I know how a conductivity meter works. Power is supplied to electrodes and these electrodes change charge. The ions in the solution will be attracted to the oppositely charged electrode. The charge they carry at the electrodes is measured.
The equivalence point may be located graphically by plotting the change in conductance as a function of the volume of titrant added.
I'm not sure this is an accurate enough answer. Is there something I'm leaving out or that could improve the answer?
I’m assuming you’re talking about acid/base titrations.
In these, you typically start with an acid (maybe strong) and slowly add a strong base until you reach the equivalence point (classical, using indicator) or are sufficiently past it (for plotting, using a conductivity meter). Of course, switching acid and base is possible but for the ease of discussion I will consider only acids to be titrated with bases.
Say for example your mixture included hydrochloric acid and acetic acid. Hydrochloric acid (strong) can be considered fully dissociated, acetic acid (weak) mainly undissociated in the initial solution. Thus, your solution contains significant concentrations of: $$\ce{H3O+\qquad Cl- \qquad CH3COOH}\tag{1}$$ Now you start adding a base such as sodium hydroxide. The most significant macroscopic observation is that the concentration of hydronium ions $\ce{H3O+}$ decreases while additional sodium ions are added. At half-way towards having titrated away all hydrochloric acid, you would have the following species in solution: $$\ce{\underset{less}{H3O+}\qquad Cl- \qquad Na+ \qquad CH3COOH}\tag{2}$$ At the equivalence point of hydrochloric acid, we can consider the significant portion of hydronium to be consumed; remember that only a small percentage of acetic acid is dissociated, meaning most of it is not. Thus, we have the species: $$\ce{Cl- \qquad Na+ \qquad CH3COOH}\tag{3}$$ Henceforth, any additional hydroxide added will, in a macroscopic and sighlty simplified view, only serve to deprotonate the acetic acid to acetate. Thus, half-way towards the equivalence point of acetic acid, we can observe the following species in significant concentrations: $$\ce{Cl- \qquad Na+ \qquad \underset{less}{CH3COOH} \qquad CH3COO-}\tag{4}$$ Finally, we will reach the second equivalence point at which all acetic acid has been consumed to acetate, meaning the solution features: $$\ce{Cl- \qquad Na+ \qquad CH3COO-}\tag{5}$$ Hitherto, we have not observed any hydroxide because it was immediately quenched. Now, however, all species that would be able to quench hydroxide have been consumed. Thus, any additional sodium hydroxide will remain sodium hydroxide, giving us the following species: $$\ce{Cl- \qquad Na+ \qquad OH- \qquad CH3COO-}\tag{6}$$
Now that the types of species in each solution has been established, which influence the conductivity and how? Water by itself does not conduct electricity but water with dissolved ions does. Therefore, we would immediately expect the solution’s conductivity to rise from point (3) to point (5) since a non-ion is converted to an ion (and sodium ions are constantly being added). This is indeed observed: a slow but steady rise in conductivity from (3) to (5).
The key, however, is to realise what happens from (1) to (3) (which also influences our understanding of (6)). Here, it may seem that hydronium ions are constantly being replaced by sodium ions, meaning that the total ion count should remain constant. Indeed, it approximately does. But hydronium (and hydroxide) is a much stronger conductor than any other ion.
Typical ions would conduct electricity by being drawn towards the anode or cathode depending on their charge. This is also, in general, what happens to hydronium (or hydroxide) ions. However, because they are defined merely by an excess (or missing) proton added to (or substracted from) a water molecule, they have the ability to move much more rapidly following the mechanism below:
$$\begin{multline}\ce{cathode^-\bond{...}H-(OH)\bond{...}H-(OH)\bond{...}H-(OH)\bond{...}H-(OH)\bond{...}H-(OH)\bond{...}H-OH2+} \\\ce{-> cathode-H\bond{...}(OH)-H\bond{...}(OH)-H\bond{...}(OH)-H\bond{...}(OH)-H\bond{...}(OH)-H\bond{...}OH2}\end{multline}\tag{7}$$
Because this proton-exchange mechanism (the Grotthuss mechanism) allows both hydronium and hydroxide travel speeds much more rapidly than diffusion control, these two ions will induce a much higher conductivity if present. Therefore, our overall graph will feature:
To find points 3 and 5, the best method is to determine sufficiently many data points of the linear sequences 2, 4 and 6 and extend the straight lines. The crossing points represent the equivalence points 3 and 5.
What I have outlined here with two acids also, naturally, works for a single acid; depending on the nature of the acid you will either not see 1/2 or not see 4.
