Ch 15, Applications of Aq Equilibria
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1 Ch 15, Applications of Aq Equilibria We will focus on 3 areas: 1) buffers (incl. Henderson-Hasselbalch Transformation) 2) titrations 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) H 2 O(l) + NaCl(aq) 2. Can you write a net ionic equation for the above? + 3. Will the above rxn have mostly products or reactants present at equilibrium? Logic? 2
3 B. Weak acid-strong base 1. Again, let s start by looking at an example: CH 3 COOH (aq) + NaOH (aq) 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) 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 ] 3
4 2. Is there any way we can get a value for K n without going into the lab and measuring it? Try finding 2 equations that sum to the one above: CH 3 COOH + H 2 O H 3 O + + CH 3 COO K a = 1.8 x 10 5 H 3 O + + OH 2 H 2 O 1/ K w = 1.0 x CH 3 COOH + OH 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. 4
5 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 component is weak enough to ignore. 2. Look at the rxn. of acetic acid with ammonia: CH 3 COOH(aq) + NH 3 (aq) NH 4+ (aq) + CH 3 COO (aq) What rxns. will sum to give us that rxn.? 5
6 + + K a = + + K b = + = CH 3 COOH(aq) + NH 3 (aq) NH 4+ (aq) + CH 3 COO (aq) K n = K a x K b x K n = 3. Will there be mostly prods or react at equilibrium? 4. Perform a similar analysis with HCN (a weaker acid) instead of CH 3 COOH on your own. Try prob (c), p
7 II. The Common Ion Effect (build up to buffers) A. In Chapter 14 we considered solns. of pure acid or base. We now consider in more detail what happens when you look at mixed systems. 1. Consider mixing acetic acid and sodium acetate: CH 3 COOH (l) + CH 3 COONa (s) Na + (aq) + H 3 O + (aq) + CH 3 COO (aq) Can we calculate [CH 3 COOH], [H 3 O + ], & [CH 3 COO ] at equilibrium? ([Na + ] ususally not of interest.) 7
8 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: CH 3 COOH + H 2 O 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 6 Step Prog. p. 604) 8
9 Try Key Conceptual Prob. 15.5, p Do Prob
10 III. Buffer Solutions A. These are really important in your body. B. A buffered solution resists ph changes (relative to a non-buffered solution) 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. 10
11 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? HCN + OH H 2 O + CN b) What happens if you add a H 3 O + to the buffer? CN + H 3 O + 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) is present. b) The amount of base that can be absorbed is related to how much acidic component (HCN above) is present. 11
12 E. Where (on the ph scale) does a buffer work? 1. Recall the K a expression: K a = [H 3 O + ] [A ] [HA] 2. This can be rearranged to obtain: [H 3 O + ] = 3. This tells us: K a [HA] [A ] 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 Conceptual Prob. 15.6, p
13 13
14 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 14
15 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]) 15
16 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 16
17 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: 17
18 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? 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. 18
19 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. 19
20 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) 2H 2 O(l) 1. See Fig for titration of HCl with known NaOH. 2. Shape of curve. Note equivalence point at ph = 7.0. Equivalence point: when stoichiometrically equal quantities of acid and base have been added to the system. 20
21 21
22 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 + OH H O + A (aq) (aq) 2 (l) 1. See Fig. 15.8, p. 620, for titration of CH 3 COOH with NaOH. 2. Note equivalence point at ph = 8.72 ( 7.0). (Indicator?) Can you use the H-H Relationship to understand why this (ph 7.0) must be so? 22
23 23
24 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. 24
25 To reinforce see Conceptual Prob 15.15, p
26 VIII. Weak Base-Strong Acid Titrations (VII) IX. Polyprotic Acid-Strong Base Titrations A. Analogous to VII, above. Fig , p
27 B. Try Prob , p. 618 on your own. 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) Ca 2+ (aq) + 2 F (aq) Can you write a K sp expression for this? 27
28 K sp = 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. 28
29 B. We will examine some of the reasons for differences in K values between different salts sp 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. 29
30 Review pp on your own. This will be particularly helpful in your understanding of the qual scheme. Fig is particularly interesting. 30
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