Question

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

Answer 1

The ionic strength of 0.0087 mol/L `"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.

`color(blue)(bar(ul(|color(white)(a/a) I = 1/2sum_(i = 1)^nc_iz_i^2color(white)(a/a)|)))" "`

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"`