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Solutions A and B are separated by a membrane that is permeable to Ca2+ and impermeable to Cl−. Solution A contains 10 mM CaCl2 , and solution B contains 1 mM CaCl2.
Assuming that 2.3 RT/F = 60 mV, Ca2+ will be at electrochemical equilibrium when

(A) solution A is +60 mV

(B) solution A is +30 mV

(C) solution A is −60 mV

(D) solution A is −30 mV

(E) solution A is +120 mV

(F) solution A is −120 mV

(G) the Ca2+ concentrations of the two solutions are equal

I eliminated A,B and E because of the +ve sign immediately, and G because it doesn't take into account the electrical potential factor. I also understand the general idea (that Ca2+ ions will cross the membrane but Cl- won't so it will stop when solution A is negative enough for the electrical gradient to be > concentration gradient) but I don't know what to do next. Using Nernst equation (here's the version in my bookenter image description here) I got that the equilibrium potential is about -60 mV (I did -61log(10/1)) But the book's answer is D which makes me think I might have a huge misunderstanding in the principle itself.

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    $\begingroup$ Can you try to format your question to be more readable? As-is it is a complete mess. Show your thinking step-by-step: which numbers actually matter to the answer, what is the equation you are using, etc. And remove the bits about your professor and your university being shut down - yes, that sort of thing is affecting all of us, but it doesn't change the on/offtopicness of questions here. $\endgroup$ – Bryan Krause Mar 23 at 16:49
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    $\begingroup$ Thanks for the edits, this is much better looking! One last thing: can you write out the entire Nernst equation, and show where you have substituted numbers for variables? You're very close. $\endgroup$ – Bryan Krause Mar 23 at 22:33
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The Nernst equation should be:

RT/(zF)*ln(Out/In) = 2.3 RT/(zF)*log10(Out/In)

You substituted for 2.3 RT/F (approximately 60 or 61 mV: use 60 because your instructor asked you to), but forgot about z. Calcium is divalent (and positive) so for calcium z=2. Substitute this and you can see how you get your answer.

(Also note the "out/in" which is how you get negative numbers when z is positive and out < in )

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