Chemistry 271, Section 22xx Your Name: Prof. Jason Kahn University of Maryland, College Park Your SID #: General Chemistry and Energetics Your Section #: Exam I (100 points total) October 6, 2010 You have 50 minutes for this exam. Exams written in pencil or erasable ink will not be re-graded under any circumstances. Explanations should be concise and clear. I have given you more space than you should need. There is extra space on the last page if you need it. You will need a calculator for this exam. No other study aids or materials are permitted. Partial credit will be given, i.e., if you don’t know, guess. Useful Equations: Ka = [H+][A–]/[HA] pH = –log([H+]) Kb = [BH+][HO–]/[B] F = ma ¸eiπ + 1 = 0 PV = nRT Kw = [H+][HO–] = 10–14 pH = pKa + log([A–]/[HA]) pH (e.p.) = (pKa1 + pKa2)/2 R = 0.08206 L·atm/mole K 0 °C = 273.15 K pKa = –log(Ka) Honor Pledge: At the end of the examination time , please write out the following sentence and sign it, or talk to me about it: “I pledge on my honor that I have not given or received any unauthorized assistance on this examination.”Chemistry 271, section 22xx Exam I, 10/6/10 2/7 Score for the page 1. (20 pts; 2 points each) True or false: Place an X in the appropriate column Statement True False The pH at the half-equivalence point is always the average of the pKa at the beginning of a titration and at the equivalence point. The histidine side chain is neutral at pH 4. At equilibrium the rates of the forward and reverse reactions are both zero. The Henderson-Hasselbach equation is not correct when [A–] is too large. When we add equations we add equilibrium constants. The buffering range for a weak acid is between the pKa and pKa + 2, but not below the pKa. Kp and Kc are equal to each other if there are no gases involved in the equilibrium. The percent dissociation of a weak acid decreases as its concentration decreases. In multiple equilibrium problems, all equilibria must be satisfied. The equilibrium constant of a reverse reaction is the reciprocal of the equilibrium constant of the corresponding forward reaction.Chemistry 271, section 22xx Exam I, 10/6/10 3/7 Score for the page 2. (24 pts) Chemical Equilibria (a; 9 pts) State LeChatelier’s principle. Define the reaction quotient Q, and give the equivalent of LeChatelier’s principle in the language of thermodynamics. (b; 3 pts) Write the Kc expression for the following reaction: BaSO4 (s) ⇌ Ba2+ (aq) + SO42–(aq) (c; 2 pts) Does the amount of dissolved barium ion depend on the amount of solid barium sulfate sitting at the bottom of the beaker? Circle one: YES NO (d; 6 pts) It turns out that the numerical value of Kc is 1.1 × 10-10. The small value for Kc for some sulfate salts is the basis of selective precipitations in quantitative analysis, in which, for example, one can use a standardized sodium sulfate solution to measure the barium ion concentration in a test sample. Calculate the concentration of barium ion in solution if the [SO42–] is 75 mM. If the initial barium ion concentration before precipitation with NaSO4 was 125 mM, what percentage of the barium ion remains in solution?Chemistry 271, section 22xx Exam I, 10/6/10 4/7 Score for the page (e; 4 pts) Why do we use NaSO4 instead of sulfuric acid for the barium precipitation (other than safety)? 3. (18 pts) Biological Applications (a; 6 pts) Draw the structure of the Lysine side chain at pH 7 and give its approximate pKa. (b; 12 pts) Explain why measuring the pH dependence of an enzymatic reaction is a useful probe into mechanism. Explain why, however, the experiment offers only a guide to identifying the active site residue type(s). As a concrete example, explain how a rate vs. pH curve might suggest but not prove that an aspartic acid or glutamic acid residue is involved in catalysis (vs. histidine for example).Chemistry 271, section 22xx Exam I, 10/6/10 5/7 Score for the page 4. (38 pts) Acid-base chemistry and multiple equilibria Consider the diprotic acid malonic acid, with pKa’s 2.83 and 5.89. (Malonic acid is an inhibitor of the important metabolic enzyme succinate dehydrogenase.) (a; 8 pts) The first pKa is very low relative to a typical carboxylic acid like acetic acid, pKa = 4.75. This means that malonic acid is a (circle one): stronger acid or a weaker acid than acetic acid. Sketch a structure that rationalizes this (three acceptable possible answers). On the other hand, the second pKa of malonic acid is higher than the pKa of acetic acid. Explain why. (b; 4 pts) Why can we ignore the second pKa when calculating the pH of a malonic acid solution? (c; 6 pts) Calculate the pH of 0.150 M malonic acid, assuming (dubiously) that “x” << 0.150 M.Chemistry 271, section 22xx Exam I, 10/6/10 6/7 Score for the page (d; 9 pts) Sketch a curve of the pH as we titrate 0.150 M malonic acid with NaOH, ignoring dilution. Specify the pH for the beginning (the number you just calculated in c), the two half-equivalence points, and the two equivalence points (the second one is at pH = 9.53).Chemistry 271, section 22xx Exam I, 10/6/10 7/7 Score for the page (e; 4 pts) We did not learn to calculate the exact pH for diprotic acid titrations, but we do know how to set up the problem. For any point on a titration of malonic acid H2M being converted to HM– and M2– with added NaOH, we know the [Na+] and we would like to calculate the concentrations of five other chemical species. One is hydroxide. Write down the other four. (Not a trick question) (f; 7 pts) We need five equations to solve the problem. Two of them are the two Ka relationships for the first and second acid dissociation equilibria. Write down the last equilibrium relationship needed and the two conservation equations that we need to have five equations. Hint: the total charge of the solution is zero. Page Score 2 /20 3 /20 4 /22 5 /18 6 /9 7 /11 Total