# How to calculate ionic strength of zwitterionic buffers?

Using the formula for the molar ionic strength from Wikipedia — Ionic strength

$$I = \frac 1 2\sum_{i = 1}^nc_iz_i^2,$$

I have calculated that $$\pu{0.1 M}$$ of HEPES $$(\ce{C8H18N2O4S})$$ would have a strength of $$\pu{1.7 M}.$$ Is this correct?

## 1 Answer

The Ionic Strength is calculated as:

$$I = \frac{1}{2}\cdot\sum{c_iz_i^2},$$

where where $$c_i$$ is the molar concentration of ion i ($$M$$, $$\pu{mol/L}$$), $$z_i$$ is the charge number of that ion, and the sum is taken over all ions in the solution.

This is your HEPES molecule ($$\ce{C8H18N2O4S}$$):

Only one proton at $$\ce{-SO3H}$$ group can dissociate, as far as I see. So in your solution you will have two different types of ions with charge equal to 1 (assuming that molecule will dissociates 100%, and than:

$$I = 0.5\cdot(0.1M\cdot1^2+0.1M\cdot(-1)^2) = 0.1M$$

So your calculations are wrong.

But also be aware that this molecule is Zwitterion! READ THIS Do zwitterions contribute to the ionic strength of a solution? (Stellwagen et. al., Anal Biochem. 2008, 373(2): 407–409) The resume of the articles:

"Hence, electrostatic effects due to zwitterions are important, even if the zwitterions do not contribute directly to ionic strength."

My calculations are primitive. They are also far from the truth, but they show how the principle works for you.

• I think calculation of the ionic strength may be complicated due to the zwitterionic character of HEPES. According to this source, zwitterions have no net contribution to ionic strength, so that HEPES produces only a very small (though likely still non-zero) effect on ionic strength. Commented Feb 28, 2014 at 13:07