Question

(a) A vessel is separated into two parts, A and B, by a bilayer lipid membrane containing Na+ channels. Part A holds an aqueous solution of NaCl with concentration 0.15 mol L-1 and of disaccharide sucrose with concentration 0.01 mol L-1 . Part B holds an aqueous solution of NaCl with concentration 0.015 mol L-1 . Calculate the electrical potential between the two sides of the membrane at 30 oC and at 50 oC. Neglect the effect of temperature on the molarity of the solutions. Assume that the membrane is not permeable for glucose and for Cl ions.

Answer 1

Here the species being transported is Na+1 because Na+1 channel is selectively permeable to Na+

the electrode potential is given by the nernst equation:

E = (RT/zF) ln ([Na+]₁ / [Na+]₂)

here E is the electric potential

z is the charge on ion thus Na = 1

R is the gas constant , R = 8.1345 J.mol^-1K^-1

F is the Faraday Constant F = 96485 C/mol

T is the temperature in Kelvin.

[Na+]₁ = 0.15 M is the concentration on side 1

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

case 1 at T =30^oC = 30 + 273.15 = 303.15 K

putting all the values

E =[(8.1345 J.mol^-1K^-1) (303.15 K) / ((1) ( 96485 C/mol)) ] ln (0.15 / 0.015)

E = (2466 / 96485) X 2.302

E = 0.0588 V

case 2 at T =50^oC = 30 + 273.15 = 323.15 K

putting all the values

E =[(8.1345 J.mol^-1K^-1) (323.15 K) / ((1) ( 96485 C/mol)) ] ln (0.15 / 0.015)

E = (2628.66 / 96485) X 2.302

E = 0.0627 V

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