Question

an artificial membrane has an area of 3.10 x 10^-15 m^2 and a thickness of 35.0 m. on one side of the membrane there is pure water, on the other side there is a 0.270 x 10^3 mol/m^3 solution of the ions. if 3.50 x 10^4 ions cross the membrane each second, what is the diffusion coefficient of the membrane for the ion?

Answer 1

The diffusion coefficient of the membrane for the ion is 2.58 x 10^-20 m^2/s.

Step-by-step explanation:

To find the diffusion coefficient of the membrane for the ion, we can use Fick's first law of diffusion, which states that the rate of diffusion is equal to the diffusion coefficient multiplied by the concentration gradient and the area of the membrane.

Given:

• Area of the membrane (A) = 3.10 x 10-15 m2
• Thickness of the membrane (d) = 35.0 m
• Concentration of ions on one side of the membrane (C) = 0.270 x 103 mol/m3
• Number of ions crossing the membrane per second (n) = 3.50 x 104

First, let's calculate the concentration gradient (∆C/∆x), which is the change in concentration per unit distance:

(∆C/∆x) = C/d

Now, let's substitute the values into Fick's first law:

n = D * (∆C/∆x) * A

Plugging in the values and solving for D (diffusion coefficient), we get:

D = n / ((∆C/∆x) * A)

Now, substitute the given values and calculate D:

D = (3.50 x 104) / ((0.270 x 103) / (35.0)) * (3.10 x 10-15)

Calculating D gives us D = 2.58 x 10-20 m2/s