# Molar conductivity calculation in a titration curve

At an exam for analytical chemistry, I was asked to calculate the molar conductivity of a solution for an ongoing titration. I just solved it the way our professor showed it to us but I have some doubts if it even is correct like this.

The question:

A diluted solution of hydrobromic acid is titrated with 2 equivalents of sodium hydroxide solution. Assume that the solution of $$\ce{HBr}$$ is diluted enough that the molar conductivity $$\Lambda_0$$ is the same as when at infinite dilution, and that volume change is negligible because $$\ce{NaOH}$$ is very concentrated. Calculate the values of molar conductivity at start of titration, equivalence point and 1 equivalent of $$\ce{NaOH}$$ beyond the equivalence point.

The given solution:

We just need to look at the tabulated molar conductivities. As there are only $$\ce{H3O+}$$ and $$\ce{Br^-}$$ ions, the molar conductivity at the start is $$\pu{349.7 S cm^2/mol} + \pu{78.1 S cm^2/mol} = \pu{427.8 S cm^2/mol}$$, respectively. At the equivalence point, as there are only sodium and bromide ions, the molar conductivity is $$\pu{50.1 S cm^2/mol} + \pu{78.1 S cm^2/mol} = \pu{128.2 S cm^2/mol}$$, and at $$\pu{1 eq}$$ above the equivalence point, we got one equivalent of $$\ce{NaOH}$$ and $$\ce{NaBr}$$, so $$(50.1 + 50.1 + 78.1 + 198)\pu{S cm^2/mol} = \pu{376.3 S cm^2/mol}$$.

My doubts:

First of all, if the value of Λ₀ is independent of concentration for our purposes, why do we need to take it times two for the end point of the titration? All what we are adding is another equivalent of $$\ce{NaOH}$$, but the solution already contains sodium ions. This would be comparable to a situation when we got $$\ce{NaCl}$$ and $$\ce{KCl}$$ dissolved near infinite dilution; would the chloride ions be counted twice?

And secondly, what does the $$\Lambda_0$$ (molar conductivity) even tell us? How would we use it to calculate the actual conductivity ($$\kappa$$) of our solution? All I know is that they are related with each other by

$$\Lambda_0 = \frac{\kappa} {c^*}$$

Again, if we for example had a mixture of $$\pu{1 mol/L}$$ $$\ce{NaCl}$$ and $$\pu{1 mol/l}$$ $$\ce{KCl}$$ (same solution) and we calculated $$\Lambda_0$$, would it basically be $$\Lambda_0 (\ce{NaCl}) + \Lambda_0 (\ce{KCl})$$? And what concentration $$c$$ would we use, equally $$\pu{1 mol/L}$$, or $$\pu{2 mol/L}$$ as there is $$\pu{1 mol/L}$$ of 2 species?

• Welcome to the site. Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers. And with \pu{}, it is easier to type units consistently here, including a space which is retained, including in proximity to a line feed. Commented Jan 13, 2022 at 21:42
• The addition of an aqueous solution of $\ce{NaCl}$ ($\pu{1 L}$ of $c = \pu{1 mol/L}$) to an aqueous solution of $\ce{KCl}$ ($\pu{1 L}$ of $c = \pu{1 mol/L}$) doesn't alter the concentration of $\ce{Cl^-}$, but halves both the concentration of $\ce{K+}$ and $\ce{Na+}$ because amount ($n$), concentration ($c$), and volume ($V$) relate by $n = c \cdot V$. Note the change in volume by mixing the two solutions. Commented Jan 13, 2022 at 21:48
• Yeah what you say makes sennse, but I meant a solution which has both 1M NaCl and KCl in it Commented Jan 14, 2022 at 0:00
• Note that 1 M concentration is quite high and there are significant nonlinear and interfering effects. Molar conductivity at 1 M is significantly different to the limit one at infinite dilution. i.e. $\Lambda_0$. // See also wikipedia.org: Variation_of_molar_conductivity_with_dilution Commented Jan 14, 2022 at 8:57
• Yeah that was an example to simplify the calculation, it should be more around 0.001M or around that Commented Jan 14, 2022 at 14:22

Let's consider the beginning of the titration, with maybe $$0.01 M ~\ce{HBr}$$. Its molar conductivity $$\Lambda_0$$(HBr) is $$350 + 78 = 428 ~ \mathrm{S~ cm^2/mol}$$. Its conductivity is : $$\sigma$$(HBr) = $$\Lambda_0$$(HBr) c = $$\Lambda_o$$(HBr) $$0.01 = 4.28·10^{-3}~ \mathrm{S/m}$$ . Please observe the change of unit ! Here we will admit that $$\Lambda_o = \Lambda$$ for the sake of simplicity !
At the equivalence point, the solution contains no $$\ce{HBr}$$ any more. But it is made of $$0.01$$ M $$\ce{NaBr}$$, if the volume change is negligible. Its molar conductivity $$\Lambda_0$$(NaBr) is $$50 + 78 = 128 ~ \mathrm{S~ cm^2/mol}$$. Its conductivity $$\sigma$$ is : $$\sigma$$(NaBr) = $$\Lambda_0$$(NaBr) · $$0.01 = 1.28·10^{-3}~ \mathrm{S/m}$$
If the double amount of $$\ce{NaOH}$$ is added, the solution is $$0.01$$ M $$\ce{NaBr}$$ + $$0.01$$ M $$\ce{NaOH}$$. The two molar conductivities are : $$\Lambda_0$$(NaBr) is $$50 + 78 = 128 ~ \mathrm{S~ cm^2/mol}$$ and $$\Lambda_0$$(NaOH) is $$50 + 198 = 248 ~ \mathrm{S~ cm^2/mol}$$. The conductivities $$\sigma$$ of the two components are :
• For $$\ce{NaBr}$$ : $$\sigma$$(NaBr) = $$\pu{1.28 10^{-3}}$$ S/m
• For $$\ce{NaOH}$$ : $$\sigma$$(NaOH) = $$\Lambda_0$$(NaOH) · $$0.01$$ = $$2.48~ 10^{-3}$$ S/m
As the conductivities are additive, the mixture containing both $$\ce{NaBr}$$ and $$\ce{NaOH}$$ has a conductivity equal to $$\pu{1.28 10^{-3} + 2.48 10^{-3} = 3.76 10^{-3}}$$ S/m