Question

1.)   A solution of 0.40 M HCl is found to have a pH of 0.52. Know that pH, when properly defined, is given in terms of activities, or pH º - log₁₀[a_H+].

a.)    Calculate the ideal solution pH, the activity of this solution and the activity coefficient.

b.)   Use the Debye-Huckel Limiting Law to calculate the pH and compare to experimental and ideal. Explain any differences.

Answer 1

HCl + H2O <==> H3O⁺ + Cl^-

pH = -log(a_H+)

In terms of activity we have to:

0.52 = -log(a_H+)

10^0.52 = 10^-log(a_H+)

a_H+ = 3.31 M

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From activity we can find:

a_H+ = x_h+ * y_h+

where:

x_h+ is the mole froction of the mixture

y_h+ is the activity coefficient

y_h+ = 3.31M/0.40M

y_h+ = 8.275

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Ideal pH

HA <==> H⁺ + A^-

0.40 0 0

-0.40 +0.40 +0.40

0.00 0.40 0.40

[H⁺] = 0.40 M

pH = -log(H⁺) = -log(0.40)

pH = 0.39

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b) The Duckel-Huckel limiting law

The Debye-Hückel Limiting Law, is applicable only for very low concentrations of solute (until unless contributions of the non-ideality become important). This law helps us in the calculation of the mean activity coefficient from basic properties of the solution:

Log γ_i = -AZ_i²√I

where;

z₊ and z_- are the charge number of respectively the cation and the anion concerned.

A is an empirical parameter, dependant upon the solvent and temperature (eg, for a solution in water at 298K, A = 0.509).

I is the ionic strength of the solution,

I =1/2 [ $\sum$ C_iZ_i²]

Cl^- charge = -1

H⁺ charge = +1

I = 1/2 (Concentration of HCl × # of H ions × (Charge of H)²) + (Concentration of HCl in M × # of Chloride ions × (Charge of Chloride)²)]

I = 1/2 [(0.40 M × 1 × (+1)²) + (0.40 M × 1 × (−1)²)] = 0.80 M / 2 = 0.40 M

ln y_H+ = -0.509 * |(+1)(-1)| * sqrt(0.40)

e^{ln y_H+} = e^0.322

y_H+ = 1.37

a_H+ = 0.40 * 1.37

a_H+ = 0.552

pH = -log(a_H+)

pH = -log(0.552)

pH = 0.25