lms solutions electrochemistry

CHAPTER 777 Solutions of Electrolytes

LAIDLER . MEISER . SANCTUARY

Physical Chemistry Electronic Edition

Publisher: MCH Multimedia Inc.

Problems and Solutions

Chapter 7: Solutions of Electrolytes Faradays Laws, Molar Conductivity, and Weak Electrolytes

7-2

Chapter 7

*problems with an asterisk are slightly more demanding

Faradays Laws, Molar Conductivity, and Weak Electrolytes

7.1. A constant current was passed through a solution of cupric sulfate, CuSO4, for 1 h, and 0.040 g of copper was deposited. Calculate the current (atomic weight of Cu = 63.5).

Solution

7.2. After passage of a constant current for 45 min, 7.19 mg of silver (atomic weight = 107.9) was deposited from a solution of silver nitrate. Calculate the current.

Solution

7.3. Electrolysis of molten KBr generates bromine gas, which can be used in industrial bromination processes. How long will it take to convert a 500.00-kg batch of phenol (C6H5OH) to monobromophenol using a current of 20 000 A?

Solution

7.4. The following are the molar conductivities of chloroacetic acid in aqueous solution at 25 C and at various concentrations c:

410c

M 625 312.5 156.3 78.1 39.1 19.6 9.8

1 2 1cm mol

53.1 72.4 96.8 127.7 164.0 205.8 249.2

Plot against c. If = 362 1 cm2 mol1, are these values in accord with the Ostwald dilution law? What is the value of the dissociation constant? (See also Problem 7.11.)

Solution

7.5. The electrolytic conductivity of a saturated solution of silver chloride, AgCl, in pure water at 25 C is 1.26 106 1 cm1 higher than that for the water used. Calculate the solubility of AgCl in water if the molar ionic conductivities are Ag+, 61.9 1 cm2 mol1; Cl, 76.4 1 cm2 mol1.

Solution

Chapter 7: Solutions of Electrolytes Faradays Laws, Molar Conductivity, and Weak Electrolytes

7-3

*7.6. The electrolytic conductivity of a 0.001 M solution of Na2SO4 is 2.6 104 1 cm1. If the solution is saturated with CaSO4, the conductivity becomes 7.0 104 1 cm1. Calculate the solubility product for CaSO4 using the following molar conductivities at

these concentrations: (Na+) = 50.1 1 cm2 mol1; 2 1 2 11 Ca 59.5 cm mol .2

+ =

Solution

7.7. The quantity l/A of a conductance cell (see Eq. 7.8) is called the cell constant. Find the cell constant for a conductance cell in which the conductance, G, of a 0.100 M KCl solution is 0.01178 S at 25 C. The equivalent conductance for 0.100 M KCl at 25 C is 128.96 S cm2 mol1. If a 0.0500 M solution of an electrolyte has a measured conductance of 0.00824 S using this cell, what is the equivalent conductance of the electrolyte?

Solution

*7.8. A conductivity cell when standardized with 0.01 M KCl was found to have a resistance of 189 . With 0.01 M ammonia solution the resistance was 2460 . Calculate the base dissociation constant of ammonia, given the following molar conductivities at these concentrations: (K+) = 73.5 1 cm2 mol1; (Cl) = 76.4 1 cm2 mol1; 4(NH )

+ = 73.4 1 cm2 mol1; (OH) = 198.6 1 cm2 mol1.

Solution

7.9. The conductivity of a 0.0312 M solution of a weak base is 1.53 104 S cm1. If the sum of the limiting ionic conductances for BH+ and OH is 237.0 S cm2 mol1, what is the value of the base constant Kb?

Solution

7.10. The equivalent conductance of KBr solutions as a function of concentration at 25 C is given in the following table. By a linear regression analysis of suitable variables, find the value of for KBr.

c/103 M 0.25 0.36 0.50 0.75 1.00 1.60 2.00 5.00 10.00 /S cm2 mol1 150.16 149.87 149.55 149.12 148.78 148.02 147.64 145.47 143.15

Solution

Chapter 7: Solutions of Electrolytes Debye-Hckel Theory and Transport of Electrolytes

7-4

7.11. Equation 7.20 is one form of Ostwalds dilution law. Show how it can be linearized (i.e., convert it into a form that will allow experimental values of at various concentrations to be tested by means of a straight-line plot). Explain how and K can be obtained from the plot.

Kraus and Callis, J. Amer. Chem. Soc., 45, 2624(1923), obtained the following electrolytic conductivities for the dissociation of tetramethyl tin chloride, (CH3)4SnCl, in ethyl alcohol solution at 25.0 C and at various concentrations c:

c/104 mol dm3 1.566 2.600 6.219 10.441 /106 1 cm1 1.788 2.418 4.009 5.336

By the use of the linear plot you have devised, determine and K.

Solution

7.12. A certain chemical company wishes to dispose of its acetic acid waste into a local river by first diluting it with water to meet the regulation that the total acetic acid concentration cannot exceed 1500 ppm by weight. You are asked to design a system using conductance to continuously monitor the acid concentration in the water and trigger an alarm if the 1500 ppm limit is exceeded. What is the maximum conductance at which the system should trigger an alarm at a constant temperature of 25 C? (Assume that the cell constant is 1.0 cm1 and that the density of 1500 ppm acetic acid solution is not appreciably different from that of pure water. The for acetic acid is 390.7 S cm2 mol1 and Ka = 1.81 105 mol dm3 at 25 C. Ignore the conductance of water.)

Solution

7.13. How far can the conductivity of water at 25 C be lowered in theory by removing impurities? The (in S cm2 mol1) for KOH, HCl, and KCl are, respectively, 274.4, 426.04, and 149.86. Kw = 1.008 1014. Compare your answer to the experimental value of 5.8 108 S cm1 obtained by Kohlrausch and Heydweiller, Z. phys. Chem. 14, 317(1894).

Solution

Debye-Hckel Theory and Transport of Electrolytes

7.14. The radius of the ionic atmosphere (1/) for a univalent electrolyte is 0.964 nm at a concentration of 0.10 M in water at 25 C ( 78). = Estimate the radius (a) in water at a concentration of 0.0001 M and (b) in a solvent of 38 = at a concentration of 0.10 M.

Solution


7-5

7.15. The molar conductivities of 0.001 M solutions of potassium chloride, sodium chloride, and potassium sulfate 2 41 K SO2

are 149.9,

126.5, and 153.3 1 cm2 mol1, respectively. Calculate an approximate value for the molar conductivity of a solution of sodium sulfate of the same concentration.

Solution

7.16. The molar conductivity at 18 C of a 0.0100 M aqueous solution of ammonia is 9.6 1 cm2 mol1. For NH4Cl, = 129.8 1 cm2 mol1 and the molar ionic conductivities of OH and Cl are 174.0 and 65.6 1 cm2 mol1, respectively. Calculate for NH3 and the degree of ionization in 0.01 M solution.

Solution

7.17. A solution of LiCl was electrolyzed in a Hittorf cell. After a current of 0.79 A had been passed for 2 h, the mass of LiCl in the anode compartment had decreased by 0.793 g.

a. Calculate the transport numbers of the Li+ and Cl ions.

b. If (LiCl) is 115.0 1 cm2 mol1, what are the molar ionic conductivities and the ionic mobilities?

Solution

7.18. A solution of cadmium iodide, CdI2, having a molality of 7.545 103 mol kg1, was electrolyzed in a Hittorf cell. The mass of cadmium deposited at the cathode was 0.03462 g. Solution weighing 152.64 g was withdrawn from the anode compartment and was found to contain 0.3718 g of cadmium iodide. Calculate the transport numbers of Cd2+ and I.

Solution

7.19. The transport numbers for HCl at infinite dilution are estimated to be t+ = 0.821 and t = 0.179 and the molar conductivity is 426.16 1 cm2 mol1. Calculate the mobilities of the hydrogen and chloride ions.

Solution

7.20. If a potential gradient of 100 V cm1 is applied to a 0.01 M solution of NaCl, what are the speeds of the Na+ and Cl ions? Take the ionic conductivities to be those listed in Table 7.3 on p. 291.

Solution


7-6

*7.21. A solution of LiCl at a concentration of 0.01 M is contained in a tube having a cross-sectional area of 5 cm2. Calculate the speeds of the Li+ and Cl ions if a current of 1 A is passed. Use the ion conductivities listed in Table 7.3.

Solution

7.22. What is the work required to separate in vacuum two particles, one with the charge of the proton, from another particle with the same charge of opposite sign? Carry out the calculations for an initial distance of (a) 1.0 nm to an infinite distance apart and (b) from 1.0 mm to an infinite distance apart. (c) In (a) how much work would be required if the charge is moved to a distance of 0.1 m? The charge on a proton is 1.6 1019 C.

Solution

*7.23. According to Bjerrums theory of ion association, the number of ions of type i present in a spherical shell of thickness dr and distance r from a central ion is

2 20 Bexp( /4 ) 4i i i cdN N z z e rk T r dr =

where zi and zc are the charge numbers of the ion of type i and of the central ion and e, 0 , , and kB have their usual significance. Plot the exponential in this expression and also 4r2 against r for a uni-univalent electrolyte in water at 25.0 C ( 78.3). = Allow r to have values from 0 to 1 nm. Plot also the product of these functions, which is (dN1/N1)dr and is the probability of finding an ion of type i at a distance between r and r + dr of the central ion.

By differentiation, obtain a value r* for which the probability is a minimum, and calculate the value for water at 25.0 C. The electrostatic potential is given to a good approximation by the first term in Eq. 7.47 on p. 280. Obtain an expression, in terms of kBT, for the electrostatic energy between the two univalent ions at this minimum distance, and evaluate this energy at 25 C.

Solution

Chapter 7: Solutions of Electrolytes Thermodynamics of Ions

7-7

Thermodynamics of Ions

7.24. The following are some conventional standard enthalpies of ions in aqueous solution at 25 C:

Ion fH/kJ mol1

H+ 0

Na+ 239.7

Ca2+ 543.1

Zn2+ 152.3

Cl 167.4

Br 120.9

Calculate the enthalpy of formation in aqueous solution of 1 mol of NaCl, CaCl2, and ZnBr2, assuming complete dissociation.

Solution

7.25. One estimate for the absolute Gibbs energy of hydration of the H+ ion in aqueous solution is 1051.4 kJ mol1. On this basis, calculate the absolute Gibbs energies of hydration of the following ions, whose conventional standard Gibbs energies of hydration are as follows:

Ion hydG k J mol1

H+ 0

Na+ 679.1

Mg2+ 274.1

Al3+ 1346.4

Cl 1407.1

Br 1393.3

Solution


7-8

7.26. Calculate the ionic strengths of 0.1 M solutions of KNO3, K2SO4, ZnSO4, ZnCl2, and K4Fe(CN)6; assume complete dissociation and neglect hydrolysis.

Solution

7.27. Calculate the mean activity coefficient for the Ba2+ and 24SO ions in a saturated solution of BaSO4 (Ksp = 9.2 1011 mol2 dm6)

in 0.2 M K2SO4, assuming the Debye-Hckel limiting law to apply.

Solution

7.28. The solubility of AgCl in water at 25 C is 1.274 105 mol dm3. On the assumption t hat the Debye-Hckel limiting law applies,

a. Calculate G for the process AgCl(s) Ag+(aq) + Cl(aq).

b. Calculate the solubility of AgCl in an 0.005 M solution of K2SO4.

Solution

7.29. Employ Eq. 7.114 to make plots of log against I for a uni-univalent electrolyte in water at 25 C, with B = 0.51 mol1 dm3/2 and B = 0.33 1010 mol1 dm3/2 m1, and for the following values of the interionic distance a:

a = 0, 0.1, 0.2, 0.4, and 0.8 nm

Solution

7.30. Estimate the change in Gibbs energy G when 1 mol of K+ ions (radius 0.133 nm) is transported from aqueous solution ( = 78) to the lipid environment of a cell membrane ( = 4) at 25 C.

Solution

7.31. At 18 C the electrolytic conductivity of a saturated solution of CaF2 is 3.86 105 1 cm1, and that of pure water is 1.5 106 1 cm1.

The molar ionic conductivities of 21 Ca2

+ and F are 51.1 1 cm2 mol1 and 47.0 1 cm2 mol1, respectively. Calculate the solubility of

CaF2 in pure water at 18 C and the solubility product.

Solution


7-9

7.32. What concentrations of the following have the same ionic strength as 0.1 M NaCl?

CuSO4, Ni(NO3)2, Al2(SO4)3, Na3PO4 Assume complete dissociation and neglect hydrolysis.

Solution

7.33. The solubility product of PbF2 at 25.0 C is 4.0 109 mol3 dm9. Assuming the Debye-Hckel limiting law to apply, calculate the solubility of PbF2 in (a) pure water and (b) 0.01 M NaF.

Solution

7.34. Calculate the solubility of silver acetate in water at 25 C, assuming the DHLL to apply; the solubility product is 4.0 103 mol2 dm6.

Solution

*7.35. Problem 7.30 was concerned with the Gibbs energy change when 1 mol of K+ ions are transported from water to a lipid. Estimate the electrostatic contribution to the entropy change when this occurs, assuming the dielectric constant of the lipid to be temperature independent, and the following values for water at 25 C: 78; = ln /T = 0.0046 K1. Suggest a qualitative explanation for the sign of the value you obtain.

Solution

*7.36. Assuming the Born equation (Eq. 7.86) to apply, make an estimate of the reversible work of charging 1 mol of Na+Cl in aqueous solution at 25 C ( = 78), under the following conditions:

a. The electrolyte is present at infinite dilution.

b. The electrolyte is present at such a concentration that the mean activity coefficient is 0.70.

The ionic radii are 95 pm for Na+ and 181 pm for Cl.

Solution

7.37. If the solubility product of barium sulfate is 9.2 1011 mol2 dm6, calculate the solubility of BaSO4 in a solution that is 0.10 M in NaNO3 and 0.20 M in Zn(NO3)2; assume the DHLL to apply.

Solution

Chapter 7: Solutions of Electrolytes Ionic Equilibria

7-10

7.38. Silver chloride, AgCl, is found to have a solubility of 1.561 105 M in a solution that is 0.01 M in K2SO4. Assume the DHLL to apply and calculate the solubility in pure water.

Solution

7.39. The enthalpy of neutralization of a strong acid by a strong base, corresponding to the process

H+(aq) + OH(aq) H2O

is 55.90 kJ mol1. The enthalpy of neutralization of HCN by NaOH is 12.13 kJ mol1. Make an estimate of the enthalpy of dissociation of HCN.

Solution

7.40. Make use of the Debye-Hckel limiting law to estimate the activity coefficients of the ions in an aqueous 0.004 M solution of sodium sulfate at 298 K. Estimate also the mean activity coefficient.

Solution

Ionic Equilibria

7.41. A 0.1 M solution of sodium palmitate, C15H31COONa, is separated from a 0.2 M solution of sodium chloride by a membrane that is permeable to Na+ and Cl ions but not to palmitate ions. Calculate the concentrations of Na+ and Cl ions on the two sides of the membrane after equilibrium has become established. (For a calculation of the Nernst potential, see Problem 8.18.)

Solution

7.42. Consider the ionizations

3 2 3 2 2 2H H N CH COO H N CH COOH H NCH COOH H+ + + ++ +

Assume that the following acid dissociation constants apply to the ionizations:

103 2 a

3a

NH NH H ; 1.5 10

COOH COO H ; 4.0 10

K MK M

+ +

+

+ =

+ =

Estimate a value for the equilibrium constant for the process

3 3 2 2H N CH COO H NCH COOH+

Solution

Chapter 7: Solutions of Electrolytes Essay Questions

7-11

7.43. The pK values for the successive ionizations of phosphoric acid are given on p. 308. Which of the four species is predominant at the following values of the hydrogen or hydroxide concentration?

a. [H+] = 0.1 M. b. [H+] = 2 103 M. c. [H+] = 5 105 M. d. [OH] = 2 103 M. e. [OH] = 1 M.

Solution

7.44. Two solutions of equal volume are separated by a membrane which is permeable to K+ and Cl ions but not to P ions. The initial concentrations are as shown below.

[K+] = 0.05 M [K+] = 0.15 M [Cl] = 0.05 M [P] = 0.15 M

Calculate the concentrations on each side of the membrane after equilibrium has become established. (See Problem 8.26 in Chapter 8 for the calculation of the Nernst potential for this system.)

Solution

Essay Questions

7.45. State Faradays two laws of electrolysis and discuss their significance in connection with the electrical nature of matter.

7.46. Discuss the main ideas that lie behind the Debye-Hckel theory, as applied to the conductivities of solutions of strong electrolytes.

7.47. Outline two important methods for determining transport numbers of ions.

7.48. Explain why Li+ has a lower ionic conductivity than Na+ and why the value for H+ is so much higher than the values for both of these ions.

7.49. Describe briefly the type of hydration found with the following ions in aqueous solution: Li+, Br, H+, OH.

7.50. What modifications to the Debye-Hckel limiting law are required to explain the influence of ionic strength on solubilities?

Chapter 7: Solutions of Electrolytes Solutions

7-12

Solutions

7.1. A constant current was passed through a solution of cupric sulfate, CuSO4, for 1 h, and 0.040 g of copper was deposited. Calculate the current (atomic weight of Cu = 63.5 g mol-1).

Solution:

Given: 11 h 3600 s, 0.040 g, 63.5 g molt m M = = = =

Required: I

To solve this problem we must use Eq. 7.6 and apply Faradays Laws of Electrolysis. Eq. 7.6 is given by,

Q It=

Rearranging to solve for I we obtain,

QIt

=

where Q, is the quantity of electricity.

Q is proportional to the mass of the element produced at the electrode. Faradays constant, given by the symbol F, relates the amount of substance deposited to the quantity of electricity, Q, passed through the solution. The charge carried by 1 mol of ions bearing z unit charges is zF, where1 F = 96 485 C mol-1.

Hence,

mQ zFM

=

Using the expression above, we can solve for the current through the solution.

zF mIt M =

The charge on copper in CuSO4 is Cu2+, therefore z = 2


7-13

12 96 485 C molI

=0.040 g

3600 s 63.5 g 1mol

1

1

0.033 765 529 3 C swhere1C s 1 A

33.8 mA

I

I

=

=

=

Back to Problem 7.1 Back to Top


7-14

7.2. After passage of a constant current for 45 min, 7.19 mg of silver (atomic weight = 107.9 g mol-1) was deposited from a solution of silver nitrate. Calculate the current.

Solution:

Given: 145 min 2700 s, 7.19 mg 0.00719 g, 107.9 g molt m M = = = = =

Required: I

This problem can be solved in a similar manner as problem 7.1, using the expression for current as, zF mIt M =

.

The charge on silver in AgNO3 is Ag+, therefore z = 1

11 96 485 C molI

=0.00719 g

2700 s 107.9 g 1mol

1

1

0.002 381 2417 C swhere1C s 1 A

2.4 mA

I

I

=

=

=



7-15

7.3. Electrolysis of molten KBr generates bromine gas, which can be used in industrial bromination processes. How long will it take to convert a 500.00-kg batch of phenol (C6H5OH) to monobromophenol using a current of 20 000 A?

Solution:

Given: 500.00 kg, 20 000 Am I= =

Required: t

To solve this problem, we must first outline the chemical reactions that are taking place.

( )( )( )

2

6 5 2 6 4

1 2Br Br (g) 2

2 2K 2 2K(s)

3 C H OH Br (g) C H (Br)OH HBr

e

e

+

+

+

+ +

Two moles of electrons are involved in the generation of each mole of bromine gas, which reacts with one mole of phenol, therefore z = 2.

Each batch consists of 500.00 kg of phenol therefore converting to the number of moles and we can determine the time required to convert all 500.00 kg of phenol into monobromophenol.

Using Eq. 7.6 and mQ zFM

=

from problem 6.1, we can solve for t,


7-16

( ) ( )6 56 5

1 1 1C H OH

1C H OH

1

6 12.011g mol 6 1.007 94 g mol 15.9994 g mol

94.11304 g mol

2 96 485 C mol

QtIzF mtI M

M

M

t

=

=

= + +

=

=

3500.00 10 g20 000 A

94.113 04 g 1mol

1where1C s 1 A1 h51 260.165 44 s

3600 s14.238 934 84 h

14.239 h

t

t

t

=

=

=

=



7-17

7.4. The following are the molar conductivities of chloroacetic acid in aqueous solution at 25 C and at various concentrations c:

410c

M 625 312.5 156.3 78.1 39.1 19.6 9.8

1 2 1cm mol

53.1 72.4 96.8 127.7 164.0 205.8 249.2

Plot against c. If = 362 1 cm2 mol1, are these values in accord with the Ostwald Dilution Law? What is the value of the dissociation constant? (See also Problem 7.11.)

Solution:

Given: 1 2 1, , 362 cm molc =

Required: plot of against c, are these values in accord with the Ostwald Dilution Law, K

Using the data above, we can create the following plot of against c.


7-18

The Ostwald dilution law is given by Eq. 7.20 as:

2( / )1 ( / )cK

=

To determine if the data given above follows the Ostwald Dilution Law, we see if we can calculate a fixed value for K, the dissociation constant of the solution.

The results are plotted in the table below.

c

10-4 M

1 cm2 mol1

K

M

625 53.1 0.001575951

312.5 72.4 0.0015625

156.3 96.8 0.001525554

78.1 127.7 0.001501592

39.1 164 0.001467205

19.6 205.8 0.001468105

9.8 249.2 0.001490406

Since the values of K are reasonably constant, we can say that data given above follows the Ostwald Dilution Law.

The value of the dissociation constant, K, can be calculated from the average of the K values obtained above.

average

3average

0.001513045

1.5 10

K M

K M=

=



7-19

7.5. The electrolytic conductivity of a saturated solution of silver chloride, AgCl, in pure water at 25 C is 1.26 106 1 cm1 higher than that for the water used. Calculate the solubility of AgCl in water if the molar ionic conductivities are Ag+, 61.9 1 cm2 mol1; Cl, 76.4 1 cm2 mol1.

Solution:

Given: + -6 1 1 1 2 1 1 2 1Ag Cl1.26 10 cm , 61.9 cm mol , 76.4 cm mol = = =

Required: solubility

The expression for molar conductivity is given by Eq. 7.9.

c

=

In order to solve this problem we can use the concentration as a measure of solubility.

+ -

+ -

AgCl

AgCl Ag Cl

Ag Cl

6 1

solubility

solubility

1.26 10 solubility

=

= +

=

+

=

1

1

cm61.9 2 1 1 cm mol 76.4 + 2 1

9 3

6 3

6 3

cm molsolubility 9.110 629 067 10 mol cmsolubility 9.110 629 067 10 mol dm

solubility 9.11 10 mol dm

=

=

=



7-20

7.6. The electrolytic conductivity of a 0.001 M solution of Na2SO4 is 2.6 104 1 cm1. If the solution is saturated with CaSO4, the conductivity becomes 7.0 104 1 cm1. Calculate the solubility product for CaSO4 using the following molar conductivities at these concentrations:

(Na+) = 50.1 1 cm2 mol1; 2 1 2 11 Ca 59.5 cm mol .2

+ =

Solution:

Given: 2 4

2 4 4

4 1 1 4 1 1Na SO 1 1Na SO CaSO

2 2

0.001 , 2.6 10 cm , 7.0 10 cmc M = = =

( ) 1 2 1 2 1 2 11Na 50.1 cm mol , Ca 59.5 cm mol2 + + = =

Required: 4for CaSOsK

To determine the base dissociation constant for CaSO4, we must first realize which chemical reactions are taking place.

( )( )

22 4 4

2 24 4

1 Na SO 2Na SO

2 CaSO Ca SO

+

+

+

+

The solubility product is therefore given by:

2 24Ca SOsK

+ = ,

To determine the concentrations of the species in the solution, we will determine the electrolytic and molar conductivities.

We are given that the initial electrolytic conductivity of the Na2SO4 solution is 2.6 104 1 cm1 which is raised to 7.0 104 1 cm1 upon saturation with CaSO4.This means the increase in electrolytic conductivity is:

4 1 1 4 1 1

4 1 1

7.0 10 cm 2.6 10 cm4.4 10 cm

=

=

The molar conductivity of the CaSO4 solution can be calculated using Eq. 7.9.


7-21

41 CaSO2 2

c

c

=

=

where c is the concentration of CaSO4 and 2c is the concentration of CaSO4.

The molar conductivity of the Na2SO4 solution is:

2 4

2 4

2 4

1 Na SO2

1 Na SO2

4 1 1

1 Na SO2

2

2.6 10 cm

c

=

=

3 32 0.001 10 mol cm ( )2 4

1 2 11 Na SO2

130 cm mol =

The molar conductivity of the Na2SO4 is the same as,

22 4 4

1 1NaNa SO SO2 2

+ = +

Therefore we can solve for 2

41SO2

to calculate the molar conductivity of the CaSO4.


7-22

24 2 4

24

24

2 24 4

4

4

1 1 NaSO Na SO2 2

1 2 1 1 2 11SO2

1 2 11SO2

1 1CaCaSO SO2 2

1 2 1 1 2 11 CaSO2

1 2 11 CaSO2

130 cm mol 50.1 cm mol

79.9 cm mol

59.5 cm mol 79.9 cm mol

139.4 cm mol

+

+

=

=

=

= +

= +

=

The concentration of CaSO4 is therefore given by:

41 CaSO2

4 1

2

4.4 10

c

c

=

=

1

1

cm

2 139.4

( )2 16 3

3 3

cm mol

1.578192 253 10 mol cm1.578192 253 10 mol dm

cc

=

=

Solving for Ks,

2

2 3 34

Ca

SO 1.0 10 mol dm

c

c

+

= = +

The concentration of SO42- is influenced by the contributions of CaSO4 and Na2SO4.


7-23

( )3 36 2 6

6 2 6

1.0 10 mol dm

4.068 883 038 10 mol dm

4.07 10 mol dm

s

s

s

K c c

K

K

= +

=

=



7-24

7.7. The quantity l/A of a conductance cell (see Eq. 7.8) is called the cell constant. Find the cell constant for a conductance cell in which the conductance, G, of a 0.100 M KCl solution is 0.01178 S at 25 C. The equivalent conductance for 0.100 M KCl at 25 C is 128.96 S cm2 mol1. If a 0.0500 M solution of an electrolyte has a measured conductance of 0.00824 S using this cell, what is the equivalent conductance of the electrolyte?

Solution:

Given: 2 1 KCl0.01178 S, 25 C, 128.96 S cm mol at 0.100G T c M= = = =

electrolyte0.00824 S, 0.0500G c M= =

Required: l/A, electrolyte

In order to determine the cell constant, we must first calculate the value for the electrolytic conductance. As we know, this can be obtained using Eq. 7.9.

30.100 10 mol

cc

=

=

= ( )3 2 1cm 128.96 S cm mol ( )10.012896 S cm =

We can now use Eq. 7.8 to solve for the cell constant, l/A.


7-25

(conductance)

0.012896 S

AGl

lA GlA

=

=

=1 cm

0.01178 S

1

1

1.094 736 842 cm

1.09 cm

lAlA

=

=

The equivalent conductance of the electrolyte in this same cell can be found using the cell constant calculated above and solve for from Eq. 7.9.

clGA

=

=

( )

electrolyteelectrolyte

1

electrolyte

0.00824 S 1.094 736 842 cm

lGA

c

=

=( )

3 30.0500 10 mol cm 2 1

electrolyte

2 1electrolyte

180.412 6316 S cm mol

180 S cm mol

=

=



7-26

7.8. A conductivity cell when standardized with 0.01 M KCl was found to have a resistance of 189 . With 0.01 M ammonia solution the resistance was 2460 . Calculate the base dissociation constant of ammonia, given the following molar conductivities at these concentrations: (K+) = 73.5 1 cm2 mol1; (Cl) = 76.4 1 cm2 mol1; 4(NH )

+ = 73.4 1 cm2 mol1; (OH) = 198.6 1 cm2 mol1.

Solution:

Given: 3KCl NH

0.01 , 189 , 0.01 , 2460c M R c M R= = = =

( ) ( )1 2 1 1 2 1K 73.5 cm mol , Cl 76.4 cm mol , + = = ( ) ( )1 2 1 1 2 14NH 73.4 cm mol , OH 198.6 cm mol + = =

Required: Kb

In order to determine the base dissociation constant for ammonia, we must first outline which chemical reactions are taking place.

( )( )

3 2 4

4 4

1 NH H O NH OH

2 NH OH NH OHbK + +

+

The base dissociation constant is therefore given by:

[ ]

4

4

NH OHNH OHb

K+ =

To solve for the concentrations of each species, we may begin by calculating the value for the electrolytic conductance of the standard KCl

solution in the cell.

+ -KCl K Cl1 2 1 1 2 1

KCl1 2 1

KCl

3KCl


149.9 cm mol

0.01 10 mol

cc

=

= = +

= +

=

= ( )3 1 2 1cm 149.9 cm mol ( )1 1

KCl 0.001 499 cm=


7-27

Recall that the resistance is inversely proportional to the conductance. The electrolytic conductivity of the ammonia solution is therefore,

3

3

3

KClNH KCl

NH

1 1NH

1890.001 499 cm

RR

=

=

2460

3

4 1 1NH 1.151670 732 10 cm

=

The molar conductivity of 4NH OH+ + is:

+ -4 4

4

4

NH OH NH OH

1 2 1 1 2 1NH OH

1 2 1NH OH


272 cm mol

= +

= +

=

Using Eq. 7.9, we can calculate the concentrations of 4NH and OH+ .

3

4

NH

NH OH

4 11.151670 732 10

c

c

c

=

=

=

1

1

cm272 2 1

7 3

4 3

cm mol4.234 083 572 10 mol cm4.234 083 572 10 mol dm

cc

=

=

Knowing that; 4NH OHc+ = = , we can solve for Kb.

4NH OH 4NH+ + OH

Cinitial 0.01 0 0 mol dm-3 Cequilibrium 0.01 c c c mol dm-3


7-28

[ ]

( )

24

4

24 3

4 3

NH OHNH OH 0.01

4.234 083 572 10 mol dm

0.01 4.234 083 572 10 mol dm

b

b

cKc

K

+

= =

=

5 3

5 3

1.872 008 786 10 mol dm

1.9 10 mol dmb

b

K

K

=

=



7-29

7.9. The conductivity of a 0.0312 M solution of a weak base is 1.53 104 S cm1. If the sum of the limiting ionic conductances for BH+ and OH is 237.0 S cm2 mol1, what is the value of the base constant Kb?

Solution:

Given: 4 1 2 10.0312 M, 1.53 10 S cm , 237.0 S cm molc = = =

Required: Kb

In order to determine the base dissociation constant for the solution, we must write out the chemical reaction that is taking place.

2B H O BH OHbK + + +

Since we are told we have a weak base, it is possible to apply Ostwalds Dilution Law and introduce the degree of dissociation, , given by Eq. 7.11.

=

B + 2H O BH

+ + OH

( )1c

c c

The base dissociation constant is therefore given by Eq. 7.18.

2

1bcK

=

And the degree of dissociation is determined by calculating the molar conductivity of 2B H O+ using Eq. 7.9.

4 11.53 10 S cmc

=

=

3 30.0312 10 mol cm 2 14.903 846154 S cm mol =


7-30

The degree of dissociation is therefore,

4.903 846154 S =2 cm 1mol

237.0 S 2 cm 1mol0.020 691334 =

Solving for Kb, we obtain the following:

( )( )23

5 3

5 3

0.0312 mol dm 0.020 691334

1 0.020 6913341.363 992 486 10 mol dm

1.36 10 mol dm

b

b

b

K

K

K

=

=

=



7-31

7.10. The equivalent conductance of KBr solutions as a function of concentration at 25 C is given in the following table. By a linear regression analysis of suitable variables, find the value of for KBr.

c/103 M 0.25 0.36 0.50 0.75 1.00 1.60 2.00 5.00 10.00 /S cm2 mol1 150.16 149.87 149.55 149.12 148.78 148.02 147.64 145.47 143.15

Solution:

Given: 25 CT = , data given above

Required: KBr

The relationship between and c is given by the Debye-Hckel-Onsager Equation, Eq. 7.53.

( )P Q c = +

In order to solve for KBr , we can plot c against , where 310c c =

( )c M ( )2 1 S cm mol 0.01581139 150.16

0.01897367 149.87

0.02236068 149.55

0.02738613 149.12

0.03162278 148.78

0.04 148.02

0.04472136 147.64

0.07071068 145.47


7-32

0.1 143.15

From the linear regression, the y- intercept will be the value of KBr

2 1KBr

2 1KBr

151.41268 S cm mol

151.41S cm mol

=

=



7-33

7.11. Equation 7.20 is one form of Ostwalds dilution law. Show how it can be linearized (i.e., convert it into a form that will allow experimental values of at various concentrations to be tested by means of a straight-line plot). Explain how and K can be obtained from the plot.

Kraus and Callis, J. Amer. Chem. Soc., 45, 2624(1923), obtained the following electrolytic conductivities for the dissociation of tetramethyl tin chloride, (CH3)4SnCl, in ethyl alcohol solution at 25.0 C and at various concentrations c:

c/104 mol dm3 1.566 2.600 6.219 10.441 /106 1 cm1 1.788 2.418 4.009 5.336

By the use of the linear plot you have devised, determine and K.

Solution:

Given: 25 CT = , data above

Required: , K

Ostwalds dilution law, given by Eq. 7.20 can be linearized in the following manner:


7-34

( )( ) ( )( )

( )( )( )

2

2

2

2

2

2

2

2

( / )1 ( / )

1 / /

/

/

/

cK

K c

K Kc

K Kc

K Kc

Kc K

=

=

=

=

=

=

From here, we can plot c against 1

and determine and K.

We can calculate from Eq. 7.9 at each concentration given. c

= . This leads to a table of values with the following:

c

104 mol dm3

106 1 cm1

1 cm2 mol-1

c

-1 cm-1

1/

cm

-2 mol

1.566 1.788 11.417625 1.78810-6 0.087584

2.6 2.418 9.3 2.41810-6 0.107527

6.219 4.009 6.446374 4.00910-6 0.155126

10.441 5.336 5.1106216 5.33610-6 0.195671

Now we obtain the following graph:


7-35

From the linear regression, the y- intercept will be K and the slope will be 2K


7-36

1 1

2 2 1

2

2 1

1 1

1 2 1

1 2 1

1 1

1

0.00111 cm0.03294 cm mol

0.03294 cm mol0.00111 cm

29.675 675 68 cm mol

30 cm mol

0.00111 cm

0.00111

KK

KK

K

K

=

=

=

=

=

=

=

=

1

1

cm29.675 675 68

2 1

5 3

2 3

cm mol3.740 437158 10 mol cm

3.7 10 mol dm

K

K

=

=



7-37

7.12. A certain chemical company wishes to dispose of its acetic acid waste into a local river by first diluting it with water to meet the regulation that the total acetic acid concentration cannot exceed 1500 ppm by weight. You are asked to design a system using conductance to continuously monitor the acid concentration in the water and trigger an alarm if the 1500 ppm limit is exceeded. What is the maximum conductance at which the system should trigger an alarm at a constant temperature of 25 C? (Assume that the cell constant is 1.0 cm1 and that the density of 1500 ppm acetic acid solution is not appreciably different from that of pure water. The for acetic acid is 390.7 S cm2 mol1 and Ka = 1.81 105 mol dm3 at 25 C. Ignore the conductance of water.)

Solution:

Given: 1 2 1 5 3max acetic acid water a1500 ppm, 1.0 cm , , 390.7 S cm mol , 1.81 10 mol dm

Ac Kl

= = = =

25 CT =

Required: G

In order to solve this problem, we must first convert the concentration from parts per million to SI units.

( ) ( ) ( )

6

1 1 1acetic acid

1acetic acid

1500 g acid1500 ppm10 g solution

1.500 g acid1000 g solution

2 12.011g mol 4 1.007 94 g mol 2 15.9994 g mol

60.052 56 g mol

1.500 g

c

c

M

M

c

= =

=

= + +

=

=acid

60.052 56 g 1

1

11 kg solutionmol

0.024 978119 2 mol kgc

=

Since the solution has the same density as water, 1.00 kg of solution has a volume of 1.0 dm3. Therefore we can assume the solution has concentration, 0.0249781192 M.c =

The conductance of a solution is given by Eq. 7.8.


7-38

(conductance) AGl

=

Since acetic acid is a weak acid, we may we use the Ostwalds Dilution Law, Eq. 7.20, to solve for .

( ) ( )( ) ( )

2

2

2

( / )1 ( / )

/ /

/ / 0

cK

K K c

K K c

=

=

=

To solve for / , we use the quadratic equation.

( )( )

2

2

2

42

4/

2

4/2

b b acxa

K K c Kc

K K cKc

=

=

+ =

( ) ( )( )( )

25 5 51.81 10 1.81 10 4 0.0249781192 1.81 10 /

2 0.0249781192

/ 0.026 939 5691and 0.026 903 3691

M M M M

M

+ =

=

We will disregard the negative value and take / 0.026 939 5691 = to solve for .

( )( )( )( )2 1

2 1

0.026 939 5691

0.026 939 5691 390.7 S cm mol

10.525 289 64 S cm mol

=

=

=


7-39

Using Eq. 7.9, we can substitute for the value of and determine the conductance of the solution.

( )

0.024 978119 2 mol

cc

AGl

AG cl

G

=

=

=

=

=( )3 2dm 10.525 289 64 S cm 1mol( ) 11.0 cm( )

3

3 3

4 2

4 2

0.262 901938 9 dm S cm0.262 901938 9 10 cm S cm2.629 019 389 10 S cm

2.63 10 S cm

GGG

G

=

=

=

=



7-40

7.13. How far can the conductivity of water at 25 C be lowered in theory by removing impurities? The (in S cm2 mol1) for KOH, HCl, and KCl are, respectively, 274.4, 426.04, and 149.86. Kw = 1.008 1014. Compare your answer to the experimental value of 5.8 108 S cm1 obtained by Kohlrausch and Heydweiller, Z. phys. Chem. 14, 317(1894).

Solution:

Given: 2 1 2 1 2 1KOH HCl KCl274.4 S cm mol , 426.04 S cm mol , 149.86 S cm mol = = =

14 8 1w exp,25 C, 1.008 10 5.8 10 S cmT K = = =

Required:

The dissociations of each salt in water are given by:

( )( )( )

1 KOH K OH

2 HCl H Cl

3 KCl K Cl

+

+

+

+

+

+

By rearranging we find that,

2

2

2

H O KOH HCl KCl

2 1 2 1 2 1H O

2 1H O

274.4 S cm mol 426.04 S cm mol 149.86 S cm mol

550.58 S cm mol

= +

= +

=

In pure water, the only species that conduct electricity are H and OH+ ions. According to H OHwK+ = , each have a concentration of

;

14 3 7 31.008 10 mol dm 1.003 999 203 2 10 mol dmwK = = .

Since this concentration is very low, we can assume that 2 2H O H O

.


7-41

101.003 999 203 2 10 mol

c

=

= ( )3 2 1cm 550.58 S cm mol ( )8 1

8 1

5.527 779 329 10 S cm

5.528 10 S cm

=

=

Compared to the experimental value of 8 15.8 10 S cm , the conductivity determined through this process produces a very similar result.



7-42

7.14. The radius of the ionic atmosphere (1/) for a univalent electrolyte is 0.964 nm at a concentration of 0.10 M in water at 25 C ( 78). = Estimate the radius (a) in water at a concentration of 0.0001 M and (b) in a solvent of 38 = at a concentration of 0.10 M.

Solution:

Given: 1 0.964 nm, 0.10 M, 25 C, 78c T= = = =

Required: see above

a) Eq. 7.50 indicates that the thickness of the ionic atmosphere is inversely proportional to the square root of the concentration. 1/2

0 B2 2

1

i ii

k Te c z L

=

1 1c

Therefore the radius in water, where the electrolyte has c = 0.0001 M, can be obtained from the ratio of proportions.


7-43

1 2

2 1

11

2 2

1

2 12

2

1 1

11

1 1

1 0.1

c c

c

c

cc

M

=

=

=

= 0.0001 M

( )

2

2

0.964 nm

1 30.484 356 64 nm

1 30.5 nm

=

=

b) Similarly, we see from Eq. 7.50 that the thickness of the ionic atmosphere is proportional to the square root of the permittivity 1

The radius in water where 38 = , can be obtained from the ratio of proportions.


7-44

( )

2

2 11

2

2

2

1 1

1 38 0.964 nm78

1 0.672 855 072 6 nm

1 0.673 nm

=

=

=

=



7-45

7.15. The molar conductivities of 0.001 M solutions of potassium chloride, sodium chloride, and potassium sulfate 2 41 K SO2

are 149.9,

126.5, and 153.3 1 cm2 mol1, respectively. Calculate an approximate value for the molar conductivity of a solution of sodium sulfate of the same concentration.

Solution:

Given: 0.001 Mc =

2 4

1 2 1 1 2 1 1 2 1KCl NaCl 1 K SO

2

149.9 cm mol , 126.5 cm mol , 153.3 cm mol = = =

Required:2 4

1 Na SO2

The molar conductivity of 2 41 Na SO2

is given by the combination of the molar conductivities of each salt. We must also subtract the molar

conductivity of KCl since we are considering the solution containing only sodium and sulfate ions.

2 4 2 4

2 4

2 4

1 NaCl 1 KClNa SO K SO2 2

1 2 1 1 2 1 1 2 11 Na SO2

1 2 11 Na SO2

126.5 cm mol 153.3 cm mol 149.9 cm mol

129.9 cm mol

= +

= +

=



7-46

7.16. The molar conductivity at 18 C of a 0.0100 M aqueous solution of ammonia is 9.6 1 cm2 mol1. For NH4Cl, = 129.8 1 cm2 mol1 and the molar ionic conductivities of OH and Cl are 174.0 and 65.6 1 cm2 mol1, respectively. Calculate for NH3 and the degree of ionization in 0.01 M solution.

Solution:

Given: 3 3 4

1 2 1 1 2 1NH NH NH Cl18 C, 0.0100 M, 9.6 cm mol , 129.8 cm mol ,T c

= = = =

1 2 1 1 2 1OH Cl

174.0 cm mol , 65.6 cm mol , 0.01 Mc = = =

Required:3NH,

In solution, ammonia reacts in following manner,

3 2 4

4 4

NH H O NH OH

NH OH NH OH+ +

+

As a result, we can obtain3NH

from the molar conductivity of NH4OH.

4 4

4

4

4

NH OH NH Cl OH Cl

1 2 1 1 2 1 1 2 1NH OH

1 2 1NH OH

1 2 1NH OH

129.8 cm mol 174.0 cm mol 65.6 cm mol

238.2 cm mol

238 cm mol

= +

= +

=

=

The degree of dissociation is defined by Eq. 7.11 which states,


7-47

19.6

=

=

2cm 1 mol1238.2 2cm 1 mol

2

0.040 302 267

4.0 10

=

=



7-48

7.17. A solution of LiCl was electrolyzed in a Hittorf cell. After a current of 0.79 A had been passed for 2 h, the mass of LiCl in the anode compartment had decreased by 0.793 g.

a. Calculate the transport numbers of the Li+ and Cl ions.

b. If (LiCl) is 115.0 1 cm2 mol1, what are the molar ionic conductivities and the ionic mobilities?

Solution:

Given: LiCl

1 2 10.79 A, 2 h, 0.793 g, 115.0 cm molI t m = = = =

Required: see above

a) To solve this problem, we can use the Hittorf method. This method gives the transport numbers according to Eq. 7.75 and Eq.7.76. amount lost from anodecompartment

amount depositedt+= And

amount lost from cathodecompartmentamount deposited

t=

We will use the number of moles to measure the amounts of the Li+ and Cl ions. To determine the total amount deposited, we use Eq. 7.6.

( )0.79 A 2 h

Q It

Q

=

=s3600 h

5688 A s5688 C

QQ

==

In problem 7.1 we found that Q zFn= since the charge carried by 1 mol of ions bearing z unit charges is zF, where1 F = 96 485 C mol-1.

Solving for n, we can determine the total amount deposited.

5688 Camount deposited

QnzF

=

=( )1 96 485 C( )1mol

amount deposited 0.058 952168 7 mol

=

The amount lost of LiCl lost in the anode compartment is given by nCl.


7-49

LiClLiCl

LiCl1 1

LiCl 6.941g mol 35.4527 g mol

mnM

M

=

= +

1LiCl

LiCl

42.3927 g mol

0.793 g

M

n

=

=42.3927 g 1

LiCl

mol

0.018 705 609 6 moln

=

The anode reaction that is occurring is 21Cl Cl2

e + , therefore 0.058 952168 7 mol of Cl- are removed by electrolysis. The amount lost

from the anode compartment is given by, net total LiCln n n=

net

net

0.058 952168 7 mol 0.018 705 609 6 mol0.040 246 559 2 mol

nn

=

=

Solving for tCl- we obtain,

Cl

0.040 246 559 2 molt t + = = 0.058 952168 7 mol

Cl

Cl

0.682 698 534

0.68

t

t

=

=

The second transport number is given by1 t+ ,

Li

Li

Li

Li

1

1 0.682 698 534

0.317 301 466

0.32

t t ttt

t

+

+

+

+

+= =

=

=

=


7-50

b) In order to determine the molar ionic conductivities we will use Eq. 7.79 which states,

andt t +

+ = =

To solve, we rearrange and useLiCl

1 2 1115.0 cm mol = .

( )( )Cl Cl

1 2 1Cl

1 2 1Cl

1 2 1Cl

115.0 cm mol 0.682 698 534

78.510 331 41 cm mol

79 cm mol

t

=

=

=

=

( )( )Li Li

1 2 1Li

1 2 1Li

1 2 1Li

115.0 cm mol 0.317 301 466

36.489 668 59 cm mol

36 cm mol

t

+ +

+

+

+

=

=

=

=

The ionic mobility is given in terms of molar ionic conductivity by Eq. 7.64.


7-51

1 2 1

Cl

4 1 2 1 1Cl

1 1 1

1 1 1

4Cl

78.510 331 41 cm mol96 485 C

8.137 050 465 10 cm mol C

where1 1 A V and 1 A 1C stherefore,1 1C s V

8.137 050 465 10 C

Fuc

uF

u

u

u

++ +

+

+

+

= =

=

=

=

= =

=

= ( )1 1 2 1 1 s V cm mol C 4 2 1 1 1

Cl

4 2 1 1 1Cl

1 2 1

Li

4 2 1 1 1Li

4 2 1 1 1Li

8.137 050 465 10 cm mol V s

8.1 10 cm mol V s

36.489 668 59 cm mol96 485 C

3.781900 667 10 cm mol V s

3.8 10 cm mol V s

u

u

u

u

u

+

+

+

=

=

=

=

=



7-52

7.18. A solution of cadmium iodide, CdI2, having a molality of 7.545 103 mol kg1, was electrolyzed in a Hittorf cell. The mass of cadmium deposited at the cathode was 0.03462 g. Solution weighing 152.64 g was withdrawn from the anode compartment and was found to contain 0.3718 g of cadmium iodide. Calculate the transport numbers of Cd2+ and I.

Solution:

Given: 2+3 1 anodeCd7.545 10 mol kg , 0.03462 g, 152.64 g,molality m m= = =

2CdI in anode

0.3718 gm =

Required: 2+Cd I,t t

When working with a Hittorf cell, we must use Eq. 7.75 and Eq. 7.76 to solve for 2+Cd Iandt t .

amount lost from anodecompartmentamount deposited

t+= And amount lost from cathodecompartment

amount depositedt=

The number of coulombs of charge will be used as a measure of each amount.

The anode compartment initially contained the following number of moles:

anode

3 17.545 10 mol kg

i

i

n molality m

n

=

= ( ) 152.64 g 3 kg10 g0.001151669 molin

=

The anode compartment finally contained,


7-53

( )

2

2

2

2

CdI in anode

CdI

1 1CdI

1CdI

112.411g mol 2 126.904 47 g mol

366.219 94 g mol

0.3718 g

f

f

mn

M

M

M

n

=

= +

=

=366.219 94 g 1mol

0.001015 236 9 molfn

=

The number of moles lost from the anode compartment is therefore,

4

0.001151669 mol 0.001015 236 9 mol1.364 318 907 10 mol

i fn n nnn

=

=

=

The total amount of Cd2+ deposited is calculated by,

2

2

2

2

2

CdCd

Cd1

Cd

Cd

112.411g mol

0.03462 g

mn

M

M

n

+

+

+

+

+

=

=

=112.411 g

2

1

4Cd

mol

3.079 769 773 10 moln +

=

Now it is possible to determine the transport number at the anode.

4

4

1.364 318 907 10 mol3.079 769 773 10 mol0.442 993 797 4

0.4430

t

t

t

=

=

=


7-54

The second transport number is given by1 t .

t 1 0.442 993 797 4t 0.557 006 202 6

t 0.5570

+

+

+

=

=

=



7-55

7.19. The transport numbers for HCl at infinite dilution are estimated to be t+ = 0.821 and t = 0.179 and the molar conductivity is 426.16 1 cm2 mol1. Calculate the mobilities of the hydrogen and chloride ions.

Solution:

Given: HCl

1 2 10.821, 0.179, 426.16 cm molt t + = = =

Required: ,u u+

The ionic mobility is given in terms of molar ionic conductivity by Eq. 7.64.

Fuc

++ ++

= =

The molar ionic conductivities are given by Eq. 7.79.

andt t +

+ = =

By rearranging and substituting equations 7.64 and 7.79, we can obtain an expression for the ionic mobility.

u

Ftu

F

=

=

Now it is possible to solve for andu u+ .


7-56

( )( )1 2 1

1 2 1 1

1 1 1

1 1 1

426.16 cm mol 0.82196 485 C

0.003 626 236 cm mol C

where1 1 A V and 1 A 1C stherefore,1 1C s V

0.003 626 236 C

u

u

u

+

+

+

=

=

= =

=

= ( )1 1 2 1 1 s V cm mol C

( )( )

3 1 2 1 1

1 2 1

4 1 2 1 1

4 1 2 1 1

3.63 10 V cm mol s

426.16 cm mol 0.17996 485 C

7.906165 725 10 V cm mol s

7.91 10 V cm mol s

u

u

u

u

+

=

=

=

=



7-57

7.20. If a potential gradient of 100 V cm1 is applied to a 0.01 M solution of NaCl, what are the speeds of the Na+ and Cl ions? Take the ionic conductivities to be those listed in Table 7.3 on p. 291.

Solution:

Given: 1 NaCl100 V cm , 0.01 M, Table 7.3V c= =

Required:Na Cl

,v v+

The ionic mobility is given in terms of the molar ionic conductivity by Eq. 7.64.

Fuc

uF

++ +

+

+

+

= =

=

From Table 7.3 we are given that,

2 1 2 1Na Cl

50.08 S cm mol and 76.31S cm mol + = =

2 1

Na

50.08 S cm molu +

=196 485 C mol

4 2 1Na

1 1 1 1

1 1

4Na

5.190 44411 10 S cm C

where1 S 1 and 1 1 A V and 1 A 1C stherefore,1S 1C s V

5.190 44411 10 C

u

u

+

+

=

= = =

=

= ( )1 1 2 1 s V cm C 4 1 2 1

Na

2 1

Cl

5.190 44411 10 V cm s

76.31S cm mol

u

u

+

=

=196 485 C mol

4 1 2 1Cl

7.909 001399 10 V cm su =


7-58

From section 7.5 we know that, speed = uV. The velocities in a gradient of 100 V cm1 are thus,

4 1Na

5.190 44411 10 Vv + = 2cm( )1s 100 V 1 cm( )2 1

Na

2 1Na

4 1Cl

5.190 44411 10 cm s

5.19 10 cm s

7.909 001399 10 V

v

v

v

+

+

=

=

= 2cm( )1s 100 V 1 cm( )

2 1Cl

2 1Cl

7.909 001399 10 cm s

7.91 10 cm s

v

v

=

=



7-59

7.21. A solution of LiCl at a concentration of 0.01 M is contained in a tube having a cross-sectional area of 5 cm2. Calculate the speeds of the Li+ and Cl ions if a current of 1 A is passed. Use the ion conductivities listed in Table 7.3.

Solution:

Given: 2LiCl 0.01 M, 5 cm , 1 Ac A I= = = , Table 7.3

Required: Li Cl

,v v+

In order to determine the speeds of the ions, we must find the potential gradient. Once we have this information, we can proceed in a similar manner as was done in problem 7.20.

The potential gradient can be calculated using Ohms Law (Eq.7.7) in conjunction with Eq. 7.9.

VRI

= and c

=

The specific conductivity of a 0.01 M solution is calculated according to:

c =

We determine the molar conductivity of LiCl using the data for the ionic conductivities of Li+ and Cl- found in Table 7.3.

LiCl Li Cl2 1 2 1

LiCl2 1

LiCl

38.66 S cm mol 76.31S cm mol

114.97 S cm mol

+

= +

= +

=

Hence,

30.01 10 mol = ( )3 2 1 cm 114.97 S cm mol ( )3 11.1497 10 S cm =

Let us consider the fact that the resistance is inversely proportional to , and we must factor in the 5 cm2 of surface area.


7-60

3 1

11.1497 10 S cm

R

= 25 cm

1173.958 423 9 cmR =

The potential gradient required to produce a current of 1 A is therefore,

( )( )11

1 A 173.958 423 9 cm

where1 1 V A

V =

=

1173.958 423 9 V cmV =

The ionic mobilities can be calculated using Eq. 7.64.

Fuc

uF

++ +

+

+

+

= =

=

From Table 7.3 we know that,


7-61

2 1 2 1Li Cl

38.66 S cm mol and 76.31S cm mol + = =

2 1

Li

38.66 S cm molu +

=196 485 C mol

4 2 1Li

1 1 1 1

1 1

4Li

4.006 840 442 10 S cm C

where1 S 1 and 1 1 A V and 1 A 1C stherefore,1S 1C s V

4.006 840 442 10 C

u

u

+

+

=

= = =

=

= ( )1 1 2 1 s V cm C 4 1 2 1

Li

2 1

Cl

4.006 840 442 10 V cm s

76.31S cm mol

u

u

+

=

=196 485 C mol

4 1 2 1Cl

7.909 001399 10 V cm su =

From section 7.5 we are given that, speed = uV.

4 1 2 1 1Li

4.006 840 442 10 V cm s Vv + = 2cm( )1s 173.958 423 9 V 1 cm( )1

Li

2 1Li

0.069 702 364 8 cm s

7.0 10 cm s

v

v

+

+

=

=

4 1Cl

7.909 001399 10 Vv = 2cm( )1s 173.958 423 9 V 1 cm( )1

Cl

1Cl

0.137 583 7418 cm s

0.14 cm s

v

v

=

=



7-62

7.22. What is the work required to separate in vacuum two particles, one with the charge of the proton, from another particle with the same charge of opposite sign? Carry out the calculations for an initial distance of (a) 1.0 nm to an infinite distance apart and (b) from 1.0 mm to an infinite distance apart. (c) In (a) how much work would be required if the charge is moved to a distance of 0.1 m? The charge on a proton is 1.6 1019 C.

Solution:

Given: vacuum: 1 2Q Q= , e = 1.6 1019 C

Required: see above

Recall that work is defined as the application of a force through a distance. This definition is given by Eq. 1.1, dw = F dl. In this case, the force we are concerned with is an electrostatic force, and the distance in a vacuum we use as r.

From Eq. 7.1, the electrostatic force is given by:

1 22

04Q QF

r =

To determine the amount of work done, we will take the integral of F with respect to r.

2

1

2

1

1 22

04

r

r

r

r

w Fdr

Q Qw drr

=

=

Since the particles have opposite charges, we will introduce a negative sign.

2

1

1 22

0

1 2

0 2 1

4

1 14

r

r

Q Qw drr

Q Qwr r

=

=

The permittivity of a vacuum has the value, 12 2 1 10 8.854 10 C J m = . This will be used when solving parts a, b and c.


7-63

a)

91 2

19

1.0 10 m, m

1.6 10 C

r r

w

= =

=

( ) 212 2 4 8.854 10 C 1 1J m ( )

1 9

11.0 10 m

192.300 8621 10 Jw

=

192.3 10 Jw =

b)

31 2

19

1.0 10 m, m

1.6 10 C

r r

w

= =

=

( ) 212 2 4 8.854 10 C 1 1J m ( )

1 3

11.0 10 m

25

25

2.300 8621 10 J

2.3 10 J

w

w

=

=

c)

91 2

19

1.0 10 m, 0.10 m

1.6 10 C

r r

w

= =

=

( ) 212 2 4 8.854 10 C 1 1J m ( ) 9

1 10.10 m 1.0 10 m

19

19

2.300 8621 10 J

2.3 10 J

w

w

=

=



7-64

7.23. According to Bjerrums theory of ion association, the number of ions of type i present in a spherical shell of thickness dr and distance r from a central ion is

2 20 Bexp( /4 ) 4i i i cdN N z z e rk T r dr =

where zi and zc are the charge numbers of the ion of type i and of the central ion and e, 0 , , and kB have their usual significance. Plot the exponential in this expression and also 4r2 against r for a uni-univalent electrolyte in water at 25.0 C ( 78.3). = Allow r to have values from 0 to 1 nm. Plot also the product of these functions, which is (dN1/N1)dr and is the probability of finding an ion of type i at a distance between r and r + dr of the central ion.

By differentiation, obtain a value r* for which the probability is a minimum, and calculate the value for water at 25.0 C. The electrostatic potential is given to a good approximation by the first term in Eq. 7.47 on p. 280. Obtain an expression, in terms of kBT, for the electrostatic energy between the two univalent ions at this minimum distance, and evaluate this energy at 25 C.

Solution:



7-65

7.24. The following are some conventional standard enthalpies of ions in aqueous solution at 25 C:

Ion fH/kJ mol1

H+ 0

Na+ 239.7

Ca2+ 543.1

Zn2+ 152.3

Cl 167.4

Br 120.9

Calculate the enthalpy of formation in aqueous solution of 1 mol of NaCl, CaCl2, and ZnBr2, assuming complete dissociation.

Solution:

Given: standard enthalpies

Required: enthalpies of formation

In order to calculate the enthalpies of formation, we will simply sum up the standard enthalpies of the ions present in the solution.


7-66

( )

Na Cl

2 Ca Cl

2

2

2 Zn Br

2

NaCl

1 1NaCl

1NaCl

CaCl

1 1CaCl

1CaCl

ZnBr

ZnBr

239.7 kJ mol 167.4 kJ mol

407.1 kJ mol

2

543.1 kJ mol 2 167.4 kJ mol

877.9 kJ mol

2

f f f

f

f

f f f

f

f

f f f

f

H H H

H

H

H H H

H

H

H H H

H

= +

=

=

= +

=

=

= +

( )2

1 1

1ZnBr

152.3 kJ mol 2 120.9 kJ mol

394.1 kJ molf H

=

=



7-67

7.25. One estimate for the absolute Gibbs energy of hydration of the H+ ion in aqueous solution is 1051.4 kJ mol1. On this basis, calculate the absolute Gibbs energies of hydration of the following ions, whose conventional standard Gibbs energies of hydration are as follows:

Ion hydG k J mol1

H+ 0

Na+ 679.1

Mg2+ 274.1

Al3+ 1346.4

Cl 1407.1

Br 1393.3 Solution:

Given: ( )H

1hyd absolute 1051.4 kJ molG +

=

Required: ( )hyd absoluteG for each ion

In order to find the absolute Gibbs energies of hydration, we can either lower the cations standard Gibbs energies of hydration, or raise the anions absolute Gibbs energies of hydration by1051.4 kJ mol1 (per charge).


7-68

( )

( )

( )

( )

( ) ( )

( )

H

H

Na

Na

2Mg

2Mg

1hyd

1hyd

1 1hyd

1hyd

1hyd

hyd

absolute 0 1051.4 kJ mol

absolute 1051.4 kJ mol

absolute 679.1 kJ mol 1051.4 kJ mol


absolute 274.1 kJ mol 2 1051.4 kJ mol

absolute

G

G

G

G

G

G

+

+

+

+

+

+

=

=

=

=

=

( ) ( )( )

( )

( )

3Al

3Al

Cl

Cl

1

1 1hyd

1hyd

1 1hyd

1hyd

1828.7 kJ mol

absolute 1346.4 kJ mol 3 1051.4 kJ mol




G

G

G

G

+

+

=

=

=

= +

=

( )

( )Br

Br

1 1hyd

1hyd



G

G

= +

=



7-69

7.26. Calculate the ionic strengths of 0.1 M solutions of KNO3, K2SO4, ZnSO4, ZnCl2, and K4Fe(CN)6; assume complete dissociation and neglect hydrolysis.

Solution:

Given: 0.1 Mc =

Required: I

The ionic strength of a solution is given by Eq. 7.103:

212 i ii

I c z=

where zi is the valency of each ion present.


7-70

( )

( )

( )

( )

3

3

2 4

2 4

4

4

2

2

3 3

2 2KNO

KNO

22 4 4

2 2K SO

K SO

2 24 4

2 2ZnSO

ZnSO

22

2 2ZnCl

ZnCl

4

KNO K NO1 0.1 M 1 0.1 M 120.1 M

K SO 2K SO1 0.2 M 1 0.1 M 220.3 M

ZnSO Zn SO1 0.1 M 2 0.1 M 220.4 M

ZnCl Zn 2Cl1 0.1 M 2 0.2 M 120.3 M

K Fe

I

I

I

I

I

I

I

I

+

+

+

+

+

= +

=

+

= +

=

+

= +

=

+

= +

=

( ) ( )

( ) ( )

( )

4 6

4 6

4

6 6

2 4K Fe CN

K Fe CN

CN 4K Fe CN

1 0.4 M 1 0.1 M 421.0 M

I

I

+ +

= +

=



7-71

7.27. Calculate the mean activity coefficient for the Ba2+ and 24SO ions in a saturated solution of BaSO4 (Ksp = 9.2 1011 mol2 dm6)

in 0.2 M K2SO4, assuming the Debye-Hckel limiting law to apply.

Solution:

Given: BaSO4: Ksp = 9.21011 mol2 dm6, 2 4K SO

0.2 M c =

Required:

When determining the mean activity coefficient, we use the Debye-Hckel limiting law given in Eq. 7.111:

310log 0.51 /moldmz z I

+ =

We may then calculate the ionic strength from Eq. 7.103 in the following manner,

212 i ii

I c z=

( )2 42 4

22 4 4

2 2K SO

K SO

K SO 2K SO1 0.4 M 1 0.2 M 220.6 M

I

I

+ +

= +

=

Finally we can solve for the mean activity coefficient which produces;

( )( )

10

0.51 2 2 0.6

2

log 0.51 2 2 0.6

100.026 291949 8

2.6 10

=

=

=

=



7-72

7.28. The solubility of AgCl in water at 25 C is 1.274 105 mol dm3. On the assumption t hat the Debye-Hckel limiting law applies,

a. Calculate G for the process AgCl(s) Ag+(aq) + Cl(aq).

b. Calculate the solubility of AgCl in an 0.005 M solution of K2SO4.

Solution:

Given: 5 31.274 10 mol dm , 25 Cs T= =

Required: see above

a. To calculate the Gibbs energy, we first need to determine the solubility product of AgCl in water. Eq. 7.121 shows that 2[Ag ][Cl ]sK

+ = , and since [Ag ] [Cl ]

+ = , we can write

2 2sK s = .

We solve for the mean activity coefficient from Eq. 7.111, using the solubility as a measure of ionic strength.

( )( ) 5

310

510

0.51 1 1 1.274 10

log 0.51 /moldm

log 0.51 1 1 1.274 10

100.995 817 261 4

z z I

+

=

=

=

=

The solubility product is then,

( ) ( )2 2510 2

1.274 10 0.995 817 261 4

1.609 526 59 10s

s

K M

K M

=

=

Solving for Gibbs energy, using ln sG RT K = , we obtain


7-73

18.3145 J KG = ( )1mol 298.15 K ( ) ( )10 21

1

ln 1.609 526 59 10

55 900.51131 J mol

55.90 kJ mol

M

G

G

=

=

b. To solve for the solubility in a solution of 0.005 c M= K2SO4, we need to calculate the ionic strength since we have a common ion present. We can calculate the ionic strength from Eq. 7.103 in the following manner,

( )

2

2 2

121 0.01 M 1 0.005 M 220.015 M

i ii

I c z

I

I

=

= +

=

We solve for the mean activity coefficient from Eq. 7.111, using the solubility as a measure of ionic strength.

( )

310

10

100.062 461988

log 0.51 / mol dm

log 0.51 1 1 0.015log 0.062 461988

100.866 04012

z z I

+

=

=

=

=

=

Solving for the solubility by rearranging Eq. 7.121,


7-74

2 2

10 2

5

5

M

M

M

1.609 526 59 10 0.86604012

1.464 91 10

1.46 10

s

s

K s

Ks

s

s

s

=

=

=

=

=



7-75

7.29. Employ Eq. 7.114 to make plots of log against I for a uni-univalent electrolyte in water at 25 C, with B = 0.51 mol1 dm3/2 and B = 0.33 1010 mol1 dm3/2 m1, and for the following values of the interionic distance a:

a = 0, 0.1, 0.2, 0.4, and 0.8 nm

Solution:



7-76

7.30. Estimate the change in Gibbs energy G when 1 mol of K+ ions (radius 0.133 nm) is transported from aqueous solution ( = 78) to the lipid environment of a cell membrane ( = 4) at 25 C.

Solution:

Given: water membraneK1 mol, 0.133 nm, 78, 4, 25 Cn r T+= = = = =

Required: G

Eq. 7.87 (given below) may be used to estimate the change in Gibbs energy.

2 2

es08

z eGr

=

Since we are given 1 mol of K+, we will multiply the expression above by L, Avogadros number.


7-77

( )

2 2

es0

2 19

es

8

1 1.602 10 C

z e LGr

G

=

+ =

( ) 2 ( )23 112 2

6.022 10 mol

8 8.854 10 C

1 2N m ( ) 90.133 10 m( )1

es

1

es

1

water

1water

1

membrane

1membrane

es membrane water

e

5 222197.4616 N m mol

where1 N m 1 J5 222197.4616 J mol

5 222197.4616 J mol78

6694.839 251 J mol

5 222197.4616 J mol4

130 549.3654 J mol

G

G

G

G

G

GG G GG

=

=

=

=

=

=

=

=

1 1s1

es

1es

130 549.3654 J mol 6694.839 251 J mol

123 854.5261 J mol

124 kJ mol

G

G

=

=

=



7-78

7.31. At 18 C the electrolytic conductivity of a saturated solution of CaF2 is 3.86 105 1 cm1, and that of pure water is 1.5 106 1

cm1. The molar ionic conductivities of 21 Ca2

+ and F are 51.1 1 cm2 mol1 and 47.0 1 cm2 mol1, respectively. Calculate the

solubility of CaF2 in pure water at 18 C and the solubility product.

Solution:

Given: 2 2

5 1 1 6 1 1CaF H O18 C, 3.86 10 cm , 1.5 10 cm ,T = = =

2+

1 2 1 1 2 11 FCa2

51.1 cm mol , 47.0 cm mol = =

Required: , ss K

The expression for the molar conductivity is given by Eq. 7.9:

c

=

It is possible to use the concentration to determine the solubility.

2

-2+2

2

2

1CaF2

1 1 FCaF Ca2 2

1 2 1 1 2 11CaF2

1 2 11CaF2


98.1 cm mol

c

=

= +

= +

=

The observed due to the salt is therefore,


7-79

5 1 1 6 1 1

5 1 1

5 1

3.86 10 cm 1.5 10 cm3.71 10 cm

3.71 10 c

=

=

=

1

1

cm98.1 2 1

7 3

4 3

cm mol3.781855 25 10 mol cm3.781855 25 10 mol dm

cc

=

=

21 CaF2

has the molar mass,

( )2

2

1 11 CaF2

1CaF

1 40.078 g mol 18.998 403 2 g mol2

39.037 403 2 g mol

M

M

= +

=

The solubility can now be determined.

139.037 403 2 g mols = ( ) 43.781855 25 10 mol( )33

2 3

dm

0.014 763 380 8 g dm

1.48 10 g dm

s

s

=

=

The solubility product is given by,

( ) ( )

22

24 3 4 3

11 3

11 3

Ca F

1 3.781855 25 10 mol dm 3.781855 25 10 mol dm2

2.704 485 84 10 mol dm

2.70 10 mol dm

s

s

s

s

K

K

K

K

+

=

=

=

=



7-80

7.32. What concentrations of the following have the same ionic strength as 0.1 M NaCl?

CuSO4, Ni(NO3)2, Al2(SO4)3, Na3PO4 Assume complete dissociation and neglect hydrolysis.

Solution:

Given: NaCl 0.1 Mc =

Required: ( ) ( )4 3 42 43 32CuSO Na POAl SONi NO, , ,c c c c

As we have previously seen, the ionic strength of a compound may be determined using Eq. 7.103.

( )

2

2 2NaCl

NaCl

12

1 0.1 M 1 0.1 M 120.1 M

i ii

I c z

I

I

=

= +

=

For each salt, we set NaCl 0.1 MI I= = to solve for c .


7-81

( )

( ) ( )( )( )

( )

( ) ( )( )( )

( )

4 4

4

4

3 32 2

3 2

3 2

2 4 2 43 3

2 4 3

2 4 3

3 4

2 2CuSO CuSO

CuSO

2CuSO

2 2Ni NO Ni NO

Ni NO

2Ni NO

2 2Al SO Al SO

Al SO

3Al SO

2Na PO Na

10.1 M 2 22

0.1 M 4

2.5 10 M

10.1 M 2 2 12

0.1 M 3

3.3 10 M

10.1 M 2 3 3 22

0.1 M 15

6.7 10 M

10.1 M 3 12

c c

c

c

c c

c

c

c c

c

c

c c

= +

=

=

= +

=

=

= +

=

=

= +( )3 43 4

3 4

2PO

Na PO

2Na PO

3

0.1 M 6

1.7 10 M

c

c

=

=



7-82

7.33. The solubility product of PbF2 at 25.0 C is 4.0 109 mol3 dm9. Assuming the Debye-Hckel limiting law to apply, calculate the solubility of PbF2 in (a) pure water and (b) 0.01 M NaF.

Solution:

Given: 9 3 94.0 10 mol dm , 25 CsK T= =

Required: see above

a) The dissolution of PbF2 is written as: 22PbF Pb 2F+ + , hence the solubility product is given by:

22Pb FsK+ = .

In order to solve for solubility, we must first neglect the effect of the activity coefficients and write, [ ][ ]2 32 4sK s s s= = .

Solving for s, we obtain,

3 9 3 9

3 3

4 4.0 10 mol dm1.0 10 mol dm

ss

=

=

We will determine the activity coefficients of Pb2+ and F- by applying the Debye-Hckel limiting law. To solve, we must first calculate the ionic strength of PbF2 from Eq. 7.103.

( )

2

2 2

3 3

121 2 2 1233.0 10 mol dm

i ii

I c z

I s s

I sI

=

= +

=

=

According to the Debye-Hckel limiting law, Eq. 7.111,


7-83

( )

3

3

310

0.51 /moldm

0.51 2 1 3.0 10

log 0.51 /moldm

10

100.879 290 334 3

z z I

z z I

+

+

=

=

=

=

In order to find the true solubility, we must factor in the activity coefficients.

[ ][ ]2 3

32

2 4

4

s

s

K s s sKs

+

+

= =

=

( )

9 3 9

3 3

3 3

3 3

4.0 10 mol dm4 0.879 290 334 3

1.137 280 8 10 mol dm

1.1 10 mol dm

s

s

s

=

=

=

b) In 0.01 M NaF, the ionic strength is essentially 0.01 mol dm-3. Calculating the activity coefficients, we obtain,

( )

30.51 /moldm

0.51 2 1 0.01

10

100.790 678 628

z z I

+

=

=

=

If s is the solubility then,


7-84

[ ]( )

( )

22

2

3

23

2 2 6

2 2 6

9 3

Pb F

Pb

F 0.01 mol dm

0.01 mol dm

0.0001 mol dm

0.0001 mol dm

4.0 10 mol

s

s

s

s

K

s

K s

K s

Ks

s

+

+

+

+

+

= = =

=

=

=

=

9 dm( )3 20.790 678 628 0.0001 mol 6 dm

5 3

5 3

8.092 076 715 10 mol dm

8.1 10 mol dm

s

s

=

=



7-85

7.34. Calculate the solubility of silver acetate in water at 25 C, assuming the DHLL to apply; the solubility product is 4.0 103 mol2 dm6.

Solution:

Given: 3 2 64.0 10 mol dm , 25 CsK T= =

Required: s

We will solve this problem in a similar manner as the previous question. We may start by neglecting the activity coefficients to find the first approximation of s.

The dissolution of silver acetate is given by the following:

2 3 2 3AgC H O Ag CH COO+ +

[ ][ ] 2

3 2 6

3

4.0 10 mol dm

0.063 245 553 2 mol dm

s

s

K s s s

s K

s

s

= =

=

=

=

The ionic strength is thus,

( )2 23

1 1 12

0.063 245 553 2 mol dm

I s s

I s

= +

= =

According to the Debye-Hckel limiting law, Eq. 7.111,

( )

310

0.51 1 1 0.063 245 553 2

log 0.51 /moldm

100.744 289 325

z z I

+

=

=

=

The second approximation of the solubility is therefore,


7-86

[ ][ ] 2ss

K s s s

Ks

+ +

+

= =

=

( )

3 2 6

2

3

2 3

4.0 10 mol dm0.744 289 325

0.084 974 419 3 mol dm

8.5 10 mol dm

s

s

s

=

=

=

We may continue to take third and fourth approximations by repeating the above steps.

( )

( )

0.51 1 1 0.084 974 419 3

3 2 6

2

3

100.710122 220 4

4.0 10 mol dm0.710122 220 4

0.089 062 912 5 mol dm

sKs

s

s

+

=

=

=

=

=

The third approximation is then; 2 38.9 10 mol dms = .

( )

( )

0.51 1 1 0.089 062 912 5

3 2 6

2

3

100.704 366 363 8

4.0 10 mol dm0.704 366 363 8

0.089 790 706 2 mol dm

sKs

s

s

+

=

=

=

=

=


7-87

The fourth approximation is then; 2 39.0 10 mol dms = .



7-88

7.35. Problem 7.30 was concerned with the Gibbs energy change when 1 mol of K+ ions are transported from water to a lipid. Estimate the electrostatic contribution to the entropy change when this occurs, assuming the dielectric constant of the lipid to be temperature independent, and the following values for water at 25 C: 78; = ln /T = 0.0046 K1. Suggest a qualitative explanation for the sign of the value you obtain.

Solution:

Given: Problem 7.30: ln25 C, 78, 0.0046TT

= = =

Required: esS , explain the sign

From Problem 7.30, we found the expression for the estimated Gibbs energy,

1

es5 222197.4616 J molG

=

For the transfer from water to lipid we can say that,

1es

lipid water

1 1/J mol 5 222197.4616G

=

From Eq. 3.119:

eses

1 1es

lipid water

hence,

1 1/ J K mol 5 222197.4616

P

P

G ST

GST

ST

=

=

=

Since lipid is temperature independent, this leads to:


7-89

water

water

1 1es

water

1 1es 2

2water

1/ J K mol 5 222197.4616

1/ J K mol 5 222197.4616

1 1 lnsince

ST

ST

T T

=

=

=

( )

1 1es

water

1 1es

1 1es

1 1es

1 ln/ J K mol 5 222197.4616

1/ J K mol 5 222197.4616 0.004678

30.796 260 56 J K mol

31 J K mol

ST

S

S

S

=

=

=

=

The entropy increases due to the release of bound water molecules when the K+ ions pass into the lipid.



7-90

7.36. Assuming the Born equation (Eq. 7.86) to apply, make an estimate of the reversible work of charging 1 mol of Na+Cl in aqueous solution at 25 C ( = 78), under the following conditions:

a. The electrolyte is present at infinite dilution.

b. The electrolyte is present at such a concentration that the mean activity coefficient is 0.70.

The ionic radii are 95 pm for Na+ and 181 pm for Cl.

Solution:

Given: Eq. 7.86, 1 mol, 25 C, 78n T= = =

Required: see above

a) At infinite dilution, the work of charging an ion is given by Eq.7.86 which states, 2 2

rev08

z ewr

=

For one mole of Na+, we multiply by Avogadros number, L and use the ionic radius of r = 95 pm.

( )

+

+

2 2

Na0

2 19

Na

8

1 1.602 10 C

z e Lwr

w

=

+ =

( ) 2 ( )23 112 2

6.022 10 mol

8 8.854 10 C

1 2N m ( )( ) 1278 95 10 m( )+

+

1Na

1Na

9372.774 952 N m mol

9372.774 952 J mol

w

w

=

=

For one mole of Cl-, we will multiply by Avogadros number, L and use the ionic radius of r = 181 pm.


7-91

( )-

2 19

Cl

1 1.602 10 Cw

=

( ) 2 ( )23 112 2

6.022 10 mol

8 8.854 10 C

1 2N m ( )( ) 1278 181 10 m( )-

1Cl

4919.412 267 J molw =

One mole of Na+Cl- at infinite dilution is thus,

+ -rev Na Cl1 1

rev 9372.774 952 J mol 4919.412 267 J mol

w w w

w = +

= +

1rev

1rev

14 292.187 22 J mol

14 kJ mol

w

w

=

=

b) These values are reduced when the electrolyte is at a higher concentration. The work of charging the ionic atmosphere is negative and equal to ln ikT . Thus for one mol of Na

+ ions, of activity + , the work of charging the atmosphere is lnRT + . Similarly, for the chloride ion, the work per mole is lnRT . For one mole of Na+Cl-

( )( )

rev

rev

rev

1rev

ln ln

ln2 ln

where 0.70

2 8.3145 J K

w RT

w RTw RT

w

+

+

= +

=

=

=

= ( )1mol 298.15 K ( )1

rev

ln 0.70

1768.37167 J molw =

The net work done is given by,

1 1rev

1rev

1rev

14 292.187 22 J mol 1768.37167 J mol

12 523.815 55 J mol

13 kJ mol

ww

w

=

=

=


7-92



7-93

7.37. If the solubility product of barium sulfate is 9.2 1011 mol2 dm6, calculate the solubility of BaSO4 in a solution that is 0.10 M in NaNO3 and 0.20 M in Zn(NO3)2; assume the DHLL to apply.

Solution:

Given: ( )3 3 211 2 6

in NaNO in Zn NO9.2 10 mol dm , 0.10 M, 0.20 MsK c c= = =

Required: s

The expression for the solubility product is given by,

[ ][ ]

2 24

2 2

Ba SOs

s

s

K

K s s

K s

+

+

= =

=

The ionic strength of the solution is calculated according to Eq.7.103.

( )

2

2 2 2 2

121 1 0.1 M 1 0.1 M 2 0.2 M 1 0.4 M20.70 M

i ii

I c z

I

I

=

= + + +

=

To find the activity coefficient, we will use the Debye-Hckel limiting law given by Eq. 7.111.

( )

3

310

0.51 /moldm

0.51 2 2 0.70

log 0.51 /moldm

10

100.019 643 2591

z z I

z z I

+

+

=

=

=

=

If the solubility s is therefore,


7-94

( )

2

11 2 6

2

4

4

9.2 10 mol dm0.019 643 2591

4.882 928 531 10 M

4.9 10 M

sKs

s

s

s

=

=

=

=


Chapter 7: Solu

lms solutions electrochemistry

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