Introduction to pH, Titration Curves, and Buffers

Richard A. Paselk

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 K_a is the acid dissociation constant. (Note that the definition of K_a is based on the Brönsted definition.) Values of K_a 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 K_a 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 K_a 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.
• pK_a = -log K_a
• 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 = pK_a + 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 (pK_a= 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 pK_a?

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:

1. 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 rowson the sheet which include the following:
   1. The pK_a.
   2. The equivalence point.
   3. The pH at +1 and -1 unit from the pK_a.
2. 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):
   1. The pK_a. (Remember H-H eqn.: pH = pK_a + log[A^-]/[HA], and log 1 = 0. So when is pH = pK_a? A line drawn parallel to the x-axis intersecting the titration curve at this point will also intersect the y-axis where pH = pK_a.)
   2. 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.)
   3. 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 pK_a.)
3. Answers to the following questions are available on the "Answers..." subpage:
   1. Do the values for the pK_a you found in the two methods agree? Which value is better for this mathematical model? Which would be more useful experimentally? Why?
   2. 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?
   3. Does the buffer region you found graphically correspond to the range found on the sheet (pK_a±1)?
   4. What is a buffer?
   5. 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?)