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4 years ago

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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

1 answer:

kow [346]4 years ago

6 0

Answer:

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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