Activity Coefficients and pH

Question

What is the activity coefficient of H+ in a solution containing 0.023 lHCl and 0.0090 Ca(ClO4)2? Activity coefficients at various ionic strengths can be found in this table.

activity coef. =?

pH=?

I'm having trouble with these types of questions

Answer 1

What is the activity coefficient of H+ in a solution containing 0.023 lHCl and 0.0090 Ca(ClO4)2? Activity coefficients at various ionic strengths can be found in this table.

First, I assume that IHCl is a typo and that both numbers are Molar concentrations.

Second, I assume that this is a 2-part question. Mixing HCl and Ca(ClO₄)₂ solutions together would be a bad idea (generates toxic, explosive ClO₂ gas!).

For activity coefficients, we can use the Debye-Huckel equation:

log γ = -0.51 * z² * sqrt(u)

1 + α * sqrt(u) / 305

where α = ionic diameter in pm = 900 for H⁺

u = ionic strength = 0.5 * ∑ C_i*z_i² , C = concentration of each ion present

γ = activity coefficient

For adjusted pH,

pH = log { [H⁺] * γ }

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0.023M HCl case

Since dilute HCl is a strong acid (essentially 100% ionized), [H+] = [Cl-] = [HCl] = 0.023

u = [ 0.023*(+1)² + 0.023*(-1)² ]/2 = 0.023

log γ(H⁺) = -0.51 * (+1)² * sqrt(0.023) = -0.0534332

1 + sqrt(0.023) * 900/305

γ(H⁺) = 10^ -0.0534332 = 0.884

Again, since dilute HCl is a strong acid (essentially 100% ionized), [H+] = [Cl-] = [HCl] = 0.023

pH = - log { [H⁺] * γ }

pH = - log (0.023 * 0.884) = 1.69

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0.0090M Ca(ClO₄)₂ case

Dissolving Ca(ClO₄)₂ in water gives

Ca(ClO₄)₂ ==> Ca⁺⁺ + 2 ClO₄^-

For the activity coefficient of H+ (i.e., H₃O⁺), we need the ionic strength u.

u = [ 0.0090*(+2)² + (2*0.0090)*(-1)² ]/2 = [ 0.0090*4 + 0.0090*2*1 ]/2

= 0.0090*6/2 = 0.018

(Note that the concentration of ClO₄^- is 2x that of the Ca⁺⁺. Also, we could include terms in u for hydronium and hydroxyl, but their contribution (at 1E-7 M) in the neutral solution would be negligible compared to the dissolved salt ions.)

Now for the activity of H⁺ from the ionization of water,

log γ(H⁺) = -0.51 * (+1)² * sqrt(0.018) = -0.0490178

1 + sqrt(0.018) * 900/305

γ(H⁺) = 10^ -0.0490178 = 0.893

Note that Ca(ClO₄)₂ is a "neutral" salt formed from a strong acid and a strong base, H₂ClO₄ and Ca(OH)₂ ), and thus forms a neutral solution in pure water. Therefore,

pH = - log { [H+]*γ(H⁺) } = - log ( 1E-7 * 0.893 } = - log (8.93 E-8 )

pH = 7.05