# Explain estimating intercellular fluxes

Formula that I am using to calculate intercellular fluxes: $$J = \frac{4D}{\pi d} \big([C]_1 - [C]_2\big),$$ where the flux through a hole is $$d$$ of a solute with a diffusion coefficient $$D$$ that is present in the signaling cell at concentration $$[C]_1$$ and in the receiving cell at $$[C]_2$$.
($$d = \pu{4nm}$$ , $$D = \pu{10^{-5} cm^2/s}$$, $$[C]_1 - [C]_2 = \pu{100 \mu M}$$)

Step 1: Standardize the units of $$d$$ and $$D$$: $$D = \pu{10^{-5} cm^2//s} = \pu{10^9 nm^2//s}$$

Step 2: Substitute values in the formula \begin{align} J &= \frac{4 \cdot \pu{10^9 nm^2}}{3.14 \cdot \pu{4 nm s}} \left(\pu{100 \mu M}\right)\\ J &= \frac{\pu{10^9 nm}}{\pu{3.14 s}}\left(\pu{10^2 \mu M}\right) \end{align}

You can compute it even further but my answer is not correct as the correct answer is $$2.4 \times 10^5$$ molecules per pore per second.
This problem is an example from a presentation. Here is the slide:

Kindly explain what I am doing wrong.

• Chem+Math Expression formatting reference: MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial // MathJax is preferred not to be used in CH SE Q titles. Commented Sep 10, 2023 at 6:49
• Welcome to Chemistry StackExchange. There are at least twi problems here. First, you do not give us the area of the hole. Second, if you want to get results in molecules instead of moles, you have to multiply with Avogadro's constant. See also this worked example. Commented Sep 10, 2023 at 7:27

I believe the equation has mistakes: Why is the diameter of the hole ($$d$$) in the denominator? This means that the larger the hole, the smaller the flux!?

Let me try to give the whole calculation. The flux $$J$$ through a hole of thickness $$l$$ of a solute with a diffusion constant $$D$$ and the concentration difference $$\Delta c$$ is given by the Fick's law:

$$J = D\frac{\Delta c}{l}.$$

Note that the flux is a quantity that tells us what is the solute current per unit area, namely $$J = \Phi/A$$.

Let's assume the thickness of the pore is $$l=\ce{4 nm}$$, then given your values of $$D=\ce{10^{-5} cm^2 s^{-1}}$$ and $$\Delta c=\ce{100 \mu M}$$ we obtain:

$$J = \ce{2.5 \cdot 10^{-2} mol m^{-2} s^{-1}}.$$ We can convert this to a number flux by multiplying with Avogadro's constant: $$J_N = J N_{\rm A} = \ce{ 1.5 \cdot 10^{22} m^{-2} s^{-1}}.$$

To calculate the ion current through one pore, we multiply flux with the area of the pore, which is $$A=\pi (d/2)^2$$ for a circular hole with diameter $$d$$:

$$\Phi_N = J_N A = J_N \pi (d/2)^2 = \underline{\ce{1.5\cdot10^5 s^{-1}}}.$$

• I don't see the diameter $d$ in that equation. Commented Sep 10, 2023 at 14:36
• @Martin-マーチン, ah, I've introduced $d$ prematurely in the text. I'll fix it. Commented Sep 10, 2023 at 14:51