# Conductometric titration curve plotting

In the conductometric titation of HCl vs NaOH, the conductance initially decreases, reaches a minimum at equivalence point, and then the slope of the curve becomes positive:

How can this be modelled like I could do for pH?

Edit: This is my attempt. Is it correct?

• The site expects that you write explicit compact summary of your prior effort to answer the question, based on your knowledge and on searching for existing related info or answers. It would prevent others to tell you what you already know or what you could easily find yourself. Jun 11, 2022 at 17:44
• I don't know where to start. Will Kohlrausch law be applicable here?
– Shub
Jun 11, 2022 at 18:10
• The start is to understand underlying chemistry.. Modelling comes after. You must know, what you are modelling. Ongoing reactions and conductivity related laws. Jun 11, 2022 at 18:16
• Just from looking at the image without distracting information like correct labels on the graph it should be almost painfully obvious: linear fit on the left and the right, where they intersect is the equivalence point. Jun 13, 2022 at 2:31
• From a technical point: please include everything necessary in the question and don't rely on external source. Links become bringen and then this question becomes useless. Hiding a link under the word source is also not appropriate. Please include a proper reference to the site it comes from. Also make sure you are allowed to use the image. Jun 13, 2022 at 2:34

Expressed in $$\pu{\Omega ^{-1} cm^2 mol^{-1}}$$, the ionic conductance of usual ions like $$\ce{Na+, K+, Ca^{2+}, Cl-, SO4^{2-}}$$ are between $$50$$ and $$80$$. But the two ions $$\ce{H+}$$ and $$\ce{OH-}$$ have a rather high conductivity : $$350$$ for $$\ce{H+}$$ and $$198$$ for $$\ce{OH-}$$. This property explain the change of the conductance observed in the titration of $$\ce{HCl}$$ solution by $$\ce{NaOH}$$.
In the beginning of the titration, [$$\ce{H+}$$] is large. So the conductance of the solution is large. When adding $$\ce{NaOH}$$, the ion $$\ce{H+}$$ is replaced by the ion $$\ce{Na+}$$ which has a poor conductivity : the conductance of the solution decreases.
At the equivalence point, the conductance is low, as it is only due to the ions $$\ce{Na+}$$ and $$\ce{Cl-}$$. If more $$\ce{NaOH}$$ is added after the equivalence point, the aded ions $$\ce{OH-}$$ aren't destroyed and stay in the solution. So its conductance increases. This is what you have observed experimentally.