A synthetic membrane that is permeable only to K ions and water is used to separ

Question

A synthetic membrane that is permeable only to K ions and water is used to separate two aqueous solutions (1 and 2) of equal volume. The initial contents of the solutions, prior to equilibration of K* across the membrane, are: Solution I Solution 2 [KCl] = 0.08 M [NaCl] = 0.05 M [KCl] = 0.05 M [NaCl]= 0.1 M Calculate the membrane potential at 25C after the system equilibrates, and make sure to specify which solution has the more positive potential. What are the (approximate) equilibrium concentrations of the three ions in Solution 1 and in Solution 2?

Explanation / Answer

Membrane Potential is defined as difference in voltage across the cell.

We use Nernst's equation to calculate membrane potential which is-

(Membrane Potential at equilibrium) Veq= (RT/zF) log([X]out/[X]in).

R= Universal Gas constant 8.314 J.K-1.mol-1 (Joules per Kelvin per mole)

T= Temperature 298 K

F= Faraday's constant 96485 C.mol-1

Z= ionic charge i.e. 1 for K+

Putting the given values in Nernst's equation, Veq= -12.068 mV

K+ ions will move across the membrane towards low concentration of K+ ions, so solution 2 will start building up positive potential and it will have more positive potential at equilibrium.

K+ ions will cross until they are evenly distributed in both solutions, so the concentration of K+ at equilibrium will be mean of two concentrations, i.e., 0.065 M. The concentration of other ions will not change because membrane is only permeable to K+

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