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Appendix a Acid-Base Calculations

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Appendix a Acid-Base Calculations Appendix A Acid-Base Calculations A. 1. IO NIC STRENGTH Ionic strength (/) is a measure of ion concentration, de­ and based on Eq. (1.2) fined by the equation (W] = (1.0 X 10-14)/0.040 = 2.5 X 10- 13 so that where c. is the concentration of ion i, z. is the charge of ' ion i, and L is "the sum ~f." pH = -log[W] = -log(2.5 X 10- 13) = 12.6 = 13 Problem: What is the ionic strength of a O.lM solu­ tion of (NH4 ) 2SO4 ? Solution: Each "molecule" of (NH4) 2SO4 yields two A.3. HE DERSON- H ASSE LBALCH NH; ions and one So~- ion. Hence, EQUATIO N [NHl l = c 1 = 0.2M; z1 = + 1 [So~ - ]= c2 = 0.1M; z2 = - 2 Problem 1: Calculate the pH of a solution made by mix­ 1 I = I/2 ~{ (0.2 X 12) + (0.1 X ( - 2)2]} = 0.3 ing 15.0mlof 1.00 X 10- MK2HP04 and25.0mlof2.00 2 X 10- MKH2P04 . Solution: Molar concentrations of the two salts in the A.2. pH final volume (15.0 + 25.0 = 40.0 ml) are: 1 forK2HP04 : 15.0 X (1.00 X 10- )/40.0 = 0.0375M Problem 1: What is the pH of 0.01 OM HCI, assuming that for KH2P04 : 25.0 X (2.00 X 10- 2)/40.0 = 0.01 25M the HCl is 100% ionized? Substituting these values into the Henderson- Hasselbalch Solution: equation and using pK~ = 7.21 yields [W] = O.OlOM = 1.0 X 10-2M pH = -log[W] = -log(l.O X 10- 2) = 2.0 (0.0375) pH = 7.21 +log (0.0125) = 7.69 Problem 2: What is the pH of 0.020M Ca(OH)2 , as­ suming that it is 100 % ionized? Problem 2: Given that lactic acid (HA = CH3 - Solution: Since each "molecule" of Ca(OH)2 yields CHOH -COOH; A- = CH3 -CHOH - coo- ) has a two OH- ions, pK~ of 3.86, calculate the percent of lactic acid present in 503 504 APPENDIX A its dissociated form (A-) at a pH that is one unit above its (0.400- x) = 0.065 mol ofK2HP04/liter pK~ (i.e., at pH 4.86). Solution: At pH 4.86, the Henderson-Hasselbalch Since you wish to prepare not I liter but only 500 ml, equation takes the form you will need one-half of the above number of moles, namely, 0.168 mol ofKH2P04 (22.8 g) and 0.033 mol of [A-] K2HP04 (5.75 g). To prepare the buffer, you would weigh 4.86 = 3.86 + log [HA] out these amounts, dissolve them in water, and dilute the solution to a final volume of 500 ml. Hence, Problem 2: How would you prepare 500 ml of a O.lOOM acetate buffer at pH 5.00 from acetic acid and NaOH? The pK~ of acetic acid is 4.76. Solution: Substituting into the Henderson-Hassel­ balch equation yields so that [A-] 5.00 = 4.76 + log[HA] [A-] = 10 = !.Q [HA] 1 Hence, For every mole of HA per liter, there exist 10 moles of A- per liter. The percentage of lactic acid present in the [A-] log [HA] = 0.24 A- form (percent dissociation) is [A-] [A-] 10 [HA] = 1.74 %A-= [A-]+ [HA] X 100 = (10 + 1) X 100 = 91% Since you wish to prepare a O.IOOM buffer, the total, In like manner you can calculate that, at a pH that is combined concentration of A- and HA must be 0.1 OOM, one unit below the pK~ value (pH= 2.86), 91% of the lac­ that is, tic acid will be present in its undissociated form (HA). You can also show that at a pH that is two units above (or be­ [A-] + [HA] = 0.100 low) the pK~ value, the percentage of lactic acid present in the dissociated (or undissociated) form rises to 99%. Each of the last two equations has two unknowns, [A-] and [HA]. You can solve two equations with two un­ knowns simultaneously. From the first equation, you have that Problem 1: How would you prepare 500 ml of a 0.400M [A-] = 1.74[HA] phosphate buffer at pH 6.50 from solid KH2PO 4 and ~HPO 4 ? The second pK~ of H3PO4 is 7.21. Substituting for [A-] into the second equation yields Solution: A 0.400M phosphate buffer contains a to­ tal of 0.400 mole of phosphate salts per liter. Hence, let­ 1.74[HA] + [HA] = 2.74[HA] = 0.100 ting x equal the number of moles of KH2PO 4 per liter, it follows that (0.400 - x) is the number of moles of so that K2HPO4 per liter. Substituting into the Henderson-Has­ [HA] = 0.0365M (0.0365 mol/liter) selbalch equation yields [A-] = 1.74[HA] = 0.0635M (0.0635 mol/liter) 6.50 = 7.21 +log (0.400- x) As you wish to prepare only 500 ml, not 1 liter, you X will need one-half of the above number of moles, name­ log (0.400- x) = -0.71 ly, 0.0182 mol of HA (acetic acid) and 0.0318 mol of A­ X (acetate). Since both buffer components are derived from acetic acid, you must start with 0.0500 mol of acetic acid (0.400- x) = 0.195 X and convert some of it to acetate by adding NaOH. From the neutralization reaction involved, x = 0.335 mol of KH2PO iliter APPENDIX A 505 0.0500- 0.0318 = 0.0182 mol you see that adding 1.0 mol of NaOH converts 1.0 mol of To sum up, you would prepare this buffer by obtain­ acetic acid to 1.0 mol of acetate. To produce 0.0318 mol ing 0.0500 mol of acetic acid (2.87 ml of concentrated of acetate, you must add 0.0318 mol of NaOH. After acetic acid, which is 17 .4M), adding 0.0318 mol of solid adding the NaOH, you will be left with the required NaOH (1.27 g), and diluting the mixture to 500 mi. amount of acetic acid: Appendix B Principles of Organic Chemistry TIONAL CR LITY Functional groups consist of two or more atoms and pos­ Many objects are asymmetric in their structure; they have sess characteristic structures and chemical reactivities. A a "handedness" like that of the left and right hands. If you given functional group generally behaves the same way in visualize your two hands placed on either side of a flat all molecules containing that group. Table B.l shows mirror, one hand will appear to be the mirror image of the some of the functional groups found in biomolecules. other. But the two hands are not identical. You cannot put one hand on top of the other, with both palms down. The two hands cannot be superimposed in space. B.., POL\ In much the same manner, biomolecules can have structural asymmetry, resulting in mirror images that Polar reactions result from the attractive force between cannot be superimposed in space. We use the term chi­ positive and negative charges (or partial charges) on mol­ rality ("handedness") to refer to the right- and left-hand­ ecules. We call the two reactants in a polar reaction nu­ edness of a molecule. At the molecular level, chirality cleophile and electrophile. A nucleophile consists of an arises when a compound contains one or more chiral atom or a group of atoms that has an electron-rich site and centers. A chiral center comprises either a chiral carbon forms a bond by donating a pair ofelectrons. By contrast, atom or some other asymmetric region in the molecule. an electrophile consists of an atom or a group of atoms A chiral carbon atom has four different substituents at­ that has an electron-poor site and forms a bond by ac­ tached to it (Figure B.l). Because tetravalent carbon is cepting a pair of electrons. We depict the electron-pair tetrahedral, these groups occupy the corners of a tetrahe­ movement by means of a curved arrow, using the con­ dron. Because of carbon's tetrahedral nature, the two vention that the electron pair moves from the tail to the mirror images of a chiral carbon cannot be superim­ head of the arrow: posed. Aside from carbon, several other atoms that form ........_ compounds having a tetrahedral structure (Si, N, P, S) can A:-- + s+ ---7 A:B exist as chiral centers under proper circumstances. Chiral Nucleophile Electrophile centers also result from molecular asymmetry that is not ----........_-- s+ s-:?" due to the presence of chiral atoms. The helical structures provide an example. A helix A:- + B: C ---7 A:B + of proteins and nucleic acids is intrinsically chiral; a left-handed helix constitutes a The second arrow in the second reaction indicates that nonidentical and nonsuperimposable mirror image of a C leaves, taking the two electrons of the B-C bond right-handed helix, much as a left-handed screw differs with it. from a right-handed screw. 507 508 APPENDIX B Table 8.1. Some Common Functional Groups in Biomolecules Compound type Structure Functional group I Alcohol R-C-OH Hydroxyl I ~0 Aldehyde R-C Carbonyl "R ~0 Ketone R-C Carbonyl Mirror "R· Figure B. l . The two enantiomers (mirror images) of a chiral carbon ~0 atom. Acid R- C Carboxyl "-oH I c is the concentration of the solution (in grams per 100 Amine R-C-N~ Amino ml). I The two mirror images of a compound containing a chiral carbon, called enantiomers, differ in their optical ~0 Amide R-C Amide rotation; they represent optical isomers.
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