-1
$\begingroup$

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 $\chi$ value should be corrected for the dilution effect. Which of these two is the correct formula?

$$\begin{equation}\tag{1} \chi_{\text{eff}} = \chi_{\text{read}} \cdot \dfrac{V_i + V}{V_i} \end{equation}$$

or

$$\begin{equation}\tag{2} \chi_{\text{eff}} = \chi_{\text{read}} \cdot \dfrac{V_i}{V_i + V} \end{equation}$$

where

  • $\chi_{\text{eff}}$ = the effective value of $\chi$
  • $\chi_{\text{read}}$ = the value of $\chi$ read by the device
  • $V_i$ = the initial volume of the solution
  • $V$ = the volume of titran reagent added
$\endgroup$
1
  • 2
    $\begingroup$ "$\chi_{\text{eff}}$ = the effective value of $χ$" — for some reason this way of defining a variable reminds me of Lem's sepulki:) (since $χ$ is not defined anywhere else in the post) $\endgroup$ – andselisk Jul 19 '20 at 19:37
0
$\begingroup$

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 $$ \begin{equation} \chi_{\text{eff}} = \chi \cdot \dfrac{V_i + V_a}{V_i} \end{equation} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.