Calculate the overall equilibrium constant, Koverall, for the reaction.

Question

HF(aq) + SO₃^2-(aq)↽−−⇀F^-(aq) + HSO^-₃(aq)

Koverall = ?

Answer 1

To calculate the overall equilibrium constant, K_overall, for this reaction, we need to realize that the overall reaction is a combination of the acid dissociation reactions for HF and HSO₃^-:

HF + H2O --> H3O+ + F- K_a (HF) = 7.2 x 10^-4

HSO₃^- + H₂O --> H₃O⁺ + SO₃^2- K_a(HSO₃^-) = 1.54 x 10^-2

To get the overall reaction:

HF + SO₃^2- --> F- + HSO₃^-

we need to reverse the acid dissociation equation for HSO₃^-to

H₃O⁺ + SO₃^2- --> HSO₃^- + H₂O, which has K = 1/K_a(HSO₃^-)

then add this to the acid dissociation equation for HF

HF + H₂O --> H₃O⁺ + F- K = K_a(HF)

H₃O⁺ + SO₃^2- --> HSO₃^- + H₂O, which has K = 1/K_a(HSO₃^-)

Crossing out the species on both sides of the question (H₂O and H₃O⁺), we arrive at our target equation,

HF + SO₃^2- --> F- + HSO₃^-

Now for the equilibrium constant for the overall reaction... Equilibrium constants are MULTIPLICATIVE, which means that when we sum 2 equations together, we multiply the respective equilibrium constants to get the overall K.

K_overall = K_a(HF) * 1/K_a(HSO₃^-) OR K_a(HF)/K_a(HSO₃^-)

We can now solve numerically for K_overall by looking up the K_a values for the 2 acids and combining them according to the above equation.

K_overall = K_a(HF)/K_a(HSO₃^-)

= (7.2 x 10^-4)/(1.54 x 10^-2) = 0.047