For a polyprotic acid with two dissociation steps, the fraction alpha H2A is given by which expression? Denominator is ([H3O+]^2+[H3O+]K1+K1K2).

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Multiple Choice

For a polyprotic acid with two dissociation steps, the fraction alpha H2A is given by which expression? Denominator is ([H3O+]^2+[H3O+]K1+K1K2).

Explanation:
In a polyprotic acid with two dissociation steps, the distribution of species is found by relating each form to the fully protonated form through the stepwise constants, then normalizing by the total amount. Let the species be H3A, H2A-, and HA2-. The equilibria are H3A ⇌ H+ + H2A- with K1, and H2A- ⇌ H+ + HA2- with K2. Express the less protonated forms in terms of the fully protonated form: [H2A-] = K1 [H3A] / [H3O+] [HA2-] = K2 [H2A-] / [H3O+] = (K1 K2 [H3A]) / [H3O+]^2 Total amount C = [H3A] + [H2A-] + [HA2-] = [H3A] [1 + K1/[H3O+] + K1K2/[H3O+]^2]. Therefore the fraction of the fully protonated form is alpha_H3A = [H3A]/C = [H3O+]^2 / ( [H3O+]^2 + [H3O+] K1 + K1 K2 ). This matches the given expression with [H3O+]^2 in the numerator. The other two fractions would be [H3O+] K1 / D and K1 K2 / D for the remaining forms, respectively.

In a polyprotic acid with two dissociation steps, the distribution of species is found by relating each form to the fully protonated form through the stepwise constants, then normalizing by the total amount. Let the species be H3A, H2A-, and HA2-. The equilibria are H3A ⇌ H+ + H2A- with K1, and H2A- ⇌ H+ + HA2- with K2. Express the less protonated forms in terms of the fully protonated form:

[H2A-] = K1 [H3A] / [H3O+]

[HA2-] = K2 [H2A-] / [H3O+] = (K1 K2 [H3A]) / [H3O+]^2

Total amount C = [H3A] + [H2A-] + [HA2-] = [H3A] [1 + K1/[H3O+] + K1K2/[H3O+]^2]. Therefore the fraction of the fully protonated form is

alpha_H3A = [H3A]/C = [H3O+]^2 / ( [H3O+]^2 + [H3O+] K1 + K1 K2 ).

This matches the given expression with [H3O+]^2 in the numerator. The other two fractions would be [H3O+] K1 / D and K1 K2 / D for the remaining forms, respectively.

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