Question

(b) A vessel separated into two parts, A and B, by a bilayer lipid membrane contained an aqueous solution of KCl with concentration 0.01 mol L-1 in part A and 0.1 mol L-1 in part B and of NaCl with concentration 0.1 mol L-1 in part A and 0.01 mol L-1 in part B. The membrane included potassium channels, which made it permeable for K+ ions. The temperature of the vessel was kept at 30 o C. (i) Calculate the electrical potential between the two sides of the membrane (membrane potential). (ii) Explain how an increase of temperature and an addition of NaCl to part B will affect the membrane potential. Assume ideal behaviour of the solutions.

Answer 1

Answer (i)

Part A of vessel:

0.01 M KCl + 0.1 M NaCl

[ K+] = 0.01 M

[Na+] = 0.1 M

[Cl-] = 0.11 M

Part B of vessel:

0.1 M KCl + 0.01 M NaCl

[K+] = 0.1 M

[Na+] = 0.01 M

[Cl-] = 0.11 M

As the membrane is permeable to K⁺, the concentration of K+ will ultimately becomes same on both sides i.e. 0.055 M

[Cl-] is already equal on both sides.

So, the membrane potential (E_m) will be due to Na+

E_m = (2.303RT/zF) log[Na+]_A/[Na+]_B

R = 8.314 J/K mol

F = 96485 C/mol

T = 30°C = 303.15 K

z = +1 C

[Na⁺]_A = 0.1 M

[Na⁺]_B = 0.01 M

Substituting values,

E_m = (2.303*8.314*303.15/1*96485) log(0.1)/(0.01)

= 0.06015 V

= 60.15 mV

Answer (ii)

Increase in temperature will increase the magnitude of membrane potential. It is because the membrane potential is directly proportional to temperature.

Addition of NaCl to part B will further decrease the concentration difference around the membrane and hence the magnitude of membrane potential will decrease.