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 12, 2011 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) Kp = Kc(RT)∆n P2/a3 = 4π2/MG x =!b ± b2! 4ac2a 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.” +1 point extra credit for filling in this boxChemistry 271, section 22xx Exam I, 10/12/11 2/7 Score for the page 1. (20 pts) Short Answer (2 pts each) pH of 1 M HNO3 pOH of 0.1 M NaOH pOH of 0.1 M HOAc (pKa 4.75) + 0.1 M NaOAc pH of 0.001 M NaOH pKa + pKb = (for Ka and Kb referring to a conjugate acid/base pair) [H+] at pH 7 [OH–] at pH 9 When Q < K, the reaction will proceed (circle one) forward. backward. The Henderson-Hasselbach relationship is always , sometimes . The pH at the first equivalence point of a polyprotic acid titration is given by .Chemistry 271, section 22xx Exam I, 10/12/11 3/7 Score for the page 2. (30 pts) Acid-Base Equilibria Consider the pH obtained upon dissolving a weak monoprotic acid HA in water, as a function of its total concentration C0 and its Ka. This is a problem you have done many times, here we are exploring a general formula. The equilibrium is of course HA ⇋ H+ + A– Ka = [H+][A–]/[HA] (a; 12 pts) Initially, assume that “x” can be ignored in the denominator and show that pH = ½ log(Ka × C0) (b; 8 pts) Now, repeat the problem but do not assume that “x” is small, i.e. use the quadratic formula to derive a general formula for the pH (c; 4 pts) Show that the more complicated expression you just obtained reduces to the simpler expression from (a) if C0 >> Ka. (This gives us a more precise description of exactly when x is negligible in the denominator.)Chemistry 271, section 22xx Exam I, 10/12/11 4/7 Score for the page (d; 6 pts) Physically, what happens to the % Dissociation of a weak acid as Ka increases or C0 decreases? Give an explanation for the C0 effect based on LeChatelier’s principle or the dynamic balance of rates. 3. (20 pts) Buffers and Titration The graph shows seven titrations at progressively lower concentrations of a weak acid titrated with strong base. The x axis is in terms of equivalents of base added relative to the acid, so the actual concentration of base added is also decreasing as C0 decreases. As usual, we ignore dilution during each individual titration, so the total acid [HA] + [A–] is constant. (a; 6 pts) What is the pKa of the weak acid being used, and how do you know?Chemistry 271, section 22xx Exam I, 10/12/11 5/7 Score for the page (b; 4 pts) Why does the pH at the equivalence point decrease as C0 decreases? A qualitative answer is fine. (c; 4 pts) The graph illustrates the critical features of buffers. How does it show us the utility of “10X” or “100X” reaction buffers in the lab? (d; 6 pts) Explain why the pH at the end (1.2 equivalents, top right corner) of the C0 = 0.5 M titration is at pH 13.Chemistry 271, section 22xx Exam I, 10/12/11 6/7 Score for the page 4. (10 pts) pH effects on Enzymes The proposed mechanism shown is the essence of catalysis by aspartyl proteases, a class of enzymes that includes HIV protease. They have classic bell-shaped pH rate profiles like those we have seen in class, with pKa’s typically at around 3 and 5. (a; 10 pts) Which residue is associated with the pKa of 3, and what is its function in the mechanism? In other words, why does the reaction fail when it is performed at a pH much below 3? Which residue has pKa = 5, and what is its function? Why does the reaction fail at basic pH?Chemistry 271, section 22xx Exam I, 10/12/11 7/7 Score for the page 5. (20 pts) Equilibrium Manipulations. We will confirm that any acid on the left column of the table can protonate any base that is below it in the right column. For example, formic acid, HCOOH, pKa = 3.75, should be able to protonate sulfite, SO32–. The pKa of hydrogen sulfite (HSO3–) is 6.97. (a; 3 pts) What is the net reaction for formic acid protonating sulfite? ! (b; 7 pts) Write down the base dissociation reaction for sulfite and calculate its pKb and Kb. (c; 3 pts) Add the acid dissociation reaction of formic acid and the base dissociation reaction of sulfite. (d; 3 pts) What other equilibrium do we need to add to give us the net reaction from part a? (e; 4 pts) Calculate the overall equilibrium reaction constant for the reaction of part a. Page Score 1 /1 2 /20 3 /15 4 /13 5 /12 6 /18 7 /12 Total