# 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. – andselisk Jul 15 '17 at 14:15