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

1 Answer
Oct 16, 2016

Answer:

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"#