Midterm Exam III. CHEM 181: Introduction to Chemical Principles November 25, 2014 Answer Key

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1 Midterm Exam III EM 181: Introduction to hemical Principles ovember 25, 2014 Answer Key 1. For reference, here are the pk a values for three weak acids: 3 pk a = 7.2 pk a = 4.8 pk a = 15 ow consider this list of compounds: 3 l 1

2 (a) An unknown acid has a pk a of 2.5 and the following IR spectrum: i. Identify this compound from the list on page 2 (redraw the Lewis structure below). This compound has a stretch in the IR spectrum and a pk a significantly more acidic than the example carboxylic acid. ii. ircle the most acidic proton of this compound. iii. Draw any resonance structures that are important in determining the acidity of this proton. iv. Label any peaks on the IR spectrum that you can assign (there will likely only be a few.) It s unclear whether the structure near 3000 cm 1 is the stretch or simply part of the feature. The stretch is important. 2

3 (b) An unknown acid has a pk a of 9.5 and the following IR spectrum: i. Identify this compound from the list on page 2 (redraw the Lewis structure below). While there must be a = bond, the pk a is too high here (it is too weak an acid) for the molecule to include a. ii. ircle the most acidic proton of this compound. iii. Draw any resonance structures that are important in determining the acidity of this proton. The major resonance structures that move the negative charge onto the oxygens are the reason this bond is acidic (much more so than the example compound.) iv. Label any peaks on the IR spectrum that you can assign (there will likely only be a few.) The stretch feature is too high in frequency (max at cm 1, but spreading out between 3000 and 3500 cm 1 ) to be 3

4 an sp 3 stretch. There is a typo on the IR peak list that says amines have a stretch at this frequency, so amine stretch was accepted along with stretch as an answer. 2. For each of the following, rank the compounds by most acidic (basic) to least acidic (basic). Give a short explanation for each ranking. (a) Write strongest next to the strongest acid and weakest next to the weakest. strongest 3 weakest The addition of a = slightly increases the acidity of the carboxylic acid, explained either with a (very poor) additional resonance structure or the difference between the sigma bond to an sp 2 instead of an sp 3 orbital. The rightmost compound is the weakest because hydrogens are almost always less acidic than hydrogens. (b) Write strongest next to the strongest acid and weakest next to the weakest. l strongest 3 weakest The phenol (on the right) is the weakest because the charge remains mostly on the in the conjugate base, with only minor resonance structures that locate the lone pair on carbon atoms. The carboxylic acids have two major resonance structures, and the addition of the electronegative l increases acidity. 4

5 (c) Write strongest next to the strongest BASE and weakest next to the weakest BASE. (It is the group that acts as the base.) weakest strongest The strongest base will have the weakest conjugate acid; the weakest base, the strongest acid. The conjugate acids are made by adding + to the ; the question of whether the is itself acidic is unrelated. The key here is that the + 3 aromatic compounds do not have minor resonance structures that double-bond the and, while the compounds their conjugate bases do: 3 vs. (and 2 others) The 2 group opposite the makes the unprotonated base even more stable relative to the protonated acid. This gives the strongest acid, and thus the weakest base. Resonance involving the 2 is not possible with the molecule in the middle position, making the conjugate acid weaker and thus the base stronger. o resonance at all happens with the rightmost compound, making the protonated acid the weakest and thus the base the strongest. (Grading was lenient on the relative strength of the third compound with respect to the other two.) 5

6 3. Match the compounds on the following page to the 1 MR spectra on this page. (All peaks are shown. Ignore very small spikes in the baseline, as well as the calibration peak at 0 ppm.) 6

7 4. Acetic acid ( 3 ) is a weak acid, and pyridine ( 5 5 ) is a weak base. K a ( 3 ) = M K b ( 5 5 ) = M If you start with 0.1 L of 0.1 M pyridine in water: (a) ow many moles of acetic acid do you have to add to convert 95% of the pyridine to its conjugate acid form ( )? Acetic acid is the strongest acid in solution (much stronger than water) and pyridine the strongest base (much stronger than water), so the reaction that matters is Ac + pyr Ac + pyr + The K for this reaction is K = [Ac ][pyr + ] [Ac][pyr] We can get the terms we need in the numerator and denominator like so K a (Ac)K b (pyr) = [Ac ][ + ] [pyr + ][ ] [Ac] [pyr] = [Ac ][pyr + ] [ + ][ ] [Ac][pyr] = K K w K = K a(ac)k b (pyr) K w = = 3.06 The initial pyridine concentration is 0.1 M, so converting 95% of it means [pyr] = M and [pyr + ] = M 7

8 These are equilibrium concentrations and let us solve for [Ac ] and [Ac]. [Ac ][pyr + ] [Ac][pyr] = 3.06 [Ac ] [pyr + ] [Ac] [pyr] = 3.06 [Ac ] [Ac] = 3.06 [Ac ] 19 [Ac] = 3.06 [Ac ] [Ac] = oncept check: i. If we added acetic acid to a strong base, all of the Ac will convert to Ac (at least until there is no strong base left; if only 95% has been used up, there can be no Ac.) ii. If we added acetic acid to a very weak base like water, then there would only be a small amount of Ac formed (because acetic acid is a weak acid), with almost everything remaining as Ac. iii. Since pyridine is a weak base, there is significant but not complete conversion of Ac to Ac. Stoichiometry says that for every pyr + created, we must also create an Ac (the reaction with water is negligible), so At equilibrium, [Ac ] [Ac] [Ac ] = [pyr + ] = M = and so [Ac] = 0.59 M The total amount of acetic acid added is equal to the combined amount of Ac and Ac at equilibrium, so and with a volume of 0.1 L [Ac] added = [Ac] eq + [Ac ] eq = n(ac) = M 0.1 L = mol 8

9 (b) What will the p of the solution be at this point? You don t need part (a) to calculate the p. Instead, K b (pyr) = [pyr+ ][ ] [pyr] = [pyr+ ] [ ] [pyr] = 19 [ ] [ ] = M [ + ] = K w [ ] = M p = 3.95 (This is essentially the same as using the enderson-asselbalch equation.) 5. is a gas that makes a weak acidic solution when dissolved in water. K a () = M The enry s law constant for the reaction is (g) (l) (aq) k = 8.3 M atm 1 A buffer solution with 0.1 M (aq) is at a p of ml of this solution is placed in a sealed 4 L bottle. What is the equilibrium pressure of (g) in this bottle at 300 K? (Assume the p remains constant.) The equilibria to consider are [ + ][ ] [] = M and [(aq)] = 8.3 M atm 1 P Because a p of 9 is constant, [ + ] = M, which means [ M][ ] [] 9 = M

10 and [ ] [] = 0.62 If all of the remains in solution, then [] + [ ] = 0.1 M [ ] [] = 0.62 [ ] = 0.62 [] [] [] = 0.1 M [] = M Use this number to calculate the gas pressure: [(aq)] P = 8.3 M atm 1 P = [(aq)] 8.3 M atm M = 8.3 M atm 1 = atm ow ask the question: does producing this much gas decrease the amount of in solution? n = P V RT ( atm)(3.9 L) = ( L atm mol 1 K 1 )(300 K) = mol With 100 ml of 0.1 M, we only had mol to start with; after producing this much gas, only mol (100 ml at M) remains in solution. The [(aq)] is reduced proportionally: [] = = M This, in turn, decreases the pressure accordingly: P = = atm 10

11 (You can iterate more times, though it is not necessary: the reduced pressure means that only mol is removed from solution, thus a more accurate concentration for is M and pressure is atm.) Alternately, using more algebra avoids iteration. Solve for number of moles of gas as a function of [(aq)]: P = [(aq)] 8.3 M atm 1 n gas = P V RT (3.9 L) = P ( L atm mol 1 K 1 )(300 K) = [(aq)] mol atm M atm 1 = L [(aq)] The total number of moles is divided between gas, aqueous, and aqueous : n tot = n gas + n (aq) + n 0.01 = [(aq)] [(aq)] [ (aq)] = [(aq)] [(aq)] [(aq)] [(aq)] = = M and P = [(aq)] 8.3 M atm 1 = atm 11

Midterm Exam III. CHEM 181: Introduction to Chemical Principles Solutions C HC N H 3 C. pka = 9.7

Midterm Exam III. CHEM 181: Introduction to Chemical Principles Solutions C HC N H 3 C. pka = 9.7 Midterm Exam III EM 181: Introduction to hemical Principles Solutions 1. For reference, here are the pk a values for four weak acids (not all resonance structures shown): N 3 N N 3 3 2 pka = 3.5 pka =