Question 15.4: Given the reactions HF(aq) ⇌ H^+(aq) + F^-(aq) Kc = 6.8 × 10...
Given the reactions
HF (aq) ⇌ H^{+} (aq) + F^{-} (aq) K_{c} = 6.8 × 10^{-4}H_{2}C_{2}O_{4} (aq) ⇌ 2 H^{+} (aq) + C_{2}O_{4} ^{2 -} (aq) K_{c} = 3.8 × 10^{-6}
determine the value of Kc for the reaction
2 HF (aq) + C_{2}O_{4 }^{2 -} (aq) ⇌ 2 F^{-} (aq) + H_{2}C_{2}O_{4} (aq)Analyze We are given two equilibrium equations and the corresponding equilibrium constants and are asked to determine the equilibrium constant for a third equation, which is related to the first two.
Plan We cannot simply add the first two equations to get the third. Instead, we need to determine how to manipulate these equations to come up with equations that we can add to give us the desired equation.
Solve
If we multiply the first equation by 2 and make the corresponding change to its equilibrium constant (raising to the power 2), we get
Reversing the second equation and again making the corresponding change to its equilibrium constant (taking the reciprocal) gives
2 H^{+} (aq) + C_{2}O^{2 -}_{4} (aq) ⇌ H_{2}C_{2}O_{4} (aq) K_{c} =\frac{1}{3.8 × 10^{-6}} = 2.6 × 10^{5}Now, we have two equations that sum to give the net equation, and we can multiply the individual K_{c} values to get the desired equilibrium constant.