Question

Calculate the equilibrium membrane potentials to be expected across a membrane at 37 ∘C, with a NaCl concentration of 0.10 M on the right and 0.01 M on the left, given the following conditions. Consider the process is spontaneous.

Part A

Membrane permeable only to Na+.

Express your answer to three significant figures and include the appropriate units.

Δψ =

Part B

State which side is positive (+)

Part C

Membrane permeable only to Cl−.

Express your answer to three significant figures and include the appropriate units.

Part D

State which side is positive (+)

A. Charge is equalized on both sides of the bilayer.

B. The left side of the bilayer is (+).

C. The right side of the bilayer is (+). Part E Membrane equally permeable to both ions. Express your answer to three significant figures and include the appropriate units. Part F State which side is positive (+) A. The right side of the bilayer is (+). B. The left side of the bilayer is (+). C. Charge is equalized on both sides of the bilayer.

Answer 1

The left side has NaCl in the concentration of 0.01 M i.e. that is the concentration of both Na⁺ ions and Cl^- ions on the left side. Same is the case with NaCl ions on the right side at a concentration of 0.1 M, To determine the membrane potentials across a membrane, the nernst equation is used. It can be explained as follows:

$Memb. Potential = \frac{RT}{zF}ln\frac{[Ion]outside}{[Ion]inside}$

So the memb. potential is expressed in Volts. In the equation,

R= Universal gas constant: 8.314 J. K^-1Mole^-1

T= Temperature in Kelvin (So in this case, T= 273.15 + 37 K =310.15 K)

z= elementary charge on the ions involved (1 in this case)

F= Faraday's constant: 96485 J. V^-1. Mole^-1

[Ion]outside= Ionic concentratiom on the outside in moles

[Ion]inside= Ionic concentratiom on the inside in moles

Here, let us consider the right side to be the 'outside' and the left side to be the 'inside'. To calculate the membrane potential on the inside i.e. left side, the equation will be as follows:

$Memb. Potential = \frac{8.314*310.15}{1*96485}ln\frac{0.1}{0.01}$

Solving the equation we get the potential as 0.061548 V or 61.548 mV (Answer part A)

Part B: The left side has positive charge because the permeability of the membrane for Na⁺ ions has increased the +ve charge on the left side. Where as the cloride ions (-ve) were initially more in concentration on the right side and are still the same. So, the net charge on the left side is positive.

Part C: The membrane potential will have the same value if the membrane becomes permeable to the cloride ions since the concentration of both Na⁺ and Cl^- ions is the same on both sides. So, the membrane potential would still be 61.548 mV

Part D: In this part, considering the membrane to be permeable for cloride ions, the charge on the right side of the bilayer would be positive since the negatively charged cloride ions got equalised through the membrane, but the positively charged sodium ions remain in higher concentration on the right side. So net charge on the right side is + (positive).