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Practice Problem Set #3

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Practice Problem Set 3. Colligative Properties and Electrochemistry 1. A fluorocarbon is believed to be either CF3(CF2)3CF3 or CF3(CF2)4CF3. If the freezing point depression of the solution of 1.0 g of the fluorocarbon in 100 g of camphor is to be used to distinguish these possibilities, what is the difference between the two in terms of their temperature drop? The cryoscopic constant of camphor is 39.0 K·kg/mol. 2. (a) The enthalpy of fusion of CCl4 is 2.5 kJ/mol at its normal melting point of −22.7 C, and the enthalpy of vaporization is 30.0 kJ/mol at the normal boiling point (77.0 C). Estimate Kf and Kb for CCl4. (b) A solution of 3.0 g of an unknown substance in 100 g of CCl4 gave a boiling point elevation of 0.60 C. Calculation the freezing point depression, the osmotic pressure of the solution at 25 C, and the molecular weight of the substance. The density of CCl4 is 1.60 kg/L. 3. The following data were obtained for the osmotic pressures of the nitrocellulose in acetone at 20 C. (1 cmH2O = 9.6210−4 atm) w (g/L) Π (cmH2O) Π/w (cmH2O·L/g)

0.56 0.28 0.50

1.16 0.62 0.53

3.66 2.56 0.70

8.38 8.00 0.95

Calculate the limiting value Π/w and hence the best estimate of the molecular wright. 4. A dilute aqueous sugar solution of total weight 1000 g contained 0.5 g of sugar(I) (M: 500 g/mol) and 0.3 g of sugar (II) (unknown M). The osmotic pressure of this solution was found to be 49.6 mmHg (across a suitable membrane separating it from pure water) at 25 C. Calculate the molecular weight M of sugar (II) and give its molecular formula. 5. A test of Donnan’s theory of membrane equilibrium was made by measuring the distribution of KCl across Cu2Fe(CN)6 membrane (Kameyama, Phil. Mag., 1925, 56, 849). To the solution on one side was added K4Fe(CN)6, the membrane being permeable to all ions present except [Fe(CN)6]4−. Three of these experiments gave the following results at equilibrium:

Exp 1 Exp 2 Exp 3

Solution I [K4Fe(CN)6] (mM) [KCl] (mM) 3.62 18.3 8.33 8.40 15.3 7.96

Solution II [KCl] (mM) 24.3 18.6 23.6

Practice Problem Set #3

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Calculate the product [K +][Cl−] for each solution and show that Donnan’s theory gives an adequate account of these experiments. 6. Consider the membrane of a plant cell that is permeable to Na+, Cl−, and H2O, but not to proteins. Suppose that at 25 C, initially there is a NaCl solution 0.050 M on each side of the membrane, and a protein P at 0.001 M concentration inside the cell, which ionized according to P = P10+ + 10 Cl−. Calculate the Donnan potential across the cell wall at equilibrium. 7. A 2 mM aqueous solution of the potassium salt of a protein (KP) is separated from an equal volume of a 6 mM aqueous KCl solution by a membrane that is permeable to all species present except the protein, and the system is left to reach equilibrium at 25 C. (a) What is the total concentration of K+ ions in the protein solution at equilibrium? (b) Calculate the Donnan potential associated with the Cl− ions at equilibrium. (c) Estimate the osmotic pressure difference across the membrane that would be observed at equilibrium. 8. The standard reduction potentials for Fe3+ and Zn2+ at 25 C are: Fe3  (aq )  e   Fe2  (aq ) 2



Zn ( aq )  2 e  Zn( s)

    0.771 V     0.762 V

(a) Determine the standard potentials for the following reactions:

i. Fe 2 (aq )  Fe3  (aq )  e  ii. 2 Fe2  (aq )  2 Fe3 (aq )  2e iii. Zn(s )  2 Fe3 (aq )  Zn 2 (aq )  2 Fe2 (aq ) (b) What is the ΔG0 for the reaction (iii)? Is this reaction spontaneous at standard conditions? (c) With the usual conventions, give a cell reaction for the cell: Zn( s) Zn 2  ( aq) Fe3  (aq ), Fe2  (aq ) Pt(s ) Such a cell is short circuited and left until current no longer flows (i.e., until equilibrium is established). If the concentration of Zn2+ ions is then 0.001 M, what is the ratio of the concentrations [Fe2+]/[Fe3+]? 9. Calculate the concentration of Pb2+ ions for which the E.M.F. of the cell is exactly zero at 25 C. The standard reduction potentials at this temperature are −0.126 V and −0.140 V for Pb2+/Pb(s) and Sn2+/Sn(s), respectively. Write down the cell reaction. Pb( s) Pb 2  ( aq) Sn 2  (0.01 M) Sn( s)

Practice Problem Set #3

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10. The equilibrium constant at 25 C for the reaction below is 1.08102, and the standard reduction potential of a Zn2+/Zn(s) electrode is −0.760 V.

3 Zn( s )  2 Cr 3  3 Zn 2   2 Cr(s ) Calculate the E.M.F. of the cell representing the given reaction when the concentration of Zn2+ and Cr3+ ions are each 10−2 M and the temperature is 25 C. What is ε0 for the Cr3+/Cr(s) electrode?

Practice Problem Set #3

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Practice Problem Set 3 − Answers. Thermodynamics 1. ΔTf = −KfbB, where Kf = 39.0 K·kg/mol, and bB = molarlity = number of moles of solute per kg of solvent. To figure out the precision needed, calculate the expected values of ΔTf for the two possible formulae: For C5F12 (M = 288 g/mol), b B 

10 10 g/kg solvent ; and  Tf  39.0    1.35 K 288 288 g/mol

For C6F14 (M = 338 g/mol), b B 

10 10 g/kg solvent ; and T f  39.0    1.15 K 338 338 g/mol

These two values differ by 0.2 K, which should be the precision of the observation.

2. (a) K f 

Kb 

MRTm2 , pure H m0 , fus , pure

MRTb2, pure H

0 m ,vap , pure



0.1538 kg/mol 8.314 J/( K  mol)  (250.45 K ) 2  32.08 K kg/mol 2500 J/mol



0.1538 kg/mol 8.314 J/( K  mol)  (350.15 K ) 2  5.26 K  kg/mol 30000 J/mol

(b) Tb  K b bB ,  bB 

Tb 0.60 K   0.114 mol/kg K b 5.26 K  kg/mol

Assume we have 1 kg of such CCl4 solution, then we have 30.0 g of the unknown substance, which is 0.114 mol from bB as calculated above. M 

30.0 g  263 g/mol 0.114 mol

T f  K f bB  32.08 K  kg/mol  0.114 mol/kg  3.66 K . Alternatively, we can use:  Tf   K f

0.60 K  32.08 K  kg/mol Tb  3.66 K .  5.26 K kg/mol Kb

Osmotic pressure: Π  cRT , where c = solute concentration in mol/L. c

0.114 mol(unknown)  0.1824 mol/L 1 kg(CCl4 )  1.60 kgCCl 4 / L

 Π  0.1824 mol/L 0.08206L atm /(K mol) 298.15 K  4.46 atm

Practice Problem Set #3

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 w  w 3. Π  limcRT  RT lim  ,  M  RT lim  . Because this formula works best at very c 0 w 0 M w 0 Π     dilute concentrations, the usual practice is plotting (Π/w) vs. w, draw a straight line, and extend it to find the intercept at w = 0. w (g/L) Π (cmH2O) Π/w (cmH2O·L/g)

0.56 0.28 0.50

1.16 0.62 0.53

3.66 2.56 0.70

8.38 8.00 0.95

Π/w vs. w 1.20

1.00

Π/w (cmH2OL/g)

y = 0.0579x + 0.4707 R² = 0.9968 0.80

0.60

0.40

0.20

0.00 0

1

2

3

4

5

6

7

8

9

w (g/L)

Π  In this case, we find the intercept: lim   0.4707 cmH 2O  L/g w 0 w  

Given 1 cmH2O = 9.62 × 10−4 atm, we have:

0.08206 L  atm/(K  mol) 293.15 K w M  RT lim    53125.4 g/mol  53 kDal  w 0 Π   0.4707 cmH2 O L/g 9.62 10 4 atm/cmH2 O 4. Πtot will be a sum of contributions from the two sugars, SI and SII, that is, Π tot  Π I  Π II  c I RT  c II RT Since the aqueous solution is very dilute, the volume of the solution should be very close to that of pure water; hence, we treat 1000 g of such solution to be 1 liter in volume.

Practice Problem Set #3

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Therefore, for SI: cI 

0.5 g/L  0.001 mol/L 500 g/mol

Π I  c I RT  0.001 mol/L 0.08206 L atm/(K  mol)  298.15 K 760 mmHg/atm  18.6 mmHg

 Π II  Π tot  Π I  49.6 18.6 mmHg  31.0 mmHg

Hence, c II 

ΠII (31.0 / 760) atm   1.67  10 3 mol/L RT 0.08206 L atm/(K  mol) 298.15 K

For 0.3 g of SII:

M 

0.3 g/L  180 g/mol 1.67  103 mol/L

In terms of molecular formula, for sugars, or carbohydrates, the general form is (CH2O)x, which means, (30 g/mol)  x  180 g/mol, x  6. Therefore, the molecular formula of sugar (II) is C6H12O6. 5. Simply calculate the product of [K +][Cl−] for both solutions and verify the Donnan equation.

[mM] Exp 1 Exp 2 Exp 3

Solution I [K4Fe(CN)6] [KCl] 3.62 18.3 8.33 8.40 15.3 7.96

Solution II [KCl] 24.3 18.6 23.6

For Solution I, [K +]I = [KCl]I + 4·[K4Fe(CN)6], [Cl−]I = [KCl]I For Solution II, [K+]II = [Cl−]II = [KCl]II 



Solution I : [K ][Cl ] Exp1 (4 3.62 18.3) 18.3 599.9 mM2 Exp2 (4 8.33 8.4) 8.4 350.4 mM2 Exp3 (4 15.3 7.96) 7.96 550.5 mM2





Solution II : [K ][Cl ] 24.3 24.3 590.5 mM 2 18.6 18.6 346.0 mM2 23.6 23.6  557.0 mM 2

From above calculations, we observe that within certain accuracy, the results satisfy the Donnan equilibrium. 6. Initially, there is more [Cl−] inside the cell, as the protein added will ionize to produce [Cl−], so the excess amount of [Cl−] will diffuse through the membrane. At the same time, [Na+] will company the diffusion of [Cl−] in order for the system to preserve the electroneutrality. The

Practice Problem Set #3

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positively charged protein molecules will not be able to diffuse to the other side since their sizes are too big to penetrate the semi-permeable membrane. Assume at equilibrium, the inside of the cell lost x M of [Na+] and [Cl−], then we have the following: [P10+] 0.001 0.001

[M] Initially Equilibrium

Inside the Cell [Na+] [Cl−] 0.05 0.06 0.05 − x 0.06 − x

Outside the Cell [Na+] [Cl−] 0.05 0.05 0.05 + x 0.05 + x

From Donnan Equilibrium,

[Na  ] in[Cl  ] in  [Na  ] out[Cl  ]out (0.05  x)(0.06  x)  (0.05  x) 2 x 2  0.11x  0.003  x2  0.10 x  0.0025 0.21x  0.0005  x  0.0024 M

 [Cl  ]in  0.0576 M, [Cl  ]out  0.0524 M   cell  

 RT  [Cl ]out ln zF  [Cl  ]in

 8.314  298.15  0.0524  3    ln   2.43  10 V  2.43 mV   0 . 0576 1 96485   

7. Follow the ideas from Q6, assume x mM of [K+] and [Cl-] diffuse across the membrane.

[mM] Initially Equilibrium

−

[P ] 2 2

Solution I [K+] 2 2+x

−

[Cl ] 0 x

Solution II [K+] [Cl−] 6 6 6−x 6−x

From Donnan Equilibrium:     [K ] I[Cl ] I  [Na ] II[Cl ] II

(2  x) x  (6  x)2 x2  2x  x2  12x  36 14 x  36  x  2.57 mM (a) [K+]I = 2 + 2.57 = 4.57 mM (b) [Cl−]I = 2.57 mM, [Cl−]II = 3.43 mM,   I II  

8.314  298.15  2.57  ln   7.4 mV 1  96485  3.43 

(c) ∆Π = Δc·RT, in which Δc is the difference in concentration of the two solutions.

 c I II  (2  4.57  2.57)  (3.43  3.43)  2.28 mM Π I II  2.28 10 3 M  0.08206 L  atm/(K  mol) 298.15 K  5.58  10  2 atm

Practice Problem Set #3

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8. (a) First, we observe that reaction i is just the reverse reaction of the standard reduction, whereas reaction ii should have the same ε0 as that of reaction i because ε0 is independent of number of moles. Therefore, we have Fe2  ( aq)  Fe3  ( aq )  e  , 2Fe2  (aq)  2Fe3 (aq )  2e  , 2Fe3 (aq )  2e  2Fe2 (aq ) ,

i. ii.

iii.  ) Zn(s )  Zn 2 (aq )  2e , Zn( s)  2Fe 3 ( aq)  Zn 2 ( aq)  2Fe 2 ( aq),

    0.771 V     0.771 V     0.771 V     0.762 V     1.533 V

(b) G 0  nF  0  2  96485 1. 533 J  295 .8 kJ (c) Zn( s)  2Fe 3 ( aq)  Zn 2  ( aq)  2Fe 2  (aq ) 0 Given    

RT ln Q , and at equilibrium, G  0, ε  0, and Q  K , we have nF

nF  0 2 96485 1.533 119.34  K  e119.34  6.74 10 51  RT 8.314  298.15 [Fe2  ]2 [ Zn2  ] K  , [Zn 2 ]  0.001 M, [ Fe 3 ]2 ln K 

[Fe 2 ]  6.74  10 51   [ Fe 3 ]  0.001

1/ 2

  

 2.60  10 27

9. The cell reaction is: Pb  Sn 2   Pb 2   Sn, ε 0  0.140  0.126  0 .014 V, n  2

  0   ln

RT [Pb2  ] RT [Pb 2  ] 0 , and [Sn 2  ]  0.01 M 0 , ln ln     2F [Sn 2  ] 2 F [Sn2  ]

[Pb 2  ] 2 96485 ( 0.014)  1 .0895,  [Pb 2  ]  0.01  e 1.0895  3.36 mM  0.01 8 .314  298.15 

10. At the given condition, K  1.08 102 

[Zn 2 ]3 , and n  6,  [Cr 3 ] 2

RT 8 .314 298.15 ln(1.08  10 2 )  0.020 V, ln K   nF 6 96485 2 3 RT 8.314  298.15 (0.01)3 [ Zn ] 0 ln 0 . 020 V ln  0 .040 V,   cell    nF [Cr 3 ] 2 (0.01)2 6 96485

0  cell

 cell

0   Cr0 cell

 Cr0

3

/Cr

3

/Cr

0   Zn 2 /Zn

0   0cell   Zn  0.020  0.760   0.740 V 2 /Zn...