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2 August, 02:46

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

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  1. 2 August, 06:29
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    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
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