Question

(b) A vessel is separated into two parts, A and B, by a bilayer lipid membrane containing Ca2+ channels. Part A holds an aqueous solution of CaCl2 with concentration 0.1 mol L-1 and part B holds an aqueous solution of CaCl2 with concentration 0.015 mol L-1 . Calculate the electrical potential between the two sides of the membrane at 20 degrees C and at 50 degrees C.

Answer 1

Here, the species being transported is Ca²⁺ because Ca²⁺ channel is selectively permeable to Ca²⁺

The electric potential is given by the Nernst equation,

Here,

E is the electric potential.

z is the Charge on ion. Thus z=2.

R is the gas constant. R = 8.3145 J.mol^-1 K^-1

F is the Faraday's Constant. F = 96 485 C/mol

T is the temperature in Kelvin.

[Ca²⁺]₁ = 0.1 M is the concentration on side 1.

[Ca²⁺]₂ = 0.015 M is the concentration on side 2.

Case 1: T= 20 ^oC = (20+273.15) K =293.15K

Putting all the values, except temperature, we have,

Case 2: T= 50 ^oC = (50+273.15) K =323.15K