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FINDING THE VALUE OF AN EQUILIBRIUM
CONSTANT II: Determining the pKa of a weak
acid by potentiometric titration.
Objectives:
 Make and standardize a 0.1 M solution of NaOH;
 Generate a titration curve by titrating a weak acid with a strong base;
 Use a pH meter;
 Determine the pKa of the weak acid.
THE SAGA OF “p”
Before you come to lab:
“p” (lower case p) is yet another
 GO back to the CH 103 web site and get the
chemistry abbreviation. It stands
“Protocol for Making 0.1 M NaOH” from the WEEK
for the negative logarithm of
6 lab)
something {-1 x log(any number)}.
Thus:
 Get the “Protocol for Standardizing NaOH by titration with
pKa is –log of the number Ka
KHP” From the Chem 103 WEEK 8 Lab
+
pH is –log [H ].
This must NEVER be written as
capital P. If you write “PKa” or
“Ph” for pKa or pH, chemists will
think you are referring to Doctors of
Philosophy (PhD’s), phenyl groups,
or, God forbid, that you are
ignorant.
The habit of using logarithms of
numbers rather than the numbers
themselves evolved for several
reasons, three of which are:
1. Logs of large numbers and – logs
of small numbers are a way to make
numbers look more accessible and
friendly. 0.00000345 is harder to
get your mind around than 5.462
(-log (0.00000345)
2. The data signal of the
instrument you will use today, the
pH meter, is proportional to the log
of the [H+]., not the [H+] directly.
3. You will be looking at
concentration ranges form 0.1 to 1 x
10-13 M H+ . That’s 12 orders of
magnitude. If you tried to graph
this out directly on a sheet of graph
paper, it would quickly get out of
control. In fact, if you made the
distance between 1 x 10-13 and 2 x
10-13 equal to 1 mm, the distance
between 0.1 and 0.2 would be
roughly 3 times from here to the
moon. When dealing with this
stretch of data, logarithms are much
more manageable.
One of the most useful and important notions in chemistry is the
concept of systems at chemical equilibrium. We are used to
thinking about chemical reactions going forward, from reactant to
product:
CH3COOH + H2O  H3O+ + CH3COO- (forward reaction)
In fact, all chemical reactions can also proceed in the reverse
direction, from products to reactants:
H3O+ + CH3COO-  CH3COOH + H2O (reverse reaction)
If you start out with ONLY reactants (in our example, acetic acid
and water) and NO products (hydronium ion and acetate ion), the
reaction will proceed “as written” in the forward direction. What
else can it do? H3O+ or CH3COO- , the reactants for the reverse
reaction, haven’t been made yet. However, as the forward reaction
keeps going, H3O+ and CH3COO- get made. Now the reverse
reaction can get started.
After a while, both reactions run at the same rate. As fast as
The forward reaction makes H3O+ and CH3COO-, the reverse
reaction shoots it back to CH3COOH and H2O. Even though both
reactions are going on, concentrations of CH3COOH, H2O, H3O+
and CH3COO- remain constant. At this point, the reaction is said
to be at equilibrium. The reactions haven’t stopped, but nothing is
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changing (concentrations, temperature, whatever). Unless something comes along to mess with
the system, like you heating it up with a Bunsen burner, or adding more acetic acid, the system
won’t change. That’s equilibrium. At equilibrium, you can write the balanced chemical equation
like this:
CH3COOH + H2O ⇄ H3O+ + CH3COO-
The double arrow (⇄) indicates that the reaction is
going in both the forward and reverse directions, and is at equilibrium. (⇌, ↔, and ⇆ are also
used as equilibrium symbols)
One of the nice things about systems at equilibrium is that they let you write equations (called
“equilibrium expressions” or “laws of mass action” describing them, using an equal sign. Equal
signs make your algebraic life much much easier.
For any common or garden variety reaction:
aA + bB
⇄ cC + dD
you can write an equilibrium expression (Law of Mass Action):
K = [C]c [D]d
[A]a [B]b
where K is the equilibrium constant. [A], [B], are the concentrations of reactants at equilibrium.
[C] and [D] are concentrations of products at equilibrium. a, b, c and d are just the
stoichiometric coefficients.
A Bestiary of Equilibrium Constants
The equilibrium expression describing the reaction
CH3COOH(aq) + H2O(l) ⇄ H3O+(aq) + CH3COO-(aq)
Equilibrium constants are just what they say they
are, a numerical constant which is equal to the
concentration of products divided by the
concentration of reactants of a reaction at
equilibrium.
That said, they are given a variety of different
subscripts to tell the reader what kind of reaction
they are describing. A partial list of equilibrium
constants:
Ka given to K’s for reactions of acids with water:
HA + H2O ⇄ A-+ H3O+ Ka = [A-][ H3O+ ]/[HA]
Kb given to K’s for reactions of bases with water:
A-+ H2O ⇄ HA + OH- Kb = [HA][ OH- ]/[A-]
Kw given to reaction of water with water:
H2O + H2O ⇄ H3O+ + OH- Kw = [H3O+][OH-]
Ksp given to K’s for reactions of slightly soluble
salts with water:
MA + H2O ⇄ M+ + A- Ksp = [M+][A-]
Keq and Kc given to K’s for reactions that aren’t any
of the above.
is:
Ka = [H3O+][CH3COO-]
[CH3COOH]
.
The subscript “a” indicates that the K is for a reaction of
an acid with water.
Water is a reactant in this reaction, but you’ll notice it’s left
out of the equilibrium expression. What’s the story? The
concentration of pure liquids and pure solids is constant.
Actually, the [H2O] is there, but since it is a constant, it is
incorporated into the equilibrium constant Ka.
Before you come to lab: If the DENSITY of water
is 1.00 g/mL and the molecular weight of water is
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18.0 g/mol, calculate the [H2O] in moles/L.
Tables of Ka’s can be found in the Handbook of Chemistry and Physics, and in Appendix C, Table C20 in Munowitz.
What do equilibrium constants tell you?
HCl(aq) + H2O(l) ⇄ H3O+(aq) + Cl-(aq)
Ka = [H3O+][Cl-] = 1 x 106
[HCl]
Ka (HCl) is HUGE. This means that if you dump HCl in a bucket of water and stir it up with a spoon,
at equilibrium, [H3O+] and [Cl-] will be very large in comparison to [HCl], which will be very small.
Which makes sense. We know HCl is a strong acid, which ionizes virtually 100% in water.
On the other hand, consider acetic acid:
CH3COOH(aq) + H2O(l) ⇄ H3O+(aq) + CH3COO-(aq)
Ka = [H3O+][CH3COO-] = 1.76 x 10-5
[CH3COOH]
Ka (HAc) is pretty small. This means that if you dump acetic acid in a bucket of water and stir it up
with a spoon , at equilibrium, [H3O+] and [CH3COO-] will be very small in comparison to
[CH3COOH], which will be relatively large.
Which makes sense. We know acetic acid is a weak acid, which hardly ionizes at all in water.
The smaller the Ka, the less product gets made before the system comes to equilibrium, and the
weaker the acid. Hypoiodic acid, HIO, for example, is a very weak acid, with Ka = 3 x 10-11. Add 1.0
mole of it to a liter of water and only 0.00055% of it will ionize. Iodic acid, HIO3 is much stronger
(though still a weak acid). Its Ka is 1.7 x 10-1. Add 1.0 mole of it to a liter of water and 34% of it will
ionize.
The Ka’s of weak acids are LESS THAN 1. Typically, you will see them listed as pKa’s [-1 x log(Ka)].
Thus, Ka(HAc) = 1.76 x 10-5, pKa = 4.75. Why bother with pKa’s? Well, for one thing, they are easier
to type!
Which brings us to this week’s lab.
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Your goal: Find the pKa (and calculate the Ka) of an unknown weak acid. Do this by titrating a
known volume of a weak acid with a known amount of sodium hydroxide. Record the progress of
the titration by measuring the pH (-log[H+]. This technique, by the way, is called a potentiometric
titration.
You will be issued a weak acid, call it HA. As you know, when the weak acid is in water, the
following reaction occurs:
HA + H2O ⇄ A- + H3O+ and Ka = [A-][ H3O+]/[HA]
Notice that H3O+ is a product, and you know how far the reaction proceeds if you know what [H3O+
is. There is an instrument which measures this, called a pH meter, which measures potentials
voltages across an electrode permeable to H3O+ (Hence the term potentiometric titration). (Read the
memoirs of the father of the pH electrode at: http://anestit.unipa.it/esiait/astruing.htm )
The pH meter measures potentials with are proportional to the log of the
[H3O+], so the output is in pH, not [H3O+]. That being the case, let’s do
some algebra on the equilibrium expression so that it has pH and no
[H3O+]
Ka = [A-][ H3O+]/[HA]
Take logs of both sides:
A pH meter and electrode
http://www.lavallab.com/wateranalysis/ph-redox/hanna-bench-meters.htm
Log Ka = log[A-] + log[ H3O+] - log[HA]
Multiply both sides by –1:
-Log Ka = -log[A-] - log[ H3O+] + log[HA]
Remember –log is “p”
pKa = -log[A-] + pH + log[HA]
Neaten things up:
pH = pKa + log[A-]/[HA]
This equation was originally derived by LJ Henderson of the Harvard Medical
School, and KA Hasselbach, a Danish physiologist, studying pH levels of
blood.
The Henderson-Hasselbach equation relates a variable you can measure easily (pH) to a value you
want, pKa. Useful. Now, if we knew [A-] and [HA], we’d be golden. That’s where the NaOH
titration comes in. Acids react with bases, and your weak acid HA will react with NaOH:
HA + NaOH  NaA + H2O, or HA(aq) + OH-(aq)  A- + H2O
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If you know how many moles of NaOH you add to your weak acid HA, you know how many moles
of A- you have left over. Volume of NaOH added is proportional to moles of NaOH is equal to
moles of A- made. If you measure pH and volume of NaOH added to your weak acid, you can
generate a titration curve:
Top
line
pH = pKa
Bottom
line
Half-way
point
Equivalence
point
The equivalence point of the titration (where exactly enough base has been added to react with all the
HA, no less and no more) is the inflection point of the graph. You can find that by drawing two
tangent lines (“top line” and “bottom line” ). Connect them with a third vertical line (blue). The
equivalence point is the midpoint of the blue line between the top & bottom lines. Knowing the
equivalence point will allow you to calculate the molecular weight of the unknown acid.
What about the pKa? Well, now that you know (From the equivalence point ) how much NaOH you
have to add to react with ALL the HA, you know also how much base you have to add to react with
exactly HALF the HA. (In the example above, 100.00 mL/2 = 50.00 mL). Exactly half-way through
the titration (HA(aq) + OH-(aq)  A- + H2O) if you started with 1 mole of HA, you would have made
0.5 moles of A- and had 0.5 moles of HA left. A-/HA = 0.5/0.5 = 1:1. This 1:1 ratio is true no matter
how many or how few moles of HA you started with. Halfway through, half is HA and half is A-.
Plug that into the Henderson Hasselbach equation:
pH = pKa + log[A-]/[HA]
pH = pKa + log1/1
pH = pKa + log1
pH = pKa + 0
Half way through the titration, pH = pKa.
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THE EXPERIMENT:
To make 6 M NaOH:
You will need:
MW NaOH = 40 g/mol.
To Make 0.1 M NaOH:
Take 240 g solid NaOH (Be careful, it’ll do
1 L bottle and stopper
1
a number on your skin and eyes! And
50 mL graduated cylinder
1
dissolve it up to 1 L in distilled water. It
******************************
will need a lot of stirring with a magnetic
6 M NaOH in stoppered
spin vane, and it will get HOT!
ehrlenmeyers in the hood
To Standardize NaOH:
250 mL Ehrlenmeyer flasks
Magnetic stirrer and spin
vane
50 mL buret
Buret reading card
4
1
1
1
Primary standard KHP (dry in oven) 2 g/student pair
2% phenolphthalein in ethanol
2 dropper bottles/riser
To analyze the unknown weak acid HA:
250 mL Ehrlenmeyer flasks
Magnetic stirrer and spin
vane
Hot plates
50 mL buret
Buret reading card
pH meter & electrode with
soaking flask
Wash bottle of DI water
250 mL waste rinse beaker
pH 4, 7 and 10 standards
KNIGHT CARD TO GET
LAPTOP COMPUTER
2
1
Demo set up: pH meter,
buret, pH meter, pH 4, 7 and
10 standards, pH meter
calibration instructions, wash
bottle, waste rinse beaker,
electrode soaking flask
1
1
1
1
1
1
1 set
1
Unknown monoprotic weak acid 1 g /student pair
Caution: NaOH is caustic and corrosive. Protect your eyes with safety glasses. NEVER pipet by
mouth!.
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THE PROCEDURE:
Before you come to lab: Make an Excel Template of the data chart on page
9. PUT IT ON A FLOPPY DISK & BRING IT TO LAB.
BRING YOUR KNIGHTCARD TO LAB!!!!
Make up your 0.1 M NaOH solution per the quick and dirty protocol in the handout “Protocol For
Making 0.1 M NaOH”.
2. Standardize your 0.1 M NaOH against KHP according to the protocol outlined in the handout
“Standardizing Your NaOH”.
3. CALCULATE AND WRITE DOWN the average [NaOH] you determined in part 2 in your
notebook!
1.
Now you have the tools in hand to analyze your unknown acid, and determine its pKa and molecular weight.
4. Obtain a solid unknown weak acid from the TA. RECORD THE UNKNOWN NUMBER! Write it
down in your notebook!
5. Rinse out your Erlenmeyer flasks from the NaOH standardization with distilled water. Don’t bother
to dry them.
6. Weigh (to the nearest 0.1 mg) 2 samples of your unknown HA. Since I the author know what the
“unknown” is I can suggest that a convenient mass is about 0.5 g. Write down your 2 masses in your
notebook!
Put one weighed HA unknown sample into each of 2 labeled Erlenmeyer flasks. Be sure to keep track of which
known mass goes into which flask!
Get your bottle of standardized NaOH and shake it up. Fill your buret with NaOH, read and record the
volume.
DOING THE TITRATION:
7. Find a pH meter. Calibrate it with the pH standards according to the instructions on the card. Set your
magnetic stirrer up under the buret & make sure the buret tip has no drops dangling from the tip. If it
does, rinse the tip with distilled water from your wash bottle into a waste beaker. Put your flask on the
stirrer, and add a clean washed magnetic spin vane (it need not be
dry) to the flask. Slowly turn on the stirrer until you get a nice
gentle vortex. Once again, stirring is key!
8. Have you recorded your buret volume? If not do it now, to the
nearest 0.02 mL.
9. Put the pH probe into the flask, in the solution and record the
pH. Don’t let the spin vane hit the electrode. Open the stopcock
and start running NaOH into your flask. Add 0.5-1 mL of NaOH,
close the stopcock, rinse the sides of the flask with DI water, and
Wash down the sides of your flask while
record the VOLUME and the pH. Keep adding 0.5 mL of NaOH
titrating.
at a time, recording pH and volume after each addition.
10. Slow down! As you move toward the endpoint, it will seem like
you’ll never get there. The pH changes very little with each
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addition of NaOH. Hang on, you are in the “buffering range” of the titration. The HA and A- resist
changes in pH here. Be patient. SLOW DOWN. When you get near the endpoint, SLOW DOWN! just
one drop of NaOH will change the pH a lot (look at the graph), and your data will be crummy if you
overshoot. You’ll get there. After the end point, add at least 10 more mL of NaOH, 0.5- 1 mL at a time.
Record your buret readings +/- 0.02 mL.
11. Repeat steps 7-10 with your other flask of previously massed and recorded unknown HA.
Calculations:
A.
B.
C.
D.
E.
Graph your data, with pH on the Y-axis and mL NaOH dispensed on the x-axis.
Determine graphically how many mL of NaOH were required to get to the equivalence point;
Calculate how many mL of NaOH were required to get a 1:1 ratio of HA and A- ;
Determine graphically the pKa of your unknown.
Calculate how many moles of NaOH were required to get to the equivalence point . You will use the
average moles/L NaOH you calculated from your standardization REM: L NaOH x (moles NaOH/L) =
moles NaOH;
F. Using the balanced equation: HA + NaOH  NaA + H2O Calculate how many moles of HA you
started with.
G. From moles HA and mass of HA, calculate the molecular weight of HA in g/mole
Sample calculation:
Data:
[NaOH]: 0.1040 M
Mass HA (UK #5021953): 0.2537 g
NaOHstart: 2.45 mL
NaOHend: 27.32 mL
E. mL of NaOH to equivalence point: 27.32 mL - 2.45 mL = 24.87 mL;
L of NaOH to equivalence point: 24.87 mL(1 L/1000 mL) = 0.02487 L
moles of NaOH to equivalence point: 0.02487 L (0.1040 mole NaOH/L) = 2.586 x 10-3moles NaOH
F. moles of HA : 2.586 x 10-3moles NaOH (1 HA/1 NaOH) = 2.586 x 10-3moles HA
G. MW HA: 0.2537 g HA/2.586 x 10-3moles HCl =98.11 g/mol
12. CLEAN UP! Rinse and refill burets with distilled water. Be sure to rinse through the tips!
Return spin vanes! Rinse all glassware with distilled water and return them! Leave pH meters with
electrodes soaking.
ALL WASTE FROM THIS EXPERIMENT GOES IN THE JUG LABELLED “ACIDS AND
BASES” IN THE HOOD.
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SAMPLE EXCEL DATA TABLE:
LAB WRITE UP:






You should have all your data from your NaOH standardization and Unknown Acid analysis in Excel.
Report each [NaOH], the average, the standard deviation and the % error.
Report the unknown number of your unknown.
Report each pKa of your unknown, and the average pKa
Report each calculated molecular weight of your unknown, and the average molecular weight.
Include a summary table of your data (Items in italics above) and an analysis of your results.
REMEMBER TO INCLUDE YOUR UNKNOWN NUMBER!