As I'm sure most of you have access to kaplan materials, I thought this would be a great place to ask. When taking this test, I got two questions wrong which I completely disagree with. I will post the question, my reasoning, and kaplan's answer. I would appreciate someone either pointing out my flawed reasoning or validating that this is in fact an error. Which of the following will occur if ammonia were added to a saturated solution of Na2SO4? A)The Ksp will decrease and additional Na2SO4 will dissolve. B)The Ksp will be unchanged and additional Na2SO4 will dissolve. C)The Ksp will increase and additional Na2SO4 will precipitate. D)The Ksp will remain unchanged and additional Na2SO4 will precipitate. Kaplan: A saturated solution of sodium sulfate will be acidic, since sulfuric acid will form when the sulfate ions are hydrated. If ammonia is added to the solution, additional sodium sulfate will dissolve because the acid protons will react with the base. The MCAT expects you to know that Ksp is a function of the structure of the salt and the temperature of the solution. Adding a stress to the system will not affect the Ksp if the temperature remains the same, so the answer is (B) . My explanation: Ksp doesn't change dependent upon reaction conditions, so A,C can be eliminated. However, since when is a solution to which you add a base acidic? SO42- is the conjugate base of sulfuric acid, not sulfuric acid itself. Therefore, when you add a BASE to water, you should get H20 + SO42- -> OH- +HSO4- I don't think it is any leap of faith to say this should be a basic solution, not an acidic one. Therefore, when you add NH3 two things should happen. 1) you get increased formation of OH-, and 2) NH3 will deprotonate HSO4- back to SO42- as ammonia is a stronger base. Now, we have a situation in which the number of sulfate ions in solution has increased above previously saturated levels. Thus it should precipitate out and the answer should be D. Next problem: For a given exothermic reaction, how are ΔH, Ea (forward) and Ea (reverse) related? A) ΔH + Ea (forward) = Ea (reverse) B) ΔH - Ea (forward) = Ea (reverse) C) ΔH = Ea (forward) + Ea (reverse) D) ΔH = Ea (forward) - Ea (reverse) Kaplan: The activation energy for the forward reaction is the amount of energy it takes to overcome the barrier between the products and the reactants. Hence, the activation energy for the reverse reaction is the amount of energy it takes to overcome the barrier between the reactants and the products. The difference between the two is the enthalpy change for the reaction. This can be written as Ea (reverse) - Ea (forward) = ΔH, or ΔH + Ea (forward) = E a (reverse), choice (A). My explanation: The problem with kaplan's answer is this: If this reaction is spontaneous, enthalpy change is negative, and Ea reverse >Ea forward. Thus, if you plug it into their equation you get something that looks like: negative number + small positive number = larger positive number wtf? They flipped the sign of dH, and thus D is my answer.