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.