Find specific conductance of $\pu{0.1 M}$ acetic acid given that its molar conductivity is $\pu{4.6 S cm^2 mol^-1}$.

I used the formula $\Lambda_\mathrm m = \kappa/c$, where $\Lambda_\mathrm m$ is the molar conductivity, $\kappa$ is the specific conductance and $c$ is the concentration of the electrolyte in $\pu{mol/L}$.

What's confusing me is the value of $c$. At first I thought it should be the given molarity, but according to my text book $c$ is the concentration of the electrolyte. Now the acetic acid molecule does not conduct electricity, rather its constituent ions do. So shouldn't the value of $c$ be the concentration of its ions which turns out to be $\pu{\sqrt{1.8} \times 10^{-3} mol/L}$ (using the equation of the acidic dissociation of acetic acid)?

Also the answer given was $\pu{4.6E-4 S cm^-1}$.


1 Answer 1


With the data provided with your problem I suspect you are only supposed to carefully use provided formula with the corresponding units of measurements. You need to convert the concentration from $\pu{mol L^-1}$ to $\pu{mol cm^-3}$ considering $\pu{1 L} = \pu{10^3 cm^3}$:

$$\kappa = \Lambda_\mathrm m c = (\pu{4.6 S cm^2 mol^-1})(\pu{0.1E-3 mol cm^-3}) = \pu{4.6E-4 S cm^-1}$$

  • $\begingroup$ So is my reasoning correct, that is, should 'c' actually be substituted as (sqrt 1.8) x 10^-3? $\endgroup$
    – Azulene
    Jul 15, 2017 at 14:07
  • $\begingroup$ @Azulene No, I have no idea where this (sqrt 1.8) x 10^-3 value comes from. I used $\pu{0.1 M}$. You are given neither the dissociation constant of acetic acid, nor the temperature. And please get to know how to use LaTeX for proper formatting of math and chemical equations. $\endgroup$
    – andselisk
    Jul 15, 2017 at 14:15

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