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### Topic: Conductometric titrations: dilution effect  (Read 193 times)

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#### Mimic

• Regular Member
•   • Posts: 25
• Mole Snacks: +0/-0 ##### Conductometric titrations: dilution effect
« on: July 19, 2020, 03:45:48 PM »
Since the conductivity of a solution changes linearly with the electrolyte concentration, during a conductometric titration the volume increases with each addition of the titrant aliquot. For this reason, each χ value should be corrected for the dilution effect.
Which of these two is the correct formula?

$$\chi_{\text{eff}} = \chi \cdot \dfrac{V_i + V}{V_i}$$

or

$$\chi_{\text{eff}} = \chi \cdot \dfrac{V_i}{V_i + V}$$

where

- $\chi_{\text{eff}}$ = the effective value of χ
- $\chi$ = the value of $\chi$ read by the device
- $V_i$ = the initial volume of the solution
- $V$ = the volume of titran reagent added

Can you help me?

#### Borek ##### Re: Conductometric titrations: dilution effect
« Reply #1 on: July 19, 2020, 04:01:38 PM »
Try to derive it, no need to guess.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### Mimic

• Regular Member
•   • Posts: 25
• Mole Snacks: +0/-0 ##### Re: Conductometric titrations: dilution effect
« Reply #2 on: July 19, 2020, 04:10:28 PM »
If χ decreases with increasing dilution, the second should be correct

#### Borek ##### Re: Conductometric titrations: dilution effect
« Reply #3 on: July 19, 2020, 05:46:42 PM »
You are still guessing, this is a thing that is easy to just calculate.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### Mimic

• Regular Member
•   • Posts: 25
• Mole Snacks: +0/-0 ##### Re: Conductometric titrations: dilution effect
« Reply #4 on: July 19, 2020, 06:26:43 PM »
Assuming we started from a initial volume $V_i$ of about 100 ml, some considerations must be made: at the end of the titration, the final volume, following the addition of the titrant reagent, will be more than 100 ml. Remembering that conductivity decreases linearly with dilution, the value provided by the instrument is influenced by the effect of dilution. Specific conductivity and concentration are directly proportional to each other
$$\chi = c \dfrac{n}{V}$$
with $c$ a proportionality constant and $n$ the total number of moles of free electrolytes in solution. The conductivity read by the instrument by adding a $V_a$ amount of titrant reagent will be
$$\chi = c \dfrac{n}{V_i + V_a}$$
while the correct conductivity in absence of dilution is given by the formula
$$\chi_{\text{eff}} = c \dfrac{n}{V_i}$$
so
$$c \cdot n = \chi_{\text{eff}} \cdot V_i$$
replacing
$$\chi = \dfrac{\chi_{\text{eff}} \cdot V_i}{V_i + V_a}$$
$$\chi \cdot (V_i + V_a) = \chi_{\text{eff}} \cdot V_i$$
we get
$$\begin{equation} \chi_{\text{eff}} = \chi \cdot \dfrac{V_i + V_a}{V_i} \end{equation}$$