Assuming we started from a initial volume [itex]V_i[/itex] 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

[tex]\chi = c \dfrac{n}{V}[/tex]

with [itex]c[/itex] a proportionality constant and [itex]n[/itex] the total number of moles of free electrolytes in solution. The conductivity read by the instrument by adding a [itex]V_a[/itex] amount of titrant reagent will be

[tex] \chi = c \dfrac{n}{V_i + V_a} [/tex]

while the correct conductivity in absence of dilution is given by the formula

[tex] \chi_{\text{eff}} = c \dfrac{n}{V_i} [/tex]

so

[tex] c \cdot n = \chi_{\text{eff}} \cdot V_i [/tex]

replacing

[tex] \chi = \dfrac{\chi_{\text{eff}} \cdot V_i}{V_i + V_a} [/tex]

[tex] \chi \cdot (V_i + V_a) = \chi_{\text{eff}} \cdot V_i [/tex]

we get

[tex] \begin{equation} \chi_{\text{eff}} = \chi \cdot \dfrac{V_i + V_a}{V_i} \end{equation} [/tex]