# Conductance of the mixture of two solutions with given resistance values [duplicate]

If $$R_1$$ and $$R_2$$ are the resistances of two solutions of equal volume what is the conductance of the mixture in the same conductivity cell?

In general, I grasped neither the approach, nor the meaning very well.

Do we add the resistances as now both the solution are there to resist, or do we add the conductance or conductivity of them as now both conduct?

As now the volume is doubled, do I divide by 2 or multiply? The area is doubled, but the ions per unit volumes is reduced to half (or maybe not, I'm not sure).

Also, how can we directly add and ignore the inter-particular interaction?

$$\frac{1}{2}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$$

• Conductance is the reciprocal of resistance, so mixing the two equal volumes results in adding the two conductances, assuming the charged species stay independent, i.e., no precipitation, etc. The factor of 1/2 is simply due to the dilution factor: the charged species are in twice the volume. Hope this helps. – Ed V Jun 29 '19 at 18:51
• @EdV But what about the Electrocostatic attractions?? – RandomAspirant Jun 29 '19 at 19:38
• I assume the charged species have concentrations that are in the Goldilocks region: not so high as to need activities and fancy theory, but not so low that water’s conductance cannot be ignored. So the ions just move around randomly. There is no need to worry about electrostatic interactions because the concentrations are not high. And precipitation, gas evolution, complex formation, etc., have all been ruled out. The equation is just a simple approximation: it would certainly fail at high concentrations. – Ed V Jun 29 '19 at 19:45
• Ohk , thanks a lot – RandomAspirant Jun 29 '19 at 19:47

There is an excellent treatment of electrolytic conductivity in Wikipedia. However, the OP’s main question involved a specific equation, pertaining to the conductance of a mixture of equal volumes of two electrolyte solutions, and this was not specifically discussed in the Wikipedia article:

$$\text{conductance} = \frac{1}{2}\left(\frac{1}{R_{1}} + \frac{1}{R_{2}}\right)$$

For simplicity, assume the electrolyte concentrations in the two solutions are in the “Goldilocks” zone: not so high that activities and fancy theory are needed, but not so low that the conductivity contribution due to water’s autoionization has to be taken into account. Neither solution is concentrated, so the ions just bounce around like they would in the unmixed solutions. This would certainly be incorrect in relatively concentrated solutions, but that is not the case at hand.

In this idealized scenario, a solution’s conductivity, i.e., specific conductance, is directly proportional to the mobile charge carrier (i.e., ion) concentrations in the solution. The conductance of a solution, denoted by $$G,$$ is directly proportional to the solution’s conductivity. The resistance of the solution, denoted by $$R,$$ is simply the reciprocal of the conductance and, likewise, the resistivity, i.e., specific resistance, is the reciprocal of the conductivity.

Now consider two electrolyte solutions that can be mixed without causing any chemical reaction. In particular, no precipitate forms, no gas is evolved, no complexes form, and so on. Measurements on the first solution provide values of $$G_1$$ and $$R_1$$, while measurements on the second solution provide values of $$G_2$$ and $$R_2$$. If equal volumes of the two solutions were well mixed, and a conductivity measurement was performed on the mixed solution, the conductivity of the mixed solution, $$G_\mathrm{mix}$$, would be

$$G_\mathrm{mix} = \frac{G_1}{2} + \frac{G_2}{2} = \frac{1}{2}(G_1 + G_2)$$

since each solution, when mixed with an equal volume of the other solution, had its constituent ion initial concentrations halved. But the total conductance is just the sum of the two conductances, $$G_1 = 1/R_1$$, and $$G_2 = 1/R_2$$, so

$$G_\mathrm{mix} = G_\mathrm{tot} = \frac{G_1}{2} + \frac{G_2}{2} = \frac{1}{2}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$$

Like all approximations, this one is only as good as the foundational assumptions upon which it is predicated.