If an acid HA has a Ka value of 1.5x10^-5, what will be the value of equilibrium

Question

If an acid HA has a Ka value of 1.5x10^-5, what will be the value of equilibrium constant for the following reaction? HA(aq) + OH-(aq)----> A-(aq) + H2O(l). Will this reaction be reactant favored or product favored? What does this tell you about the "completeness" of the titration reactions used in this labratory?

Hint: You need to use Rxns 1 and 3 and Eqs 1 and 3 from the Background material.

HA(aq) + H2O(l)---> A-(aq) + H3O+(aq) (Rxn 1)

Ka= [A-][H3O+]/[HA] (Eq 1)

H2O(l) + H2O(l)---> H3O+(aq) + OH-(aq) (Rxn 3)

Kw= [H3O+][OH-] (Eq 3)

Explanation / Answer

Ka = 1.5 x 10^-5 = [A-][H+]/ [HA]

the equilibrium constant of the reaction
HA + OH- <=> A- + H2O is
K = [A-]/ [HA][OH-]
multiply numerator and denominator by [H+]

K = [A-][H+] / [HA][OH-][H+] = Ka / Kw = 1.5 x 10^-5 / 1.0 x 10^-14 = 1.5 x 10^9
product favored

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