Idea Transcript
Chapter 21 Buffers and the titration of Acids and Bases Shuffle order make 21-3 last! 21-1 The Henderson-Hasselbalch Equation What happens when you mix an acid and its own conjugate base? Using an acetic acid buffer as an example you can see that It won’t react with itself because you get no net reaction. CH3COOH + CH3COO-6CH3COO- + CH3COOH Recognizing that the acid and the conjugate base do not react with each other, let’s calculate the pH of .15M CH3COOH .10 M CH3COOK First, what happens to CH3COOK? Ionizes to CH3COO- and K+ Do I need to worry about K+? We will use the ICE table for the reaction CH3COOH + H2O 6 CH3COO-
+ H3O+
Since I have already shown that there is no net reaction CH3COOH + H2O Initial .15 — Ä -X Equil .15-X
6
CH3COO.10 +X .1+X
+ H3O+ 1x10-7 ~ 0 +X X
plugging in to the equilibrium expression 1.8x10-5 = X(.1+X)/(.15-X) We will assume that X is small compared to .1 or ,15 so 1.8x10-5 = X(.1)/.15) X =2.7x10-5 pH = 4.57
This is actually the long way to do this problem, there is another equation, called the Henderson Hasselbalch equation that get you your answer more quickly Key Equation Henderson-Hasselbalch Equation: pH = pKa + log(A-/HA)
Sample Problem K = 1.8x10-5 pKa = -log(1.8x10-5) = 4.74 pH = 4.74 + log (.1/.15) = 4.74 + -.18 =4.56 (of a touch due to round off error)
Restrictions on HH equation You can get the HH equation by using logs on your Keq equation, but I won’t go over that in class In deriving this equation there are a few assumptions that you should be aware of Key conditions for HH Equation 1. Ka between 10-4 to 10-11 2. Ratio of A-/HA from .1 to 10 3. [ ] between 10-3 to 1.0 Let’s try another problem What is the pH of a solution of .02 M NaH2PO4 and .05M Na2HPO4 First, what do we have in solution Na+ & H2PO4- ; 2Na+ & HPO42We can ignore the Na+ right? You should recognize that H2PO4- is the acid and HPO42- is the base Ka of H2PO4- = 6.2x10-8, pKa = 7.21 Can we use the HH equation? Ka OK, Concentrations OK, ration2.5, so OK too pH = 7.21 + log (.05/.02) = 7.21 + log(2.5) =7.61 21-2 Buffers Key Definitions: A buffer is a solution that resists changes to pH. It is usually composed of an acid and its conjugate base or a base and its conjugate acid. How does it work? The conjugate base will react with any added acid, and the acid will react with any added base. An important practical example is you blood. Your blood can absorb significant amounts of acids or bases with very little change in pH.
Buffer calculations are a simple extensions HH equation we just did. Example What is the pH of a solution that is 0.5 M HF and 0.5 M NaF. Note that HF is an acid, and NaF ionizes to Na+ ,a spectator ion, and F- the conjugate base of HF. HF is a weak acid so it doesn’t ionize very much, but we can assume that NaF ionizes completely thus at the start we have HF W Initial .5M (no ionization)
H+ + 0M (no ionization
F.5M (NaF completely ionized)
Ka = 7.2x10-4, pKa = 3.14 pH = 3.14 + log (.5/.5) pH = 3.14 + 0 pH = 3.14 II. Now that we found the pH of the solution, let’s see how it respond to the addition of a base. Let’s say we had 1 L of this solution and added .01 mol of solid NaOH. What will the pH be now? Solid NaOH will dissociate completely to make Na+ (aq) and OH-(aq). The OH- is a strong base and will try to take protons off any potential proton donor molecules What molecules will it take protons from? OH- + H2O W H2O + OH- Well that wasn’t too productive OH- + HFW H2O + F- A classic acid/base reaction this reaction will go to completion until either the OH- or the HF is used up. Let’s see which happens Total HF = 0.5 mol/L X 1.0 L = 0.5 mol Total OH- = [NaOH] = .01 mol NaOH is the limiting reactant and will be used up first With this in mind let’s look at the total stoichiometry before and after the addition of NaOH Before Conc Mole HF .5M .5mol F.5M .5 mol NaOH added .01 mol Change due to NaOH
HF = .5mols -.01mols = .49 mole in 1 L = .49 M F- = .5 mols + .01 = .51 mol in 1 L = .51M Once have gotten this figured we can use the HH equation again: pH = 3.14 + log (.51/.49) =3.14 + .02 pH = 3.16 pH change by .02 units This looks like the pH didn’t change much. How much do you think it would have changed if we added the .01 mol of NaOH to 1 L of water? Initial pH? pH 7 With this in mind your take home problem is to see how much the pH would change if we add .01 mol of HCl? (answer, should drop to 3.12) A buffer can only hold the pH relatively constant until either the acid or conjugate base in the buffer is used up. The number of millimoles of acid or base that can be added before the buffer is overwhelmed is called the buffer capacity Also please note that buffers resist change in pH when they are diluted. Say you had a solution at pH 4 made with 1x10-4 HCl, and you diluted it 1:10. What would be its pH? M1V1 = M2V2 1x10-4(1) = X(10) ; X =1x10-5, pH=5 Now say it was pH of 4 with a acetate buffer: .85M HoAC, .15M NaOAc 4 = 4.74 + log (.15/.85) = 3.99 But now diluting 1:10 4 = 4.74 + log [(.15/10)/(.85/10)] = 3.99 See how the concentration term drops out of the log! 21-3 Indicators - Shuffled to later 21-4 Strong Acid - Strong base titrations Titration curves are found experimentally by following the pH of a solution as you add titrant The following detailed analysis is something I cover in depth in Analytical chemistry, and is a bit beyond what I think is appropriate at this level. I may sketch what the curves look like, and point out where the equations you have learned are applied, but skip over the rigorous tedious details of every single calculation.
One common type of titration is when you titrate an acid with a base. If the concentration of the base is known you can use this titration to determine the how much of an unknown acid you have (or vice versa). Up till now when we did a titration we only worried about where the endpoint occurs, that is , where we had added an amount of base equivalent to the amount of acid being titrated. Now however, since we are beginning to appreciate what pH is , and how it can give us clues as to what is going on in solution, let’s study what happen to the pH of the sample in and acid base titration. The easiest way to monitor what happens to pH in a titration is to make a pH curve or titration curve ( this is a plot of pH vs amount of titrant added) and looks something like this or this or this At this point you have the tools to predict what the titration curve of virtually any acid/base titration looks like. Let’s give it a try firs with a strong acid and base Let’s titrate 40 mL of 0.5 M HCl with 0.5 M KOH A. Initial pH When we start the titration the only thing we have in the flask is the 0.5 M HCl So what is it’s pH? 0.5 M HCl = 0.5M H+, 0.5M Cl-, Cl- does not contribute to pH, pH = .31 B. After addition of base but before equivalence point What happens after I add say 10 mL of base Species 40 mL 0.5 M HCl .04(.5) = .02 mol H+ = .02 mol Cl-(makes no contribution to pH can ignore 10 mL 0.5 M KOH .01(.5) =.005 mol OH=.005 mol K+(no contribution to pH ignore) 60 ml total H2O .0000001 M H+ (negligible can ignore) .0000001 M OH-(negligible can ignore) Reaction? Initial RXN Final
H+ + .02 -.005 .015
OH- = .005 -.005 0
H2O constant constant constant
So still have excess H+, determine pH from that [H+] = .015 mol / vol; vol = 40 mL + 10 mL = 50 mL = .015/.05 = .3; pH = .52
Similarly for 20 mole H+ = .02 - .02(.5) = .01; [H+] = .01/(.04+.02) = .17M ; pH = .78 For 30 mL mole H+ = .02 - .03(.5) = .005; [H+] = .005/(.04+.03) = .07M ; pH = 1.15 C. Equivalence point What about at the equivalence point where we have added equivalent amount of acid and base First where is equivalence point? equal amount of acid and base , Started with 40 ,L of .5 M acid = .02 mole so need place where add .02 mol of base mole = molarity X vol .02 = .5 (X) X = .04 L = 40 mL At this point all H+ that came form acid has been reacted with OH- from base Species in solution? K+, Cl- H2O, and H+ and OH- from autoionization of H2O K+ and Cl- don’t contribute to pH so only autoionization of H2O. so pH = 7! D. Points after equivalence point Say 60 mL NaOH 60 mL NaOH = .06(.5) or .03 mole OHReaction? Initial RXN Final
H+ + .02 -.02 0
OH- = .03 -.02 .01
H2O constant constant constant
So can see pH determined by excess OH[OH-] =.01 mol / (.04 + .06) = .1M; pOH=1, pH=13 Sketch titration on board Note how would change if did titration of a strong base with an strong acid? Curve go other way, calculation look the same, determine whether acid or base is in excess, then calculate actual concentration based on mole ad volume
21-5 Weak Acid - Strong base titrations & 21-6 pH = pKa at midpoint Let’s see what will change if we do the titration of a weak acid with a strong base Let’s do 40 mL of 0.5 M Lactic acid (Ka 1.38x10-4) with our .5 M KOH solution (Lactic acid CH3CH2COOH) A. initial pH Well by know you should be familiar with this calculation so I’ll cut to the chase X= [H+]=[A-] Ka = 1.38x10-4 = X2/(.5-X); assume .5-X = X 1.38x10-4 = X2; X= 1.17x10-2=[H+]; pH= 1.93 Note, we start off at a higher pH this makes sense. weak acid doesn’t ionize doesn’t make as much H+ [H+] lower pH higher. B. Before equivalence point What happens after we add say 10 mL of KOH? Before reaction 40 mL 0.5 Lactic A = .04(.5) = .02 HA (lactic acid) = Negligible A-( doesn’t ionize much) 10 mL 0.5 M KOH = .01(.5) = .005 mol OH= .005 mol K+ Autoionization of water .04(1x10-7) mole H+ negligible .04(1x10-7) mole OH- (negligible) Reaction? Initial RXN Final
OH- + .005 -.005 0
HA W .02 -.005 .015
A- + 0 +.005 .005
H2O constant Constant constant
We can calculate pH from the Ka equation or the HH equation Using the HH equation and the trick I showed you earlier is probably the quickest pH = pKa + log (A-/HA) pKa =3.86 pH = 3.86 + log .005/(.04+.01) .015/(.04+.01) Again can volumes cancel so can calculate directly with moles pH = 3.86 + log .005/.015) =3.86 -.48 =3.38
20 mL HA = .02 - .5(.02) = .01 A- = .5(.02) = .01 pH = 3.86 + log (.01/.01) = 3.86 Notice, this is an important point in your titration curve. ½ way to the equivalence point pH = pKa 30 mL HA = .02-.5(.03) = .005 A- = .5(.03) = .015 pH = 3.86 + log (.015/.005) = =3.86 + .48 = 4.34 40 mL Equivalence point The EP is defined as the spot where you have an equivalent amount of acid or base, it doesn’t matter whether are strong or weak so will occur at the same volume. The pH at this point will be different however. What Species do we have in solution at the equivalence point? K+ , All lactic acid (HA) converted to A-, and water Notice that our acid is now in it’s conjugate base form, and is now acting as a weak base! Thus our pH here is going to be determined by the weak base What is the KB of this weak base? (You should remember Kb = Kw/Ka = 1x10-14 / 1.38x10-4 = 7.25x10-11 The reaction we ant to look at is Kb = [BH][OH-]/ [B] mole base = moles of acid we started with = .5(.04) = .02 moles [B] = .02/(.04 + .04) = .025 M giving us Kb = [BH][OH-]/ [B] if X = [OH-] = [B] Kb = XX/ .025-X Assume .025-X = .025 Kb = X2 7.25x10-11 = X2
X = [OH-] = 8.51x10-6 ; pOH = 5.07; pH = 8.93 Note that in our titration of a weak acid we have converted our weak acid into a base, to the equivalence point has a basic pH (>7) D. Points after EQ PT At this point have excess OH-, calculate the same as before Let’s compare these curves Points 1. pH of weak acid start higher 2. pH of weak acid has buffer region 3. Equivalence points have different pH (neutral vs basic) What happens if use a different acid say one with a pKa of 10-3 instead of 10-4? Should be able to guess that stronger acid shifts down, in fact ½ to EQ pH = pKa 21-7 Weak base - Strong acid titrations (May not need to do because similar to weak acid/strong base) 35 mLs of 0.1 M methyl amine (kB = 4.38x10-4) with 0.09 M HNO3 Note: titration curves are easier if you find you EP first. This helps you recognize which equation to use so let’s do it. Where is the EP? Ep is where Acid = base Base = .035 X .1 = .0035 moles Vol of acid equivalent? Vol X Molarity = mol ? X .09 = .0035 ? = .0035/.09 = .03889 L = 38.89 mL Now let’s go on to the titration A. initial initially all B in solution Kb = [BH+[OH-] / [B] X= [OH-] = [BH+] Kb = X2 /(0.1-X) Assume 0.1-X . 0.1 4.38x10-4 = X2, X= 2.09x10-2 = [OH-], pOH = 1.67, pH = 12.33
B. Up to equivalence point (from 0 to 38.89 mLs Pay attention to what is in solution say we are at 10 mL of acid Started with .035(.1) of .0035 mol base At this point have added .01(.09) or .0009 mol acid Acid will react with base B + H+ 6 Initial .0035 .0009 RXN -.0009-.0009 Final .0026 0
BH+ 0.0 +.0009 .0009
If base ionizes B would be a touch lower and BH would be a bit higher. by factor X Kb = [BH+][OH-] / [B] [BH+] = .0009/(.04+.01) = .018 [B] = .0035 / (.04 + .01) = .07 = (.018-X)X / (.07+X) Will assume that X is small to .018-X = .018 and .07+ X =.07 KB = 4.38x10-4 = .018(X)/.07 X = .07(4.38x10-4/.018 = 1.7x10-3 = [OH-] pOH = 2.77, pH = 11.23 And so on up to equivalence point EP 38.89 mL at this point will have .0035 mol of base and .0035 mol of acid The acid will convert all of the base to it’s conjugate acid so will have .0035 mol of conjugate acid in 40 + 38.89 ml = .0035/.07889 = .0444 Ka of conjugate acid? Ka = Kw/Kb = 1x10-14 / 4.38x10-4 = 2.28x10-11 Ka = X2 / .0444-X Assume .0444-X = .0444 2.28x10-11(.0444) = X2 =; X = [H+] = 1x10-6; pH = 6 (this was an acid reaction so got [H+] not [OH-])
21-3 Indicators Cutting to the chase. An indicator is just another acid-conjugate base pair. The only difference is that the acid has a different color than the conjugate base. Key Concepts 1. The pKa of the indicator should match the pH at the equivalence point of your titration 2. Since the indicator is an acid/conjugate base pair, it is a buffer! So you want to use as little of it as possible so it won’t change the pH or the equivalence point of what you are titrating.