How do I express concentration of a substance at the interface between two membranes of different diffusivities and lengths at steady state?

Question

I am lost on this question from my textbook:

"A solute diffuses across two membranes, A and B. The membranes' thickness are $\Delta x_A$ and $\Delta x_B$ respectively. The membranes' diffusivities are $D_A$ and $D_B$ respectively. The concentration of the solute decreases linearly from $C_A$ at the left surface of A to $C_i$ at interface, then to $C_B$ at the right surface of B. Given a steady state diffusion process, derive an expression for $C_i$ in the terms of the other parameters of this system."

The answer sheet as provided in the textbook indicates that the correct solution is: $$C_i = {{D_A C_A\over \Delta x_A}+{D_B C_B\over \Delta x_B}\over {D_A + D_B\over \Delta x_A + \Delta x_B}}$$

I tried to solve this by making the following equation:

$${C_A - C_i\over {\Delta x_A \over D_A}} = {C_i - C_B\over {\Delta x_B \over D_B}}$$

This was wrong because it can't be re-arranged into the correct equation as given by the textbook. My prof hinted that since the system is at steady state, the rate of diffusion through A is equal that of B. How do I solve this?

Answer 1

The answer sheet as provided in the textbook indicates that the correct solution is: $$C_i = {{D_\text{A} C_\text{A}\over \Delta x_\text{A}}+{D_\text{B} C_\text{B}\over \Delta x_\text{B}}\over {D_\text{A} + D_\text{B}\over \Delta x_\text{A} + \Delta x_\text{B}}}$$

The textbook solution is wrong. Let test it by a trivial example, using numerical values $D_\text{A}=D_\text{B}=2$, $Δx_\text{A}=Δx_\text{B}=1$ for 2 identical membranes.

Then, $C_i= \frac{2C_\text{A} + 2 C_\text{B}}{2}=C_\text{A} + C_\text{B}$.

This is obviously wrong and the textbook provides a wrong equation.

---

Your initial equation for equal diffusion rates for the steady state is correct.

$${C_\text{A} - C_i\over {\Delta x_\text{A} \over D_\text{A}}} = {C_i - C_\text{B}\over {\Delta x_\text{B} \over D_\text{B}}}$$

If we express $C_i = f(C_\text{A},C_\text{B})$:

$$C_\text{A}\frac{\Delta x_\text{B}}{D_\text{B}} - C_i\frac{\Delta x_\text{B}}{D_\text{B}} = C_i\frac{\Delta x_\text{A}}{ D_\text{A}} - C_\text{B}\frac{\Delta x_\text{A}}{ D_\text{A}} $$

we get:

$$ C_i = \frac{C_\text{A}\frac{\Delta x_\text{B}}{D_\text{B}} + C_\text{B}\frac{\Delta x_\text{A}}{ D_\text{A}}}{\frac{\Delta x_\text{A}}{ D_\text{A}} + \frac{\Delta x_\text{B}}{D_\text{B}}} $$

This fits the general expectation the result should be in the form of a weighted average:

$$z = \frac{ax+by}{a+b}$$

Using our small test for identical membranes:

$$ C_i = \frac{C_\text{A}\frac{1}{2} + C_\text{B}\frac{1}{2}}{\frac{1}{2} + \frac{1}{2}}=\frac{C_\text{A} + C_\text{B}}{2}$$

what is obviously correct.

---

If we multiply the numerator and the denominator of the equation by $\frac{D_\text{A}D_\text{B}}{\Delta x_\text{A}\Delta x_\text{B}}$, we get the alternative version of the equation. It is almost identical to the textbook solution, but it has the denominator the sum of the fractions and not the fraction of the sums.

$$ C_i = \frac{C_\text{A}\frac{D_\text{A}}{\Delta x_\text{A}} + C_\text{B}\frac{ D_\text{B}}{\Delta x_\text{B}}} {\frac{ D_\text{A}}{\Delta x_\text{A}} + \frac{D_\text{B}}{\Delta x_\text{B}}} $$

Using our small test for identical membranes:

$$ C_i = \frac{2C_\text{A} + 2C_\text{B}}{2 + 2}=\frac{C_\text{A} + C_\text{B}}{2}$$

what is obviously correct.