Introduction to pH, Titration Curves, and Buffers
Richard A. Paselk
- This exercise is designed to review a few concepts regarding
pH, acids and bases, and the behavior of weak acids in aqueous
solution. First let's recall some definitions
- Brönsted acid: we will be using the Brönsted definition
for acids and bases - an acid is a proton donor, while a base
is a proton acceptor. Recall the corollary that acids and bases
therefore exist as conjugate acid base pairs. Note that when
an acid by this definition gives up a proton it becomes a base,
since the reverse reaction would be accepting a proton:
- Thus the acetic acid in the first reaction becomes its conjugate
base acetate ion, while the base, hydroxide ion, becomes
its conjugate acid, water. In the reverse reaction the
nomenclature also reverses. Note that a molecule such as water
can be both an acid, donating a proton to become its conjugate
base hydroxide ion, or it can be a base, accepting a proton to
become its conjugate acid, a hydronium ion.
- The strengths (ability to donate protons) of acids vary considerably.
For the general acid HA we can write:
- Where Ka is the acid dissociation constant.
(Note that the definition of Ka is based on
the Brönsted definition.) Values of Ka
can vary tremendously (1015 to 10-60) -
after all anything with at least one proton can be considered
an acid under some circumstances with this definition. The common
definition of a strong acid is an acid which dissociates completely
in a 1 M solution. The common strong acids in aqueous solution,
such as sulfuric, nitric and hydrochloric acids have Ka
values (for the first dissociation in the case of sulfuric) of
102 to 109. Thus they all dissociate completely
(first dissociation only for sulfuric) in aqueous solution, though
they will have different strengths in some other solvents. Most
common organic acids are weak in aqueous solution, having Ka
values of 10-5 to 10-15. Note that whether
an acid is strong or weak is dependent of the solvent system!
Strong acids have weak conjugate bases, and vice-versa.
- pH = -log [H+]. Remember that a lo pH means a
high concentration of protons.
- pKa = -log Ka
- For aqueous solution [H+][OH-]= 10-14;
therefore pH + pOH = 14, where pOH = -log[OH-].
- For reactions involving a strong acid or base we can assume,
for practical purposes, that all of the strong acid or base added
to a mixture will react until the base or acid originally present
in solution is completely consumed. (Of course this is an approximation,
all reactions actually approach an equilibrium condition, so
that, in theory, there is always some reactant and some product
present.) For example, if we start with a solution containing
0.100 mole of acetic acid and add 0.050 moles of sodium hydroxide
the resulting mixture will contain 0.05 moles acetic acid, 0.050
moles sodium acetate and 0.000 moles sodium hydroxide (actually
about 10-10 moles, which is 0.000 for our two significant
figure calculation).
- The equilibrium equation for a mixture of a weak acid and
its conjugate base can be rewritten by taking logs of both sides
and rearranging to give the Henderson-Hasselbalch equation: pH
= pKa + log [A-]/[HA]
Example: Modeling a Titration Reaction Using a Spreadsheet
- With this background let's look at a typical reaction of
a weak acid, acetic acid (pKa= 4.76) with a
strong base in aqueous solution. We are going to model this reaction
using a spreadsheet program, Excel (Office 98 Mac version). Other
spreadsheets will work equally well, but some of the commands
etc. will differ. Look at the launcher control panel and click
(using the left mouse button) on Academic computing, a new set
of icons will appear in the applications window. Click on the
Microsoft Excel button,
, and Excel
will launch a workbook:
-
- We now want to set up five columns, labeling them in cells
A1 - E1 as shown below:
-
- Next we want to enter data in cells A2 - C2: the initial
quantity of acid (we'll use 0.1 mole), the pK of the acetic acid
(4.76) and the initial amount of base added (0.001 mole
we will skip the zero point because of the calculation problems
of having a zero in the fraction in the log term):
-
- Next we'll do a little trick to enter lots of points (we
will want more than a hundred additions of base to create smooth
titrations curves!). Click and hold the mouse button down in
cell A2 (the background will invert to black), then drag the
mouse over to cell B2 (it will invert), and then drag it down
to cell 150 and release the mouse button. The top of you workbook
should look like this:
-
- Now press the command (apple / "flower" key) and
D keys. The columns should fill with the data in the top cells:
-
- Now we want to add some base in small increments, 0.001 mole
each. Click on cell C2 to select it and drag down 10 or more
cells (they will invert). Then go to the menu bar, click on Edit,
go down to Fill, and the to Series:
-
- You should then get a window which looks like this:
-
- Enter the values shown, 0.001 and 0.01 in the step value
and stop value boxes. If you now click OK your workbook
should look like this:
-
- Now we are ready to have the spreadsheet do some calculations
with these data. Click on cell E2 and enter the following expression
=A2-C2 (this can also be done by entering = then clicking on
cell A2 then entering + then clicking on cell C2 and hitting
the enter key). Now click and hold on cell E2 and drag down to
cell E11, then press command (apple) D. Column E should fill
with numbers as shown below. Pretty cool!
-
- Next we want to enter the Henderson-Hasselbalch equation
in cell D2 so that it uses the data in columns C and E: =B2+log(C2/E2)
{ =pK+log(A-/HA)}. Again click drag and fill. Your chart should
now look like this:
-
- We are now ready to plot the data. First select the data
in columns C and D by clicking and dragging starting with cell
C2 and going to cell D11, then go to the menu bar and find the
Insert menu drag down to Chart... and release:
-
- You will see a dialog box like this:
-
- Click on the XY (Scatter) Chart and click Next
to get:
-
- Click Next again and fill the chart title, and label
the axis (pH and moles base added), then click Next and
you will see:
-
- Select As new sheet and click Finish. Your plot should
look like:
-
-
- We now know enough about Excel to do the titration experiment.
Go back to the workbook (look at the bottom of the Excel window
and click on sheet 1). Again using increments of 0.001
mole add base until a total of 0.50 mole is reached (this should
correspond to cell 51). You will need to use the Fill
function to generate the base added column. You can then fill
down for the calculated quantities. If you included all of these
cells on your first plot, even thou empty, it should update,
otherwise insert a new chart on a new page. Your chart should
now look like:
-
- Note that if you need to modify your axis labels, grid lines,
etc. you can go to Chart Options... under the Chart
menu on the menu bar.
-
- Look at the values of [HA] and [A-] on the worksheet
(row 51). They should be identical. What is the pH? How does
it compare to pKa?
-
- According to this model, where we are using the Henderson-Hasselbalch
equation to represent our solution, is the pH affected by the
volume of the solution? (Remember concentrations will be moles/vol,
and both concentrations are affected by the total volume = initial
+ added.) It turns out that the pH values of buffers (mixtures
of acids and their conjugate bases or vice-versa) are essentially
unaffected by concentration change for dilute solutions (<
1 M to < 0.1 mM or so).
-
- Next add sufficient base to completely titrate the acid (complete
titration takes equal numbers of moles of the acid and the base,
in our case 0.100 mole of each). For the plot do not include
the last point, a zero in the equation causes problems for
Excel (and after all, the real system would not have a concentration
of zero). You should now get a plot like this:
-
- Of course for a real titration we would have a pH value for
the equivalence point, the problem is that the Henderson-Hasselbalch
equation is no longer applicable: terms neglected by this equation
because they are negligible when both the acid and conjugate
base are both present in significant amounts must now be considered.
-
- But rather than do the math, let's think about the situation.
We have been assuming that our weak acid reacts completely with
added strong base. Under this assumption at the equivalence point
the concentrations of weak acid and added hydroxyl ion will be
zero! Therefore the pH should be determined strictly by the dissociation
of water: there will be 10-7 M hydrogen ion and 10-7
M hydroxide ion and the pH should be 7.00.
-
- So far so good, and this would be the end of it for the titration
of a strong acid with a strong base. However, for our titration
of a weak acid the solution contains more than just water, it
also contains a weak base, acetate ion, C2H3O2-.
The acetate ion thus reacts with the water to make the solution
a bit more basic, in this case the pH = 8.72, assuming the initial
concentrations of acid and base were equal. (If you wish to explore
the determination of the pH at the equivalence point further,
it can be found in most General Chemistry texts, e.g. in Russell's
General Chemistry (2nd ed.) pp 583-4, Ex. 15-18, and in
Zumdahl's Chemistry (4th ed.) pp 731-2. E.)
-
- So now we can go back to our spreadsheet and just enter the
value of 8.72 for the pH at equivalence (cell E104, where moles
base added = 0.1).
-
- As additional base is added the concentration of base will
simply be the moles of base divided by the total volume. Thus
we can calculate the pH from pH = 14.00 - pOH = 14.00 -(-log[OH-])
= 14.00 + log(base/(acid + base))
-
- Returning to the spreadsheet (sheet 1) enter the value for
the equivalence pH (8.72) and the new pH equation =14+LOG((C102-0.1)/(A102+C102))
in the next pH cells. Then add another 0.01 moles of base by
0.001 mole increments. The new cells in the spreadsheet should
look like this:
- Finally, add another 0.1 mole of base by 0.01 mole increments
to get:
- Now make a new plot (or extend your old one) to get the complete
titration curve, but this time select the option with a smooth
curve fit:
-
- The plot should then look like this:
- (I made the fonts for the axis and title larger by double
clicking on them in the chart, a dialog box will then come up
allowing font changes.).
Practice Assignment
- For this exercise you should have each of
the following:
- A printout of the sheet for the final plot
(you will find the Print commands under the File
menu on the menu bar). Highlight and label the cells in the
rows on the sheet which include the following:
- The pKa.
- The equivalence point.
- The pH at +1 and -1 unit from the pKa.
- A printout of the final plot with labeled
axis and title. Find and show the following on your plot. Use
graphical methods (ruler and pencil or pen):
- The pKa. (Remember H-H
eqn.: pH = pKa + log[A-]/[HA], and
log 1 = 0. So when is pH = pKa? A line drawn
parallel to the x-axis intersecting the titration curve at this
point will also intersect the y-axis where pH = pKa.)
- The equivalence point. (Draw a line parallel
to the y-axis intersecting the axis where the amount of base
added = the initial amount of acid present. It will intersect
the the titration curve at the equivalence point.)
- The buffer region. (Draw two line segments
parallel to the y-axis such that they intersect both the titration
curve and the x-axis, and which include the approximately linear
region of the titration curve on either side of the pKa.)
- Answers to the following questions:
- Do the values for the pKa
you found in the two methods agree? Which value is better for
this mathematical model? Which would be more useful experimentally?
Why?
- Do the values for the equivalence point you
found in the two methods agree? Which value is better for this
mathematical model? Which would be more useful experimentally?
Why?
- Does the buffer region you found graphically
correspond to the range found on the sheet (pKa±1)?
- What is a buffer?
- Why does the pH level off at high pH values?
Is this another titration and buffer region? (Hint: What's the
pH of 0.1 molar base? 1.0 molar?)
© R A Paselk
Last modified 23 November 2004