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

Bilayer lipid membrane containing Na+ channels, membrane is not permeable for glucose and for Cl ions.

Since , only (Na+) and (Cl-) contributes to electrical potential between the two sides of the membrane.

So, Na+ ions moves freely across membrane , thus after some time we will have :

(Na+)A = (Na+)B

and   (Cl-)A = 0.15 M and (Cl-)B = 0.05 M

Potential difference across membrane (A-->B) : it will be due to Cl- only as , Na+ are in equilibium :

Δ = RT / F ln [ (Cl-)B ] / [ (Cl-)A ]

Δ = RT / F ln [ 0.05 ] / [ 0.15 ]

At , 30 C ( 303 K)

Δ = ( 8.314 J/K-mol *303 K ) / 96500 C/mol ln [ 0.05 ] / [ 0.15 ]  

            ( 1C-V = 1J)               Δ = - 29 mV                      ( 1V = 1000 mV)

At , 50 C ( 323 K)

Δ = ( 8.314 J/K-mol *323 K ) / 96500 C/mol ln [ 0.05 ] / [ 0.15 ]  

                    Δ = - 31 mV                      ( 1V = 1000 mV)