Determination of the Dissociation Constant
and Molar Mass for a Weak Acid
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Abstract: We will determine Ka and the molar mass for an unknown weak acid
by using a pH meter to record the pH at intervals during the titration with
sodium hydroxide. The titration curve
and its first derivative will be plotted to establish the equivalence point.
Introduction
The strength of an acid is defined by its ability to
donate a proton to a base. For many
common acids, we can quantify acid strength by expressing it as the equilibrium
constant for the reaction in which the acid donates a proton to the standard
base, water, as shown in the equations below:
HA
+ H2O Û H3O+
+ A-, 
for H3CCOOH:
H3CCOOH + H2O Û H3O+ + H3CCOO - 
The equilibrium constant for a reaction of this type
is called the Acid Dissociation Constant, "Ka", for the acid HA.
A convenient method for
determining Ka is to measure the pH of a solution of the acid after a strong
base has been added to half neutralize it.
We can calculate the amount in millimoles (mmol) of base added by
multiplying the volume in mL by the concentration in mmol per mL, which is the
same as the molar concentration (in moles per liter) since both the numerator
and denominator are divided by 1000:
n(mmol)
= V(mL) x C (mmol/mL)
For example, if 10 mL of
0.10 M NaOH is added to 20 mL of 0.10 M acetic acid, the solution will
initially contain 10 mL x 0.10 mmol/mL = 1 mmol of NaOH, and 20 mL x 0.10
mmol/mL = 2 mmol of acetic acid.
OH- +
H3CCOOH
Û H2O + H3CCOO-
init 1
mmol 2 mmol
change -1 mmol -1 mmol +1 mmol
final 0
mmol 1 mmol +1 mmol
since the reaction goes essentially to completion,
the 1 mmol of NaOH will be completely consumed, converting 1 mmol of acetic
acid to acetate ion. Then [HA] = [A-] =
1 mmol/30 mL = 0.033 M. The Ka
expression is simplified because equal terms cancel out:
= 
Thus at the point of half-neutralization, Ka = [H3O+]
or pKa = pH, where
pKa
= - log Ka, just as
pH
= - log [H3O+].
The Henderson-Hasselbalch equation,
gives the same result, since the second term is log
1, or 0.
The treatment for a diprotic
acid is essentially the same: A
diprotic acid has two Ka's:
H2A
+ H2O <=> H3O+ + HA-
HA-
+ H2O <=> H3O+ + A2-
The two Ka's can be determined separately using the
same methods suggested here for monoprotic acids.
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Titration
Curves
In order to determine the pH
at half-neutralization for an unknown acid, we will measure the pH with a pH
meter after each addition of base, and create a "Titration Curve"
which resembles the one in the figure below.
The experimental setup is shown at right. A 200 mL beaker contains the acid dissolved in distilled water
and a magnetic stirbar, so that the magnetic stirrer motor below can be used to
continually stir the mixture during the titration. The pH electrode is supported by a clamp on the left, while the
burette is held by a burette clamp in the right of the beaker, with its tip
well below the lip of the beaker to prevent titrant loss by splashing.
At the beginning of the
titration, from 0 mL to about 20 mL of added base, the pH changes very little
with each addition, and 1-2 mL of base can be added at a time. But after about 20 mL, as the titration
approaches the equivalence point, the
pH changes drastically with very small additions of base. Here it is necessary to add small volumes of
base (»0.1 mL) at a time, in order
to get sufficient points to define the shape of the curve.
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The equivalence point is the
volume at which the slope is greatest, or at which the inflection point occurs1,
where the line changes from upward curvature to downward curvature. In this
titration the equivalence point occurs when about 22 mL of base is added, and
the pH is 9.
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The equivalence point can be
determined visually, or an interesting method developed in calculus can be
used. The "first derivative"
is actually the slope of the line at a given point. It can be estimated numerically by calculating the slope of a
line in the region of the point, and the LIMSport
spreadsheet KA_DET uses this method. If
the first derivative of the titration curve is plotted, it looks like the figure
shown here.
The first derivative has a
maximum at the equivalence point where the slope of the titration curve is
greatest, indicating large pH changes
with small volume additions. By the
way, the second derivative is the slope of the first derivative, so it will be
zero at the equivalence point, where the slope of the first derivative is
horizontal.
The first derivative is
DpH/DV = (pH)2-(pH)1/
(V2-V1), but this is the average slope between points 1
and 2, and cannot be paired with either (pH)1 or (pH)2
for plotting, as we are forced to on a spreadsheet. To solve this problem, three pH values are used, and the
derivative is paired with the middle (pH)2 one in the plot. It is calculated from the pH and V values
half way between the central value and the next higher and lower values:
Once the equivalence has
been determined, the pH can be read half way to the equivalence point, and from
it the Ka of the acid, as described above.
If the acid were diprotic,
there would be two inflection points in the titration curve above, and the
pKa's would be equal to the pH at points halfway between 0 and the first
equivalence point, and halfway between the two equivalence points.
Preliminary Titration
Since
the acid is an unknown, it is impossible to predict what volume of base will be
required to titrate a given mass. To
work efficiently while using the pH meter, we want to add larger volumes of
base until just before the pH changes rapidly near the equivalence point, and
only then add small volumes so that the inflection point can be determined
accurately. It usually saves time if a fast, approximate standard titration
with an indicator is done first to determination the mass of acid which will
require 25 mL of base solution.
Having the equivalence point
occur at 25 mL is appropriate for 50 mL burettes, because it gives good
precision without using excessive volumes of base. Then, during the pH meter
titration, base can be added rapidly until the equivalence point is approached.
For example, you might start with 2 mL increments for the first 16 mL, then 1
mL increments up to 23 mL, gradually reducing to 0.1 or even 0.05 mL increments
very near the equivalence point. The approximate titration can be done with
phenolphthalein to determine the endpoint, or it can be done by using the pH
meter to determine the equivalence point, but by adding 1 mL of base at a time
and determining the pH after each addition.
Determination
of the Molar Mass of the Acid
Since
the molar mass is the amount in moles of a given mass of acid,
M(g/mol)
= m(g)/n(mol),
it can be determined for a monoprotic acid by
titrating a known mass to determine the amount of H+ present. Since one mole of H+ reacts with
one mole of OH- in the
titration,
n (H+).=
n (OH-) = C(M)
x V(L) = C(mol/L) x V(L) = C(mmol/mL)
x V(mL)
Note that for a monoprotic acid like HCl, the amount
of acid equals the amount of hydrogen ion titrated ( 
), but for a diprotic
acid like H2SO4, the amount of acid is ½ the amount of
hydrogen ions titrated, 
.
Prelaboratory
Assignment
Procedural
1. From the
titration curve shown earlier, estimate the change in pH that would result from
the addition of 1mL of .1 M NaOH to 25 mL of .1 M acetic acid (a) at the beginning
of the titration; (b) after 21 mL had been added; and c) after 30 mL had been
added.
2. Why is it
necessary to do an indicator-based titration before the pH meter
titration? How does the pH meter
titration differ?
3. Does it
matter how much water is added to dissolve the unknown acid? Why or why not?
Theoretical
4. Determine
the mass of acid which should be used
in the titration if it is found in a preliminary titration that 0.100 g of the
acid requires 8.0 mL to titrate. Hint:
5. In the
titration curve shown it the introduction, the equivalence point occurs at 22.0
mL, so the half-equivalence point is at 11.0 mL. From the plot, estimate the pKa and Ka for the acid.
6. What is
the molar mass of a monoprotic acid if 0.200 g of the acid requires 22.5 mL of
0.15 M NaOH in a titration?
7. How would
you write the Excel equation to calculate
a. the mass of acid requiring 25 mL of NaOH
from the mass and volume of titrant in step 4?
Assume that the mass of acid is in cell A9 and the volume of NaOH
required in A11.
b. the equivalent weight of the acid in step
11? Assume that the volume of NaOH is
in F40 and the concentration in F44.
1.
Place a 250 mL beaker on the balance and tare it.
2.
Add about 0.2 g of one of the unknown acids, and record the unknown
number and the mass on the spreadsheet below.
3.
Add 25-50 mL of distilled water
to the acid in the beaker, swirl to dissolve the acid, add a few drops of
phenolphthalein indicator, and titrate to the endpoint with 0.1 M NaOH
solution. If the acid dissolves slowly,
you may start the titration before it is all dissolved. Be sure not to approach the end point until
all acid dissolves, however. Added base
converts the acid to its more soluble, ionic conjugate base. You may do this titration with the pH meter
if you wish, adding 1 mL at a time and proceeding much more rapidly than you
will for the final titration below.
4.
Determine how much acid would require 25 mL of base.
5.
Dry off the outside of a 250 mL beaker, place it on the balance, and
6.
Add the mass of acid calculated in part A, and record the mass.
7. To
the acid in the 250 mL beaker, add 50 mL of distilled water, add a magnetic
spinbar (and universal indicator if you wish), and activate the stirrer to
dissolve the acid.
8.
Calibrate the pH meter with the standard buffer solution, then rinse the
electrode and immerse it in the beaker on the stirrer. Position the burette so that titrant can be
easily added.
9.
Record the pH, then add 1 mL of 0.1xxx M NaOH solution at a time,
recording the pH after each addition, until the pH changes more that 0.2-0.3
units when 1 mL of NaOH is added. At
this point, decrease the volume of NaOH added so that the change in pH is small
enough to yield a good plot. After the
rapid changes near the equivalence point, the volume of NaOH may again be
increased to 1 mL per additon. Make at
least 10 additions after the equivalence point.
10. Plot the pH (Data Series 1), 1st Derivative
(Data Series 2) and 2nd Derivative (Data Series 3) vs. Volume NaOH (X Values).
To
help visually determine the equivalence point, activate the graph, click on the
x-axis values, and in the format box which appears, select the scale tab and
set the maximum and minimum values 1 mL above and below the equivalence
point. If grid lines are added (right
mouse button) every 0.1 mL, the equivalence point can be determined
precisely. If it is still difficult to
read the equivalence point, only the Schwartz Plot can help.
Record
the exact concentration of the NaOH solution, and calculate the MW of the acid
from its mass and amount calculated from titration data.
12.
Read the pH at half equivalence and record it. From this value, calculate the Ka of the acid.