Chapter 15, Applications of Aqueous Equilibria
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1 Chapter 15, Applications of Aqueous Equilibria We will focus on 3 areas: 1) titrations 2) buffers (incl. the Henderson- Hasselbalch Transformation), 3) solubility equilibria. 1
2 I. Neutralization Reactions A. Strong acid-strong base 1. Let s start by looking at an example: HCl(aq) + NaOH(aq) W H 2 O(l) + NaCl(aq) 2. Can you write a net ionic equation for the above? + W 3. Will the above rxn have mostly products or reactants present at equilibrium? Logic? B. Weak acid-strong base 1. Again, let s start by looking at an example: 2
3 CH 3 COOH (aq) + NaOH (aq) W H 2 O (l) + CH 3 COONa (aq) Is Na + doing anything interesting? No, so leave it out! Remember your previous experience w/ spectator ions? CH 3 COOH (aq) + OH! (aq) W H 2 O (l) + CH 3 COO! (aq) How far will this proceed toward products? We can define a neutralization constant, K n, as: [CH 3 COO! ] K n = [CH3 COOH] [OH! ] 2. Is there any way we can get a value for K n without going into the lab and measuring it? 3
4 Try finding 2 equations that sum to the one above: CH 3 COOH + H 2 O W H 3 O + + CH 3 COO! K a = 1.8 x 10!5 H 3 O + + OH! W 2 H 2 O 1/ K w = 1.0 x CH 3 COOH + OH! W H 2 O + CH 3 COO! K n = K a (1/K w ) K n = 1.8 x Therefore, the rxn. goes essentially to completion. C. Strong acid-weak base. Same as B. above. D. Weak acid-weak base 1. In B, we could ignore Na + because it has essentially no acid-base properties. In this of problem, neither 4
5 component is weak enough to ignore. 2. Look at the rxn. of acetic acid with ammonia: CH 3 COOH(aq) + NH 3 (aq) W NH 4 + (aq) + CH 3 COO! (aq) What rxns. will sum to give us that rxn.? + W + K a = + W + K b = + W = CH 3 COOH (aq) + NH 3(aq) W NH 4 +(aq) + CH 3 COO! (aq) K n = K a x K b x 5
6 K n = 3. Will the system contain mostly products or reactants at equilibrium? 4. Perform a similar analysis with HCN (a weaker acid) as the acid instead of CH 3 COOH on your own. Try prob (a & d?), p II. The Common Ion Effect (build up to buffers) A. In Chapter 14 we looked at solns. of pure acid or base. We now consider what happens when you look at mixed systems. 6
7 1. Consider mixing acetic acid and sodium acetate: CH 3 COOH (aq) + CH 3 COO! (aq) W CH 3 COOH (aq) + CH 3 COO! (aq) Can we calculate [CH 3 COOH], [H 3 O + ], & [CH 3 COO! ] at equilibrium? ([Na + ] usually not of interest.) 2. Use the same approach we developed in Chapter 14 (Fig. 14.7). Main difference: [CH 3 COO! ] initial 0. a) Step #1, Identify reactive (interesting) species: CH 3 COOH Na + H 2 O CH 3 COO! acid inert acid/base base b) Step #2-3, Identify principle reaction: 7
8 CH 3 COOH + H 2 O W H 3 O + + CH 3 COO! K a = 1.8 x 10!5 c) Step #4, Set up the table. d) Step #5, Substitute values into the K a expression e) Steps remaining, do the algebra. (See Fig. 15.2) Do Prob Try Key Concept Prob. 15.5, p
9 III. Buffer Solutions A. These are very important in your body. B. A buffered solution resists ph changes upon addition of acid or base. C. How do we make a buffer? 1. Mix a weak acid with its conjugate base. 2. Mix a weak base with its conjugate acid. D. How does a buffer work? 1. Think back to the buffer we looked at in Prob a) What happens if you add a OH! to the buffer? 9
10 HCN + OH! ÿ H 2 O + CN! b) What happens if you add a H 3 O + to the buffer? CN! + H 3 O + ÿ H 2 O + HCN 2. Buffer capacity. There is a limit to how much acid or base the buffer can absorb. a) The amount of acid that can be absorbed is related to how much basic component (CN! above) of the buffer is present. b) The amount of base that can be absorbed is related to how much acidic component (HCN above) of the buffer is present. 10
11 E. Where (on the ph scale) does a buffer work? 1. Recall the K a expression: [H 3 O + ] [A! ] K a = [HA] 2. This can be rearranged to obtain: K a [HA] [H 3 O + ] = [A! ] 3. This tells us: a) The [H 3 O + ] (and therefore ph) is determined by the ratio of acid and conjugate base. b) The ph of effective buffering depends on K a. Do Key Concept Prob. 15.6, p
12 IV. Henderson-Hasselbalch Transformation Some concepts are much more clear if you look at them from a specific point of view. H-H Transformation makes some aspects of buffers more clear. This is a transformation because you are just rearranging the K expression. a 12
13 A. Derivation. Let s start with the K a expression: K a = Distributive law to get: Take log of both sides: Rearrange: [H O + ] [A! ] 3 [HA] [A! ] K a = [H 3 O + ] [HA] log K a = log [H 3 O + ] + log([a! ]/[HA])!log [H 3 O + ] =!log K a + log ([A! ]/[HA]) 13
14 Finally, substitute ph and pk a definitions: ph = pk a + log ([A! ]/[HA]) This is the H-H equation. (Note: pk = ) a B. What use is this, anyway? Let s see what happens when we mix equimolar quantities of buffer components (HA and A! ). ph = pk a + log (x/x) Because log 1 = 0, ph = pk a 14
15 1. Buffers are most effective buffering against both H + & OH! addition when buffer ph = pk a of that HA. 2. Look at the [base] ' [acid] ratios on p Try Prob , p Look at the Normal Values section of: a) Do any of: ph'paco 2 'PaO 2 'SaO 2 ' HCO 3! relate to variables in the H-H transformation? Which relate to acid-base chemistry? 15
16 ph ph = pk a + log([a! ]'[HA]) P a CO 2 P a O 2 SaO 2 HCO 3! b) Is HCO 3! acting like an acid or a base? See pk a values above, think of CO 2 leaving the body. 16
17 V. ph Titration Curves A. Titration: quantitative analysis method in chemistry. Based on chemical rxns. To do one, you need to know: 1. The stoichiometry for the reaction. 2. The concentration of the known component. 3. The volume of known component added. B. If you know these things, you can calculate the quantity of unknown present in a sample. C. You can also get pk a information from a titration. 17
18 VI. Strong Acid-Strong Base Titrations A. You get only quantitative information w/ these. B. The interesting component of a strong acid is H 3 O +, for a strong base it is OH!. 100% H 3 O + (aq) + OH! (aq) ÿ 2H 2 O(l) 1. See Fig for titration of HCl with known NaOH. 2. Shape of curve. Note equivalence point at ph =
19 VII. Weak Acid-Strong Base Titrations A. You get quantitative & K a information w/ these. B. Again, a weak acid reaction with a strong base goes to completion: 100% HA (aq) + OH! (aq) ÿ H 2 O (l) + A! 1. See Fig for titration of CH 3 COOH with NaOH. 2. Note equivalence point at ph 7.0. (Indicator?) Equivalence point: point in titration where stoichiometrically equal amounts of acid and base have been added. Seen as steep inflection point in graph. 19
20 3. Comment on the shape (vs. ph location) of the Fig curve for different weak acids. If you understand this figure, you are in good shape re. acid-base chemistry and buffers. To reinforce see Key Concept Prob 15.15, p. 607 VIII. Weak Base-Strong Acid Titrations (VII) IX. Polyprotic Acid-Strong Base Titrations A. Analogous to VII, above. Fig , p B. Try Prob , p. 612 on your own. 20
21 X. Solubility Equilibria A. Examples of biological solubility problems: 1. tooth decay 2. Atherosclerosis 3. Kidney stones (calcium oxalate) B. Consider the equilibrium: CaF 2 (s) WCa 2+ (aq) + 2 F! (aq) Can you write a K sp expression for this: K sp = 21
22 1. K sp is called the solubility product. 2. Different salts have different (sometimes very different) K sp values. Qual scheme? Try Prob c), p XI. Measuring K sp, Calculating Solubility from K sp A. Two ways to approach this problem: 1. Add increasing concentrations of components of interest until you see a ppt. (Example?) 2. Form a saturated soln. and then measure concentrations of ions in soln. B. We will examine some of the reasons for 22
23 differences in K sp values between different salts later. Try Prob , p XII. Factors That Affect Solubility A. The common-ion effect What happens if you add MgCl to a soln. of 2 MgF (aq)? See Prob , p B. ph Effects Look for component that reacts w/ H 3 O + or OH!. 23
24 Review pp on your own. This will be particularly helpful in your understanding of the qual scheme. Fig is particularly interesting. 24
25 Final thoughts on acid-base/buffer items: Change in acid-conjugate base ratios as a function of ph ph H 2 O H 3O , OH Aspirin pka: 3 HA 10 1 A H 2 CO 3 pka: 6.37 HA A - R O C O pka = H 2 O R O C O - + H 3 O + H Aspirin, HA form Aspirin, A - form O H C H O O Carbonic acid, HA form pka = H 2 O Carbonic acid, A - form If the HA form of aspirin crosses readily membranes, where in your g.i. tract will aspirin be absorbed? H O O C O - + H 3 O + 25
26 Can we start to consider the molecular basis for respiratory acidosis? Change in protein activity as a function of ph ph HProtein 1000 Protein pka = 6.0 HProtein + H 2 O Protein - + H 3 O + inactive active Most proteins have acid/base groups. Assume here that a protein has no activity in its HProtein form & is 100% active in the Protein - form. Would the activity be higher at ph 7.4 or 6.0??? What would a graph of activity vs. ph look like? 26
27 Draw the activity vs. ph graph below. 100 ph Activity Relationship 80 activity (%) ph 27
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