 19 0.5 4 0.569 0.304 0.199 0.046 20 21 Spreadsheet Document ation 22 C5=10^(-0.51*$B5^2*SQRT(C$4)/(1+3.3*$A5*SQRT(C$4)))

G

0.1 0.826 0.796 0.773 0.762 0.754 0.745 0.445 0.401 0.377 0.364 0.351 0.178 0.095 0.063 0.020

SKOOG - SOLUCIONÁRIO CAPÍTULO 10.pdf

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