Question

Given: H2O -> H+ + OH- Keq = 1.00 x 10^-14

And: HC2H3O2 -> H+ + C2H3O2- Keq = 1.00 x 10^-5

Calculate the equilibrium constant for: C2H3O2- + H2O -> OH- + HC2H3O2

Answer 1

H₂O -----> H⁺ + OH^- Keq = 1.00 x 10^-14   =====1)

HC₂H₃O₂ ---------> H⁺ + C₂H₃O^2- Keq = 1.00 x 10^-5   ======2)

Reversing the reaction 2

  C₂H₃O^2- + H⁺ ---------> HC₂H₃O₂ ====3)

K'eq = 1/Keq = 1/ 1.00 x 10^-5 = 1 X 10⁵

Adding 1 and 3rd reactions equations

H₂O -----> H⁺ + OH^-  Keq = 1.00 x 10^-14  

C₂H₃O^2- + H⁺ --------->   HC₂H₃O₂ ====3) K'eq = 1.0 X 10⁵

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HC₂H₃O₂ + H₂O -----> OH^- + HC₂H₃O₂   Net reaction

Keq = K' eq X Keq = 1.0 X 10⁵ X 1.00 x 10^-14 = 1.0 X 10^-9