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1998 Exam I - Top 5 Missed Questions

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    Here are five questions from Exam I in 1998 that caused some people condsiderable problems.

    Take a look at these before Exam II as well as the final exam. Be sure to try working the problems before looking at the answers.

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    Here we go:

    1. Question #19: What is the concentration of H+ ions in a 0.025 M Ba(OH)2 solution?

      • We start ALL problems by writing a balanced equation. Never, never, never forget to do this -- it is the KEY to this problem!

        Ba(OH)2 is a strong base so it completely dissociates in solution:

        Ba(OH)2 rt arrow Ba2+ + 2 OH-

        Notice I have balanced not only the atoms, but the charge as well. I could set up an ICE chart at this point, but there is no need. I know from the reaction equation that every mole of BaOH2 dissociates to give 1 mole of Ba2+ and TWO moles of OH-.

        Therefore, if the original concentration of BaOH2 was 0.025 M, it must dissociate into 0.025 M Ba2+ and 0.050 M OH-. But, read the question. It asks for the [H+], not [OH-]. How can we determine the value we need? We know that [H+][OH-]=Kw, so:

        [H+](0.050)=1 x 10-14

        [H+] = 2.0 x 10-13 M

        REMEMBER: ALWAYS CHECK THE QUESTION AT THE END TO BE SURE YOU'VE ANSWERED WHAT WAS ASKED.

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    2. Question #15: At 500 degrees C, Kp = 4.9 x 10-5 for the exothermic reaction:

      3 N2(g) + 3 H2(g) eq arrow 2 NH3(g)

      The value of Kp at 1000 degrees C is:

      • We check to make sure our equation is balanced. Good, it is. Now we have to ask what happens to an exothermic reaction when we raise the temperature.

        LeChatlier tells us that if we do something to a system it will try to counteract our action. Therefore if we add heat to the reaction (raise the temperature), the reaction will try to lower it.

        The reaction can do this by moving in the endothermic direction. If the reaction is exothermic as written, the enothermic direction is to the left.

        So all we have to do is ask what effect that has on K. The equilibrium constant for this reaction is:

        K = [NH3]2/[N2][H2]3

        If we make more reactants and fewer products, then the value of K must decrease. The only answer which meets this criteria is answer B, less than 4.9 x 10-5.

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    3. Question 5: When 0.700 g of an unknown molecular compound dissolves in 20.0 g of acetic acid, the freezing point of acetic acid is lowered by 1.24 degrees C. Calculate the molar mass of the unknown. (Kf(acetic acid) = 3.90 degrees C/M).

      • Obviously, this is a freezing point depression problem. The only equation we have that can address this kind of question is our freezing point depression equation:

        deltaT = iKfm

        Where i is the van't Hoff factor, K is given and m is the molality. We aren't told what the unkown is, so how do we know what i is? Simple. We are told that it is a molecular compound. Only ionic compounds dissociate. Therefore each molecule gives one particle in solution so i = 1.

        We plug in the values we do know and solve for the molality, m:

        1.24 C = (1)(3.90 C/m)m

        m = 0.318 molal

        Recall that molality is moles of solute per kg of solvent. We are given the mass of the solvent so we can easily solve for the moles:

        m = moles solute/ kg solvent

        0.318 mol/kg = x mol/0.020 kg

        x = 0.00639 moles

        If we know the moles used and the g used, we know the molar mass:

        0.700 g = (y g/mol)(0.006359 mol)

        y = 110 g/mol

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    4. Question 10: For the equilibrium:

      PH3(g) + BCl3(g) eq arrow PH3BCl3(s)

      Kp = 19.2 at 60 degrees C. Calculate Kc for this reaction at 60 degrees C.

      • We start by asking if there are any equations that relate Kp and Kc. There sure are. Try:

        Kp= Kc(RT)deltan

        Let's start with deltan, the change in number of moles of gas. There are TWO moles of gas on the left and NONE on the right (always watch out for solids in your equilibrium expression). We go from 2 moles of gas to 0 so the change is -2 (products minus reactants). Many people errred and say it was +2....

        R is a constant. T is given in degrees Kelvin for almost every equation we use so the temperature is 60 + 273 = 333 K. Now we just plug in and solve:

        19.2 = Kc[(0.0821 Latm/molK)(333 K)]-2

        Kc = 1.4 x 104

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    5. Question 24: At 20 degrees C, Ka = 4.9 x 10-10 for the reaction:

      HCN(aq) eq arrow H+(aq) + CN-(aq)

      What is Kb at 20 degrees C for the reaction:

      CN-(aq) + H2O(l) eq arrow HCN(aq) + OH-(aq)?

      • We start by checking that both equations are balanced correctly. They are. We are given a Ka for the reaction and are asked to find Kb.

        The two equations we are given are related by KaKb = Kw. We know Kw = 1 x 10-14, so all we have to do is solve!

        (4. 9 x 10-10)Kb = 1 x 10-14

        Kb = 2.0 x 10-5

        What if we weren't sure that this relationship was valid? We could have added to two equations together to check this. If we add the left side and right side of each reaction and cancel terms we'd get:

        H2O(l) eq arrow H+(aq) + OH-(aq)

        Therefore, this relationship must be true!

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    Learn from your mistakes so you don't repeat them! Watch how the logic flows so you can learn how to approach new or unfamiliar problems.

    Be sure to try not only the TopClass exercises provided here, but the extra ones at a variety of other universities.


    1998 Exam I - Top 5 Missed Questions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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