How do you calculate the ionic strength of 0.0087 M NaOH?

Oct 16, 2016

The ionic strength of 0.0087 mol/L $\text{NaOH}$ is 0.0087 mol/L.

Explanation:

The molar ionic strength, $I$, of a solution is a function of the concentration of all ions in that solution.

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} I = \frac{1}{2} {\sum}_{i = 1}^{n} {c}_{i} {z}_{i}^{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

where

${c}_{i}$ is the molar concentration of ion $i$
${z}_{i}$ is the charge on that ion
and the sum is taken over all ions in the solution.

I = 1/2["0.0087 mol/L" × ("+1")^2 + "0.0087 mol/L" × ("-1")^2] = "0.0087 mol/L"