Buffer Solutions and Titrations
The interplay of acids and bases and their conjugates is at the heart of the topics covered in this section. Think of them as a love square. These topics are critical for understanding the properties of a whole mess of biological and environmental solutions. We'll start with building a qualitative understanding of what a buffer is and then get into some titrations.
To understand how a buffer works consider the following scenario.
First, we have a generic buffer solution consisting of a weak acid. The generic equilibrium is:
Now let's suppose you added some strong base to this solution, say NaOH. First the NaOH would essentially fully dissolve leaving a bunch of free OH- around:
NaOH → Na+ + OH-
This would result in an increase in [OH-] and the pH would shoot up. But wait. There's more.
The weak acid solution already has free H+ around that will react with the free OH- to form H2O. Remember that the equilibrium constant for the dissociation of water is really small so the association of H+ and OH- will be strongly favored. The initial increase in OH- would be immediately followed by a decrease in OH- as the free H+ from the weak acid eats up the OH- to form H2O.
The decrease in free H+ that results from absorbing the OH- would ordinarily result in a pH increase as the concentration of H+ is lowered. However, the weak acid equilibrium responds to the lowering H+ levels by shifting to the product side.
This is Le Chatlier's principle, which tells us when an equilibrium is stressed by a change in the concentration of just one of the equilibrium species, the equilibrium will shift to alleviate the stress.
In this case, as the H+ levels decrease the equilibrium shifts to the product side to compensate for the reduction in [H+]. This results in the production of more H+ to fill the void left by the previous reaction of the H+ with OH- to make water. The net result is that the free [H+] does not actually change much and therefore, the pH remains relatively unchanged. This is how a buffer solution of a weak acid is able to absorb OH- molecules and buffer against the addition of a strong base.
The buffer capacity, or amout a weak acid or base buffer solution can absorb added H+ or OH-, is determined by the concentration of the weak acid. You can imagine that if we had a very dilute buffer solution of a weak acid or base and we added bucket loads of strong acid or base, the pH would eventually change significantly. At this point, the buffer loses its buffer quality. Bummer.
The Henderson-Hasselbalch equation is a handy relationship to use when dealing with buffers and their capacities. This equation tells us that the pH of a buffer solution has a value close to the pKa (-logKa) of the weak acid (or base). We can derive this equation starting with our favorite equilibrium:
Recall that the equilibrium constant for this generic equation for a weak acid is:
Rearranging and taking the log of both sides gives:
Two of these terms we have seen before. -log(Ka) is the acid's pKa and –log([H+]) is the pH of the solution.
Behold, the Henderson-Hasselbalch equation:
This equation tells us that when [A-] = [HA], the pH of the buffer solution will equal the pKa of the weak acid making up the buffer. This equation is also useful in determining how extensively dissociated a weak acid solution is based on the pH of the solution. If the pH of the solution is the pKa value of the weak acid, we know the weak acid is halfway (50%) dissociated. (This is also known as the half-equivalence point, which we'll get to later.)
If the pH of the solution is 2 pH units above the pKa of the weak acid, we know that the weak acid is 99% dissociated.
must equal 2 and 
must therefore equal 100.
The addition of a strong base to a strong acid ends in a neutralization reaction. The neutralization resulting from the addition of strong base to strong acid is also a titration.
As strong base (NaOH for example) is added stepwise to a strong acid solution (HCl for example), the pH of the acid solution gets higher and higher. This happens while an equivalent amount of base is added to the point where it is able to soak up all the H+ from the acid. Eventually, the neutralized solution contains nothing but water and the very weak conjugate acid and base, Na+ and Cl-, respectively.
The amount of strong base needed to get to neutral pH 7 corresponds to the amount of acid that was in the solution initially. If we had a 1 L solution of 1 M strong acid it would take 1 mole of base to neutralize the solution.
If we were to plot the pH of the solution (on a logarithmic scale) versus the volume of base added to the acid solution, we would have a titration curve as shown in the figure below.
In the figure above, the equivalence point for the titration of a strong acid with a strong base is always pH 7. The curve takes a characteristic steep slope near this equivalence point due to the semi-logarithmic nature of the plot where the most sudden changes in pH occur right before, during, and after we hit the equivalence point.
Now we'll look at what happens when a weak acid solution, which is a buffer solution, gets titrated with a strong base. This time the situation is a bit more complicated so we'll take it one step at a time.
Step 1) Let's start with the weak acid buffer solution before the addition of any strong base.
The pH of the solution will depend on the concentration of weak acid and the acid's pKa. For example, a 0.25 M solution of acetic acid will have a pH of 2.7 just as we calculated in the previous section on The Equilibria of Acid/Base Solutions.
Step 2) Now suppose a little strong NaOH base was added to this solution.
Here's what happens: The NaOH dissociates entirely, resulting in the release of OH-. The OH- is a strong base so it will suck up any H+ that is around. In fact, since just a small amount of OH- has been added in comparison to the amount of weak acid in solution, the OH- will drive the dissociation of the weak acid until all the OH- is absorbed into water. This is the process we discussed earlier in the section on how a buffer is able to absorb, or buffer, the addition of OH-.
The net reaction during this part of the titration looks like this:
CH3COOH + OH- → H2O + CH3COO-
Step 3) At some point the concentration of OH- from the NaOH being added will increase to a point where it has reacted with half of the initial concentration of weak acid.
This means that half of the initial acetic acid will be in the acid form (CH3COOH) and the other half will be in the conjugate base form (CHOO2-).
The Henderson-Hasselbalch equation tells us that at this point, where the [CH3COO-]/[CH3OOH] ratio equals l [log(1) = 0], the pH of the solution will equal the pKa of the weak acid, in this case 4.7.
Sort of Step 4) There is another cool little tidbit about this halfway point that you'll want to remember. At this point, when the pH equals the pKa, the buffering strength or capacity of the solution is at its best.
This means that the buffer is able to best absorb either H+ or OH- that is added to the solution because the concentrations of the acid and conjugate base are at their highest. Any OH- that is added will be absorbed by the acid and any OH- that is added will be absorbed by the conjugate base.
If we want to make a solution that will be resistant to the addition of acid or base at pH 8 for example, we would ideally use a buffer that has a pKa of 8.
Step 5) As more and more NaOH is added the pH continues to only rise a little because there is still plenty of CH3COOH around to supply the free H+ to absorb the OH-.
Step 6) Eventually, enough OH- will be added that almost all of the CH3COOH runs out.
At this equivalence point, [OH-] = [CH3COOH] and [Na+] = [CH3COO-]. Because Na+ is not an acid or base and CH3COO- is a weak base this solution will be basic. To solve for the pH we need to use the Kb of CH3COO- and its concentration. This is equal to the initial concentration of CH3COOH before it was titrated.
Note: For any weak acid-strong base titration, the pH at the equivalence point will be greater than 7 because the conjugate of the weak acid is a base.
Step 7) Once even more OH- is added, the pH of the solution continues to rise.
Soon, the principle contributor of OH- to solution is no longer the conjugate base of the weak acid as it was near the equivalence point. Instead, the dissociation of NaOH is the principle reaction contributing to the concentration of OH- because the contribution to the OH- concentration from the conjugate base is relatively tiny.
The concentration of OH- added beyond the equivalence point can be used to calculate the pH using [OH-]beyond the equivalence point × [H3O+] = Kw to solve for the [H3O+] concentration.
The titration curve shown in the figure below illustrates the titration of the weak acid buffer solution we just stepped through. Notice that around the half-equivalence region (the weak acid's pKa value) the pH doesn't change much and only slightly increases as a greater volume of base is added.
As more base is added, the pH increases steeply to the equivalence point where the amount of OH- added equals the amount of initial weak acid. After this inflection, the change flattens out again as the pH reflects the excess OH- added.
Of course, a weak base solution can also be titrated with a strong acid. The process is the opposite of the weak acid titration we just discussed, but all the principles are the same.
Become a Buffer Buff
A buffer solution does the following: it absorbs free H+ (or OH-) from acids (or bases) that are added to it. This provides a buffer against changes in the pH of the solution. You can think of a buffer solution as a burly tough guy that won't budge when someone tries to push him around. He's sort of like a bouncer at a party. A buffer solution won't budge much from its pH even when we add in an acid or base. How did these buffer solutions get so buff? It's all about the equilibrium between the acids and conjugate bases in the solution.To understand how a buffer works consider the following scenario.
First, we have a generic buffer solution consisting of a weak acid. The generic equilibrium is:
Now let's suppose you added some strong base to this solution, say NaOH. First the NaOH would essentially fully dissolve leaving a bunch of free OH- around:
NaOH → Na+ + OH-
This would result in an increase in [OH-] and the pH would shoot up. But wait. There's more.
The weak acid solution already has free H+ around that will react with the free OH- to form H2O. Remember that the equilibrium constant for the dissociation of water is really small so the association of H+ and OH- will be strongly favored. The initial increase in OH- would be immediately followed by a decrease in OH- as the free H+ from the weak acid eats up the OH- to form H2O.
The decrease in free H+ that results from absorbing the OH- would ordinarily result in a pH increase as the concentration of H+ is lowered. However, the weak acid equilibrium responds to the lowering H+ levels by shifting to the product side.
This is Le Chatlier's principle, which tells us when an equilibrium is stressed by a change in the concentration of just one of the equilibrium species, the equilibrium will shift to alleviate the stress.
In this case, as the H+ levels decrease the equilibrium shifts to the product side to compensate for the reduction in [H+]. This results in the production of more H+ to fill the void left by the previous reaction of the H+ with OH- to make water. The net result is that the free [H+] does not actually change much and therefore, the pH remains relatively unchanged. This is how a buffer solution of a weak acid is able to absorb OH- molecules and buffer against the addition of a strong base.
The buffer capacity, or amout a weak acid or base buffer solution can absorb added H+ or OH-, is determined by the concentration of the weak acid. You can imagine that if we had a very dilute buffer solution of a weak acid or base and we added bucket loads of strong acid or base, the pH would eventually change significantly. At this point, the buffer loses its buffer quality. Bummer.
The Henderson-Hasselbalch equation is a handy relationship to use when dealing with buffers and their capacities. This equation tells us that the pH of a buffer solution has a value close to the pKa (-logKa) of the weak acid (or base). We can derive this equation starting with our favorite equilibrium:
Recall that the equilibrium constant for this generic equation for a weak acid is:
Rearranging and taking the log of both sides gives:
Two of these terms we have seen before. -log(Ka) is the acid's pKa and –log([H+]) is the pH of the solution.
Behold, the Henderson-Hasselbalch equation:
This equation tells us that when [A-] = [HA], the pH of the buffer solution will equal the pKa of the weak acid making up the buffer. This equation is also useful in determining how extensively dissociated a weak acid solution is based on the pH of the solution. If the pH of the solution is the pKa value of the weak acid, we know the weak acid is halfway (50%) dissociated. (This is also known as the half-equivalence point, which we'll get to later.)
If the pH of the solution is 2 pH units above the pKa of the weak acid, we know that the weak acid is 99% dissociated.
must equal 2 and must therefore equal 100. Baby Step to Titrations
A titration consists of adding incremental amounts of base to an acid (or vice versa) in order to determine properties of the acid or base making up the solution like concentration and pKa.The addition of a strong base to a strong acid ends in a neutralization reaction. The neutralization resulting from the addition of strong base to strong acid is also a titration.
As strong base (NaOH for example) is added stepwise to a strong acid solution (HCl for example), the pH of the acid solution gets higher and higher. This happens while an equivalent amount of base is added to the point where it is able to soak up all the H+ from the acid. Eventually, the neutralized solution contains nothing but water and the very weak conjugate acid and base, Na+ and Cl-, respectively.
The amount of strong base needed to get to neutral pH 7 corresponds to the amount of acid that was in the solution initially. If we had a 1 L solution of 1 M strong acid it would take 1 mole of base to neutralize the solution.
If we were to plot the pH of the solution (on a logarithmic scale) versus the volume of base added to the acid solution, we would have a titration curve as shown in the figure below.
In the figure above, the equivalence point for the titration of a strong acid with a strong base is always pH 7. The curve takes a characteristic steep slope near this equivalence point due to the semi-logarithmic nature of the plot where the most sudden changes in pH occur right before, during, and after we hit the equivalence point.
Now we'll look at what happens when a weak acid solution, which is a buffer solution, gets titrated with a strong base. This time the situation is a bit more complicated so we'll take it one step at a time.
Step 1) Let's start with the weak acid buffer solution before the addition of any strong base.
The pH of the solution will depend on the concentration of weak acid and the acid's pKa. For example, a 0.25 M solution of acetic acid will have a pH of 2.7 just as we calculated in the previous section on The Equilibria of Acid/Base Solutions.
Step 2) Now suppose a little strong NaOH base was added to this solution.
Here's what happens: The NaOH dissociates entirely, resulting in the release of OH-. The OH- is a strong base so it will suck up any H+ that is around. In fact, since just a small amount of OH- has been added in comparison to the amount of weak acid in solution, the OH- will drive the dissociation of the weak acid until all the OH- is absorbed into water. This is the process we discussed earlier in the section on how a buffer is able to absorb, or buffer, the addition of OH-.
The net reaction during this part of the titration looks like this:
CH3COOH + OH- → H2O + CH3COO-
Step 3) At some point the concentration of OH- from the NaOH being added will increase to a point where it has reacted with half of the initial concentration of weak acid.
This means that half of the initial acetic acid will be in the acid form (CH3COOH) and the other half will be in the conjugate base form (CHOO2-).
The Henderson-Hasselbalch equation tells us that at this point, where the [CH3COO-]/[CH3OOH] ratio equals l [log(1) = 0], the pH of the solution will equal the pKa of the weak acid, in this case 4.7.
Sort of Step 4) There is another cool little tidbit about this halfway point that you'll want to remember. At this point, when the pH equals the pKa, the buffering strength or capacity of the solution is at its best.
This means that the buffer is able to best absorb either H+ or OH- that is added to the solution because the concentrations of the acid and conjugate base are at their highest. Any OH- that is added will be absorbed by the acid and any OH- that is added will be absorbed by the conjugate base.
If we want to make a solution that will be resistant to the addition of acid or base at pH 8 for example, we would ideally use a buffer that has a pKa of 8.
Step 5) As more and more NaOH is added the pH continues to only rise a little because there is still plenty of CH3COOH around to supply the free H+ to absorb the OH-.
Step 6) Eventually, enough OH- will be added that almost all of the CH3COOH runs out.
At this equivalence point, [OH-] = [CH3COOH] and [Na+] = [CH3COO-]. Because Na+ is not an acid or base and CH3COO- is a weak base this solution will be basic. To solve for the pH we need to use the Kb of CH3COO- and its concentration. This is equal to the initial concentration of CH3COOH before it was titrated.
Note: For any weak acid-strong base titration, the pH at the equivalence point will be greater than 7 because the conjugate of the weak acid is a base.
Step 7) Once even more OH- is added, the pH of the solution continues to rise.
Soon, the principle contributor of OH- to solution is no longer the conjugate base of the weak acid as it was near the equivalence point. Instead, the dissociation of NaOH is the principle reaction contributing to the concentration of OH- because the contribution to the OH- concentration from the conjugate base is relatively tiny.
The concentration of OH- added beyond the equivalence point can be used to calculate the pH using [OH-]beyond the equivalence point × [H3O+] = Kw to solve for the [H3O+] concentration.
The titration curve shown in the figure below illustrates the titration of the weak acid buffer solution we just stepped through. Notice that around the half-equivalence region (the weak acid's pKa value) the pH doesn't change much and only slightly increases as a greater volume of base is added.
As more base is added, the pH increases steeply to the equivalence point where the amount of OH- added equals the amount of initial weak acid. After this inflection, the change flattens out again as the pH reflects the excess OH- added.
Of course, a weak base solution can also be titrated with a strong acid. The process is the opposite of the weak acid titration we just discussed, but all the principles are the same.