Calculating conductivity of a solution given molar conductivities and concentration

Question

From my understanding the conductivity of a solution ($K$) is given by the sum of the conductivities of ions in solution:

$$K = c(K_{m,1}+K_{m,2})$$

Where $c$ is the concentration, and $k_{m1}$ and $k_{m2}$ are the respective molar conductivities of the two ions. For example, with a solution of $\pu{0.1 mol L-1}$ $\ce{NaOH}$:

\begin{align} K &= 0.1(K_{m,\ce{Na+}} + K_{m,\ce{OH-}})\\ K_{m,\ce{Na+}} &= \pu{50 S cm2 mol-1}\\ K_{m,\ce{OH-}} &= \pu{197.9 S cm2 mol-1}\\ K &= 0.1(247.5)\\ K &= \pu{24.75 S m^-1} (?) \end{align}

However I think the units are throwing me off here as my calculations aren't making much sense - I was led to believe the conductivity was much lower.

Could you please offer some help with these calculations?

Answer 1

Just be explicit with all the units, including for the molar concentration, and be sure they cancel out properly. You have $\pu{0.1 mol L-1}$ for the solution concentration, but your $K$ values are in $\pu{S cm2 mol-1}$.

So convert your molar concentration to cubic centimeters (there are $\pu{1e3 cm3}$ in one liter) and you get $\pu{1e-4 mol cm-3}$.

Then when you multiply both values and your units cancel nicely to give you

$$\pu{1e-4 mol cm-3} \times \pu{247.5 S cm2 mol-1} = \pu{2.475e-2 S cm-1}$$