# Specific conductance of 0.1 M acetic acid

Find specific conductance of $\pu{0.1 M}$ acetic acid given that its molar conductivity is $\pu{4.6 S * cm2 * mol^{-1}}$.

I used the formula $\Lambda_m = \kappa/c$, where $\Lambda_m$ 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 $\pu{4.6 \times 10^{-4} S * cm^{-1}}$.

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 $\pu{1 L = 10^{3} cm3}$:
$$\kappa=\Lambda_m \times c = \pu{4.6 S * cm2 * mol^{-1}} \times \pu{0.1 * 10^{-3} mol * cm^{-3}} = \pu{4.6e-4 S*cm^{-1}}$$
• @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. Jul 15 '17 at 14:15