For a polyprotic acid, the fraction alpha A2- is given by which expression? Denominator is ([H3O+]^2+[H3O+]K1+K1K2).

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Multiple Choice

For a polyprotic acid, the fraction alpha A2- is given by which expression? Denominator is ([H3O+]^2+[H3O+]K1+K1K2).

Explanation:
Distributions of species in a polyprotic acid depend on the stepwise dissociation constants and the hydronium concentration. For a triprotic acid, the doubly deprotonated form HA2− arises from two successive deprotonations, so its amount relative to the fully protonated form H3A is set by Ka1 and Ka2 and by how much H3O+ is present. Start with the relationships from the equilibria: [H2A−] = Ka1 [H3A]/[H3O+] [HA2−] = Ka2 [H2A−]/[H3O+] = Ka1 Ka2 [H3A]/[H3O+]^2 The total amount of the three considered species is proportional to [H3A] + [H2A−] + [HA2−] = [H3A] [1 + Ka1/[H3O+] + Ka1 Ka2/[H3O+]^2]. The fraction of the doubly deprotonated form is then alpha_A2− = [HA2−] / ([H3A] + [H2A−] + [HA2−]) = (Ka1 Ka2/[H3O+]^2) / (1 + Ka1/[H3O+] + Ka1 Ka2/[H3O+]^2). Multiply numerator and denominator by [H3O+]^2 to get alpha_A2− = Ka1 Ka2 / ([H3O+]^2 + [H3O+] Ka1 + Ka1 Ka2). Using K1 for Ka1 and K2 for Ka2 and recognizing [H3O+] as [H3O+], this matches the given expression. This form reflects two deprotonations feeding the A2− species while higher deprotonation (A3−) is neglected in this approximation.

Distributions of species in a polyprotic acid depend on the stepwise dissociation constants and the hydronium concentration. For a triprotic acid, the doubly deprotonated form HA2− arises from two successive deprotonations, so its amount relative to the fully protonated form H3A is set by Ka1 and Ka2 and by how much H3O+ is present.

Start with the relationships from the equilibria:

[H2A−] = Ka1 [H3A]/[H3O+]

[HA2−] = Ka2 [H2A−]/[H3O+] = Ka1 Ka2 [H3A]/[H3O+]^2

The total amount of the three considered species is proportional to [H3A] + [H2A−] + [HA2−] = [H3A] [1 + Ka1/[H3O+] + Ka1 Ka2/[H3O+]^2]. The fraction of the doubly deprotonated form is then

alpha_A2− = [HA2−] / ([H3A] + [H2A−] + [HA2−])

= (Ka1 Ka2/[H3O+]^2) / (1 + Ka1/[H3O+] + Ka1 Ka2/[H3O+]^2).

Multiply numerator and denominator by [H3O+]^2 to get

alpha_A2− = Ka1 Ka2 / ([H3O+]^2 + [H3O+] Ka1 + Ka1 Ka2).

Using K1 for Ka1 and K2 for Ka2 and recognizing [H3O+] as [H3O+], this matches the given expression. This form reflects two deprotonations feeding the A2− species while higher deprotonation (A3−) is neglected in this approximation.

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