Determine the ionic strength, μ, for each of the solutions. Assume complete dissociation of each salt and ignore any hydrolysis reactions.

(a) A solution of 0.00580 M HCl

(b) A solution of 0.00227 M CaBr2

(c) A solution of 0.000752 M Mg (NO3) 2 and 0.000776 M La (NO3) 3

The answers are: (a) 0.0058 M; (b) 0.00681 M; (c) 0.006912 M

Explanation:

Ionic strenght = μ = 1/2 ∑ c z²

Where c is the concentration of each ion (in M) and z is the charge of the ion. So, to calculate the ionic strenght of a solution you have to know the concentration of each ion and their charges. For this, the dissociation equilibrium is required.

a) HCl → H⁺ + Cl⁻

Conc H⁺ = 0.00580 M

Conc Cl⁻ = 0.00580 M

μ = 1/2 x ((Conc H⁺) (+1) ²) + ((Conc Cl⁻) (-1) ²

μ = 1/2 x ((0.00580 M) + (0.00580 M))

μ = 0.00580 M

b) CaBr₂ → Ca²⁺ + 2 Br⁻

Conc Ca²⁺ = 0.00227 M

Conc Br⁻ = 2 x 0.00227 M = 0.00454 M

μ = 1/2 ((Conc Ca²⁺) (+2) ² + ((Conc Br⁻) (-1) ²)

μ = 1/2 (0.00227 M (4)) + (0.00454 M)

μ = 0.00681 M

c) The solution has two dissociation equilibria:

Mg (NO₃) ₂ → Mg²⁺ + 2 NO₃⁻

La (NO₃) ₃ → La³⁺ + 3 NO₃⁻

Conc Mg²⁺ = 0.000752 M

Conc NO₃⁻ = 2 x (0.000752 M) + 3 x (0.000776 M) = 0.003832 M

Conc La³⁺ = 0.000776 M

μ = 1/2 ((Conc Mg²⁺) (+2) ²) + ((Conc NO₃⁻) (-1) ²) + ((Conc La³⁺) (+3) ²)

μ = 1/2 (0.000752 M (4)) + (0.003832 M) + (0.000776 M) (9))

μ = 0.006912 M