General Chemistry: Lec XV

Chem 1200

Angel C. de Dios

Reactions VIII

Zoom

Here is the case where we add the conjugate base:

Here is the case where we add a strong base. Think of this as providing the conjugate base in situ.

The above equation is called the Henderson-Hasselbalch equation. It is not really a new equation - it is the equilibrium expression for Ka recast in a logarithmic fashion.

Here is another example.

Here is how the buffer relates to the weak acid - strong base titration curve. At the half equivalence point, we converted half of the acetic acid into acetate. Thus, at this point, the amount of acetic acid equals the amount of acetate in solution. Since [A-]=[HA], pH = pKa.

One should notice that although we are not in the region of either low pH (strong acid) or high pH (strong base), the titration curve near pKa is nearly flat, illustrating that the solution is able to resist changes in pH even upon adding a strong base or strong acid. This is how a buffer works.

The pH values for a weak acid-strong base titration can be calculated easily if we are inside the buffer region, that is pKa -1 < pH < pKa + 1. As long as the pH is within one unit away from pKa, we can use the Henderson-Hasselbach equation, because the concentration of either A- or HA is much greater than H3O+. ([H3O+] is the x that we are dropping to simplify the equilibrium expression).

In the following, titration curves are shown for different weak acids. These are initially 0.100 M weak acid solutions. That is why at the start of the titration, the [H3O+] is approximately the square root of Ka times c (with the exception of the acid with pKa = 2, in this case, the approximation is not accurate) Thus, the pH is approximately (pKa divided by 2) + 0.50, The 0.50 comes from -log of square root of c (where c = 0.100). If these solutions were initially 1.000 M, then the pH at the starting point will be simply pKa/2. In all weak acid cases (except the one with pKa = 2), pH = pKa at the half-equivalence point, and this does not depend on the initial concentration of the weak acid.

So, we can calculate the pH at the starting point (since this is just an aqueous solution containing a weak acid). We can also calculate the pH near the half equivalence point or the buffer region, using the Henderson-Hasselbalch equation.

At the equivalence point, we can calculate the pH as well because all we have is an aqueous solution of the conjugate base of the weak acid (This is a weak base problem, [OH-] = square root of Kb times c. And past the equivalence point, we now have excess base, so we only need to take account excess OH-.

Let us work with more buffer problems:

The above works only with the original equilibrium expression noting that [A-] is 0.040 M and [HA] is 0.080 M (the initial concentrations of these species). It does not bother to use the Henderson-Hasselbalch equation.

Here is another example. This time, the buffer is made of a weak base and its weak conjugate acid. It uses the Ka expression.

The next example illustrates how a buffer works.

There is, of course, a limit in the pH range at which a buffer can work. A buffer is only effective if both members of the conjugate pair are substantially present, and from the Henderson-Hasselbalch equation, this clearly requires that the pH is not too far from pKa.

Remember:

pH = pKa + log ([base]/[acid])

Both members of the conjugate pair are necessary for a buffer to work. The weak acid will counteract any strong base that is added and the weak base will counteract any strong acid that is added. Both [base] and [acid] are substantial only when the pH is near pKa.

Given the pH we would like to maintain, we can determine the ratio of the buffer ingredients, as illustrated in the following examples:

How strong the buffer is depends on the amounts of the conjugate weak pair. This is another reason why a buffer is only good for a pH that is close to its pKa. If the desired pH, for example, is 1 unit away from the pKa, this means, one member of the pair will be ten times less than the other. The buffer in this case, will then be weak towards one end of the pH scale.

As an illustration, if the pH desired is 3.74 and we are using an acetic acid - acetate buffer (pKa = 4.74) then we need the amount of acetic acid to be ten times the amount of acetate. This buffer is already weak on the acetate side such that additional strong acid might be able to convert all of the acetate into acetic acid, rendering us with an aqueous solution of just acetic acid. When this happens, we no longer have a buffer.

Lactic acid, like other carboxylic acids, is weak, thus, buffers need not be optimum in order to maintain the pH.

For the remainder of this topic, we will cover the titration of a polyprotic acid.

For a diprotic acid, we expect twice the amount of base for complete neutralization. There are two equivalence points. With a diprotic acid H2A, at the first equivalence point, we convert H2A completely into HA-, (At the first equivalence point, pH = (pKa1 + pKa2)/2, and at the second equivalence point, we now have just A2-. There are two half-equivalence points as well. The first one is at pKa1 and the second one is at pKa2 (These are labeled Midpoint 1 and 2 in the following graph).

Here is a specific example, the titration of alanine:

The first half-equivalence point is at pKa1, the second half-equivalence point is at pKa2. The first equivalence point is at (pKa1 + pKa2)/2. In biochemistry, this is called the isolectric point or pI, when the zwitterion is the dominant species. A zwitterion is overall neutral.

Here are the steps in calculation the pH at various points in the titration of a weak diprotic acid. (The only exception to this is the strong diprotic acid H2SO4, which we saw earlier.)

For alanine, Ka1 is actually quite large so the quadratic formula is necessary to obtain ac accurate result.

After we start adding a strong base, we now have a buffer. And at the first half-equivalence point, we have pH = pKa1.

At the first equivalence point of the titration of a diprotic acid, we now have just HA-, which is amphiprotic. We have seen this before, the pH at this point is midway between pKa1 and pKa2.

Past the first equivalence, we go to another buffer region, but the buffer here is (HA-/A^2-).

At the second equivalence point, all we have in solution is A^2- so this is just an aqueous solution of a weak base. [OH-] = square root of Kb times c. Kb is Kw/Ka2 (because A^2- is the conjugate base of HA-).