COLLEGE OF SAINT MARY   FALL 2008       BIOPHYSICAL CHEMISTRY      INSTRUCTOR: DR. PETER ILICH

    HOMEWORK # 05  -- SOLUTIONS & ANSWERS

TOPIC:    ION ACTIVITY     MEMBRANE POTENTIAL    GALVANIC CELLS

Copyright © 2008/2010 by Peter Ilich

A pickling solution.

1.    (25 pts) A small amount of concentrated HCl is added to a millimolal (NOTE: "m", not "M") solution of GdCl₃ to bring the HCl concentration of the solution to 0.01 m.   (a) Use the Debye-Hückel Limiting Dilution Law to calculate the activities of both cations in the solution.  (b) What is the actual pH of the GdCl₃ solution?  [Hint: Refer to the WEB lecture pages.]

STRATEGY & ASSUMPTIONS:

      For the problems involving concentrations, activities and equilibrium constants of relatively diluted electrolyte solutions we use the Debye-Hückel Limiting Dilution Law.   We will use this strategy to first calculate the "average" ionic activity coefficient to find out the proton activity in the 0.01 M solution of HCl.

CALCULATION:

We assume that HCl at this concentration is completely dissociated and the concentrations of H⁺ and Cl^- are 0.01 M.

K_eq = [H⁺][A^-] / [HA]

We will then use the Debye-Huckel Limiting Dilution Law to calculate the "average" activity coefficient of the ions:

lg g(H⁺) = -0.509 * z(H⁺)² * I^1/2

and the expression for the ionic strength of the solution, I:

I = 0.5 * {c(H⁺)*z(H⁺)² + c(Cl^-)*z(Cl^-)² + c(Gd⁺³)*z(Gs⁺³)² + 3* c(Cl^-)*z(Cl^-)²}

Note, 0.01 m equals 0.0099969 M (Mw HCl = 36.461, d(36 % HCl)  = 1.179 g cm^-3) and can be approximated by 0.01 mol/L.  Note that for the concentration of ionic strength we must include ALL ions present in the solution:

I = 0.5*{{0.01 * 1² + 0.01^* 1² } + {0.001 * 3² + 3* 0.001^* 1²} = 0.01 + 0.006 = 0.016

and

lg g(H⁺) = -0.509 * 1² * (0.016^)1/2 = -6.44*10^-2 = -0.0644

which gives for the ion activity coefficient,

g_(H+)= 10^-0.0644 = 0.862  -- close to one

For gadolinium we have:

lg g(Gd⁺³) = -0.509 * 3² * (0.016^)1/2 = -5.79*10^-2 = -0.0579

g_(Gd+3) = 10^-0.0579 = 0.263  -- much smaller than one

Now we can calculate the actual pH of the HCl solution:

pH = -lg a(H⁺) = - lg {g(H⁺) * [H⁺]} = -lg g(H⁺) -lg [H⁺] = -0.0644 - lg 0.01  = + 1.94

It's all electrical

2.    (25 pts) A 0.0028 M solution of charged peptide, Z_M = 21, dialyzed against 0.0813 M CsCl, shows Donnan potential of + 9.10 mV.   At what approximate polypeptide concentration will the membrane potential be twenty times lower?  [Hint: Refer to the WEB lecture pages.]

STRATEGY & ASSUMPTIONS:

Use the standard formulas for Donnan potential [Check the WEB notes!] to calculate "r", the ratio of the concentration of the inside and outside concentration of Cs⁺ and Cl^- ions and from there estimate the concentration of the macromolecule, C_M.

CALCULATION:

                                                             V = - RT/ZF * ln r

and "r" is given as:

r(1) = exp [-V * ZF / RT] = exp[-0.0091 * 1 * 96472 / 8.3142 * 298.12] = exp[-0.342] = 0.7105 ≈ 0.71

For the twenty times lower potential, "r" will be:

r(2) = exp[-0.0091/20 * 96472 / 8.3142 * 298.15] = exp[-0.342/20] = 0.983

Now you use the expression [Check the WEB notes on Donnan potential]:

r = - (Z_Mc_M)/2c +  [(Z_Mc_M)/2c)² + 1]^1/2

Take a look at this expression; at the initial conditions the RESOLVENT, [[(Z_Mc_M)/2c)² + 1]^1/2  = 1.0634 - close to one.  For a twnety time lower Donnann concentration the concntratino of the macromolecule, CM(2) 2ill have to be lower and the resolvent will be muc closer to one.  So we can approximate the [(Z_Mc_M)/2c)² + 1]^1/2  by 1, to write:

r(2) = - (Z_M(2)* c_M)/2c +  1

and from there we get:

Z_M(2) ≈ - [0.983 -1] * 2c / Z_M ≈ 1.3*10^-4 M.

Habeas corpus: DUI

3.    (25 pts) The "Breathalyzer test" -- used by law enforcement agents to detect the presence of ethanol vapors in the "expired" air of vehicle drivers -- is based on the oxidation of ethanol to acetaldehyde, E^ө₂₉₈ (CH₃CHO/CH₃CH₂OH) -0.197 V, and the reduction of (orange) bichromate, Cr₂O₇^-2,  to (green) Cr⁺³ salt. [Check the textbook for the standard potential.]  (a) Write down an acceptable stoichiometric equation for this reaction.  (b) Calculate the E^ө, DG^ө and K_eq for this reaction.  (c) Is the reaction  spontaneous?

ANSWER:

(a) 3 CH₃CH₂OH  + 2 Na₂Cr₂O₇ + 8 H₂SO₄  + 3 CH₃COOH + 2 Cr(SO₄)₃ + Na₂SO₄  +  11 H₂O

(b) E^ө₂₉₈ = 1.397 V           Δ G^ө₂₉₈ = -781.357 kJ/mol            K_eq = 3.89*10¹³¹

(c) Yes!

 

You don't need pH paper

4.    (25 pts) A hydrogen electrode and a normal calomel electrode give an electromotive force, E = 0.435 V, at certain solution at 25 °C.  (a) What is the pH of the solution? (b) What is the value of a_H+?

ANSWER:

pH = 2.789        a(H⁺) = 1.625*10^-3