Section 9.1 discusses Donnan equilibrium, in which the presence of an impermeant ion on one side of a membrane, along with other ions that can pass through, causes a potential difference to build up across the membrane. This potential difference exists even though the bulk solution on each side of the membrane is electrically neutral.Today I present two new homework problems based on one of Donnan’s original papers.
Donnan, F. G. (1924) “The Theory of Membrane Equilibria.” Chemical Reviews, Volume 1, Pages 73-90.Here’s the first problem.
Section 9.1Stop and solve the problem using the methods described in IPMB. Then come back and compare your solution with mine (and Donnan’s).
Problem 2 ½. Suppose you have two equal volumes of solution separated by a semipermeable membrane that can pass small ions like sodium and potassium but not large anions like A−. Initially, on the left is 1 mole of Na+ and 1 mole of A−, and on the right is 10 moles of K+ and 10 moles of A−. What is the equilibrium amount of Na+, K+, and A− on each side of the membrane?
In equilibrium, x moles of sodium will cross the membrane from left to right. To preserve electroneutrality, x moles of potassium will cross from right to left. So on the left you have 1 – x moles of Na+, x moles of K+, and 1 mole of A−. On the right you have x moles of Na+, 10 – x moles of K+, and 10 moles of A−.
Both sodium and potassium are distributed by the same Boltzmann factor, implying that
[Na+]left/[Na+]right = [K+]left/[K+]right = exp(−eV/kT) (Eq. 9.4)
where e is the elementary charge, V is the voltage across the membrane, k is Boltzmann’s constant, and T is the absolute temperature. Therefore
(1 – x)/x = x/(10 – x)
or x = 10/11 = 0.91. The equilibrium amounts (in moles) are
left right
Na+ 0.09 0.91
K+ 0.91 9.09
A− 1.00 10.00
The voltage across the membrane is
V = kT/e ln([Na+]right/[Na+]left) = (26.7 mV) ln(10.1) = 62 mV .
Donnan writes
In other words, 9.1 per cent of the potassium ions originally present [on the right] diffuse to [the left], while 90.9 per cent of the sodium ions originally present [on the left] diffuse to [the right]. Thus the fall of a relatively small percentage of the potassium ions down a concentration gradient is sufficient in this case to pull a very high percentage of the sodium ions up a concentration gradient. The equilibrium state represents the simplest possible case of two electrically interlocked and balanced diffusion-gradients.Like this problem? Here’s another. Repeat the last problem, but instead of initially having 10 moles of K+ on the right, assume you have 10 moles of Ca++. Calcium is divalent; how will that change the problem?
Problem 3 ½. Suppose you have two solutions of equal volume separated by a semi-impermeable membrane that can pass small ions like sodium and calcium but not large anions like A− and B−−. Initially, on the left is 1 mole of Na+ and 1 mole of A−, and on the right is 10 moles of Ca++ and 10 moles of B−−. What is the equilibrium amount of Na+, Ca++, A− and B−− on each side of the membrane?Again, stop, solve the problem, and then come back to compare solutions.
Suppose 2x moles of Na+ cross the membrane from left to right. To preserve electroneutrality, x moles of Ca++ move from right to left. Both cations are distributed by a Boltzmann factor (Eq. 9.4)
[Na+]left/[Na+]right = exp(−eV/kT)
[Ca++]left/[Ca++]right = exp(−2eV/kT) .
However,
exp(−2eV/kT) = [ exp(−eV/kT) ]2
so
{ [Na+]left/[Na+]right }2 = [Ca++]left/[Ca++]right
or
[ (1 –2 x)/(2x) ]2 = x/(10 – x)
This is a cubic equation that I can’t solve analytically. Some trial-and-error numerical work suggests x = 0.414. The equilibrium amounts are therefore
left right
Na+ 0.172 0.828
Ca++ 0.414 9.586
A− 1 0
B−− 0 10
The voltage across the membrane is
V = kT/e ln([Na+]right/[Na+]left) = (26.7 mV) ln(4.814) = 42 mV .
I think this is correct; Donnan didn’t give the answer in this case, so I’m flying solo.
Frederick Donnan. From an |
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