What is the pH difference (△pH) across a membrane at 310 K if the membrane potential...
What is the pH difference (△pH) across a membrane at 310 K if the membrane potential is -0.15 V and the overall Gibb's free energy change across the membrane is -19 .0 kJ mol-1? (R = 8.315 J/mol-K; F = 96,485 J/V-mol). Please show me all the steps of how to solve this. Thank you!
Homework Answers
Consider movement of H+ in and out of membrane due to difference in H+ concentration
Free energy is available for movement of protons from concentration gradient across membrane due to electron transport.
∆G =∆G0+ RT lnQ
∆G = ∆G0 + RT ln [H+ in]/[H+ out]
Because membrane is in equilibrium so ∆G° is 1
∆G = -2.303RT log [H+ in]/[H+ out]
[H+ in]/[H+ out]= ∆pH
Therefore ∆G = -2.303 RT ∆pH
∆pH = ∆G/ -2.303*RT
∆pH = -19000/-2.303*8.314*310
∆pH = 19000/5935. 6
∆pH = 3.20
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