How does molar conductivity relate to the salinity of a solution?

Aug 30, 2017

In general, it is not easy to say... At very low concentrations (below $\text{0.01 molal}$), the relationship is that the conductivity is approximately a direct proportionality (linear).

But at "usual" concentrations, I believe the influence of the salinity (salt concentration) tapers off due to increased ion pairing, decreasing the effective number of ions in solution and thus the conductivity.

In that case it may be more of a square root, or even logarithmic direct proportionality as concentrations get beyond $\text{0.1 molal}$ or so.

Well, salinity of water is a type of concentration. It is commonly given as:

$\text{g salt"/"kg sea water}$

And so, it is proportional to the molar concentration, in $\text{mols salt"/"L sea water}$.

If we define that as $c$, then for a solution of ONE strong electrolyte, the molar conductivity is given for low concentrations (less than $\text{0.01 molal}$) as:

${\Lambda}_{m} = {\Lambda}_{m}^{\infty} - \left(A + B {\Lambda}_{m}^{\infty}\right) \sqrt{c}$

where:

• ${\Lambda}_{m}$ is the molar conductivity in ${\text{S"cdot"m"^2cdot"mol}}^{- 1}$.
• ${\Lambda}_{m}^{\infty}$ is the molar conductivity in the limit of infinite dilution. This is obtained by extrapolating a tangent line on a ${\Lambda}_{m}$ vs. $c$ graph to a theoretical ${\Lambda}_{m}$ at $c \approx 0$.
• $A$ and $B$ are constants from the Debye-Huckel theory of electrolytes. For further detail on how to calculate $A$ and $B$, see here.

As $c \to 0$, ${\Lambda}_{m}^{\infty}$ $\text{>>}$ $\left(A + B {\Lambda}_{m}^{\infty}\right) \sqrt{c}$, and ${\Lambda}_{m} \propto c$.

However, this relationship is muddled at higher concentrations, due to ion pairing occurring at higher concentrations, decreasing the influence of concentration on conductivity.