# How is the formula of mean activity coefficient derived?

The mean activity coefficient is defiend as follows:

$$\gamma_\pm = (\gamma_+\gamma_-)^{1/2}.\tag{1}$$

If Debye-Hückel equation

$$-\log\gamma_i = 0.5z_i^2\mu^{1/2}\tag{2}$$

is used, then the mean activity coefficient has the form

$$-\log\gamma_\pm = 0.5|z_+z_-|\mu^{1/2},\label{eqn:3}\tag{3}$$

where $$\mu$$ is the ionic strength, $$z_i$$ is the charge of species $$i,$$ $$\gamma_i$$ is the activity coefficient.

But I'm confused about the equation \eqref{eqn:3}. It can be found as equation \eqref{eqn:3-38} in Snoeyink and Jenkins' Water Chemistry [1]:

[…] developed for ionic strengths of less than approximately $$\pu{5E-3}$$ and can be stated as

$$-\log\gamma_i = 0.5Z_i^2\mu^{1/2}.\tag{3-34}$$

Because anions cannot be added to a solution without an equivalent number of cations (and vice versa). it is impossible to determine experimentally the activity coefficient of a single ion. Therefore, Eqs. 3-34, 3-35, and 3-36 cannot be verified directly. However, it is possible to define, and measure experimentally, a mean activity coefficient, $$\gamma_\pm,$$ as,

$$\gamma_\pm = (\gamma_+\gamma_-)^{1/2}.\tag{3-37}$$

The Debye-Hückel and Güntelberg relationships can be extended to the mean activity coefficient thus:

$$-\log\gamma_\pm = 0.5|Z_+Z_-|\mu^{1/2},\label{eqn:3-38}\tag{3-38}$$

Below is my derivation process:

\begin{align} -\log\gamma_+ &= 0.5z_+^2\mu^{1/2};\tag{4.1}\\ -\log\gamma_- &= 0.5z_-^2\mu^{1/2},\tag{4.2} \end{align}

so

\begin{align} -\log\gamma_\pm &= -\log[10^{-0.5z_+^2\mu^{1/2}}\times 10^{-0.5z_-^2\mu^{1/2}}]^{1/2} \tag{5.1}\\ & = -\log[10^{-0.5\mu^{1/2}(z_+^2+z_-^2)}]^{1/2} \tag{5.2}\\ & = 0.25\mu^{1/2}(z_+^2+z_-^2). \tag{5.3} \end{align}

This result is different from $$0.5|z_+z_-|\mu^{1/2}.$$ Can someone tell me which step is wrong in my process?

### Reference

1. Snoeyink, V. L.; Jenkins, D. Water Chemistry; Wiley: New York, 1980. ISBN 978-0-471-05196-1.
• The definition of $\gamma_{\pm}$ is the geometric mean $\sqrt{\gamma_+\gamma_-}$, you have used the sum so obtain a different answer. Jul 25, 2020 at 6:35
• @porphyrin I thought that I used the geometric mean since $\gamma_+\gamma_- = 10^{-0.5z_+^2\mu^{1/2}}*10^{-0.5z_-^2\mu^{1/2}}$. Then I put them into the $\sqrt{\gamma_+\gamma_-}$ and further into $-\log\gamma_\pm$. But the result is different.
– T X
Jul 25, 2020 at 12:02
• Ah, I misread, my mistake. Jul 25, 2020 at 12:11
• This formula is used for Monovalent salts where the ratio of cations to anions is 1:1. In such a scenario, $z_+$ = $z_-$ so your formula and their formula are same. Jul 27, 2020 at 6:56

The problem is you were starting with an expression for the mean activity coefficient for salts where both the cation and anion were monovalent, and then attempting to derive, from this, a general expression for salts of all empirical formulas. To correct this, you need to start with the general expression for the mean activity coefficient of salts of any emprical formula.

If the empirical formula of the salt is of the form $$\ce{A_pB_q}$$, where A is the cation and B is the anion, then the general formula for the mean activity coefficient is:

$$\gamma_{\pm} = \sqrt[^{p+q}]{\gamma^p_+\gamma^q_-}$$

[Adapted from: https://en.wikipedia.org/wiki/Activity_coefficient, where I've taken the Wikipedia expression and substituted $$\gamma_+$$ for $$\gamma_A$$, and $$\gamma_-$$ for $$\gamma_B$$, to correspond to the nomenclature you are using.]

Charge balance dictates that $$|z_+| = q$$ and $$|z_-| = p$$. [E.g., in $$\ce{A_2B_3}$$, the charge on A must be 3+, and that on B must be 2- (or some integer multiple of those).]

Substituting, we have:

$$\gamma_{\pm} = \sqrt[^{|z_-|+|z_+|}]{\gamma^{|z_-|}_+\gamma^{|z_+|}_-}$$

If you use that in place of eqn. (3-37) in your post, and work through your calculations again, you will get the result show in eqn. (3-38), which is a general result for any values of p and q.

• Thanks! I completed the calculations here: $\gamma_{\pm} = \sqrt[^{|z_-|+|z_+|}]{\gamma^{|z_-|}_+\gamma^{|z_+|}_-} = \sqrt[^{|z_-|+|z_+|}]{10^{-0.5\mu^{1/2}(z^2_+|z_-|+z^2_-|z_+|)}} = 10^{-0.5\mu^{1/2}(|z_+||z_-|(|z_-|+|z_+|))/(|z_-|+|z_+|)} = 10^{-0.5\mu^{1/2}(|z_+||z_-|)}$ Then take the logarithm of $\gamma_{\pm}$ and multiplied by -1, it becomes eqn. (3-38).
– T X
Jul 27, 2020 at 14:35
• @TX Happy to help! I left the calculation for you to complete because (a) I figured you'd have no trouble completing it, since your original calculations were correct based on what you'd started with (you just started with the wrong expression), and (b) it would allow you to see for yourself that the calculation does give the right answer. Jul 27, 2020 at 19:01