# Acid/Base Titration: An Example

As a student, when I was taught titration, it was accompanied by a long laundry list of scenarios: titration of a strong acid with a strong base, titration of a strong base with a strong acid, titration of a weak acid with a strong base, and titration of a weak base with a strong acid. Each of these scenarios were followed by a graph that corresponded to changes in pH, like the following:

Initially, these terms and images tended to confuse me. How was I supposed to know which graphs to use in what scenario? And how were these curves supposed to help me when I was being asked to come up with the pH and concentration of a solution at very specific instances? As it turns out, knowing the structure and meaning of the components of a titration curve can be essential in solving acid/base titration problems, no matter what type of solution you are titrating.

For example, say you had a problem that went something like this:

There may be multiple parts involved, but take a deep breath. You can do this, a little bit at a time. First, let’s ignore the calculation section of this question and just see if we can identify which type of titration we are performing, and what curve we will use to model it. Initially, we have a weak base, and we are adding \(\mathrm{HNO}_{3}\), a strong acid. Therefore, we are starting with a pH a little over 7, since we have a basic solution as our analyte, but not so close to 12, since it is weak. As we add acid, the concentration of H+ in our solution increases, and pH, being equivalent to -log [H+] will go down. Looking at the curves above, we can see that the curve that most closely corresponds to this scenario is:

Even though we have no specific values recorded up here, the curve is valuable in that we can use it to identify what is happening as a we add a certain amount of titrant. We will be referring back to this image at different points as we work through this problem.

### Part A

To solve this part, we need to gain some perspective on where we are on the titration curve. We are adding a very small amount of our titrant (10.0 mL) to a very large amount of our base (200 mL = 0.200 L). Looking at our titration curve above, this is modeled on the far left of the graph by a sharp dip, followed by a relatively stable pH. This situation is analogous to a buffer solution, since we have a small pH change in a solution containing both acidic and basic components. Therefore, we can use the Henderson-Hasselbach equation, which is used to find the pH of buffers, for this portion of the titration curve:

$$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log (\text { moles of acid moles of base })$$

Remember that the acid and base being referred to here is weak base and its conjugate acid, NOT the strong acid that was added. This is because as it goes into solution, it dissociates and neutralizes as much of the base as it can before running out, to get a concentration of 0. And log of 0 will break your calculator, which is not a good thing.

```
Mol weak base initially: (0.200L)(0.10 M) = 0.02 mol weak base
Mol HNO₃ added: 10 mL = 0.01L
(0.01 L)(0.20 M) = 0.002 mol HNO₃
Mol HNO₃ reacted: 0.002 mol HNO₃
Mol weak base reacted: 0.002 mol weak base
Mol weak base remaining 0.02-0.002 = 0.018 mol weak base
Mol conjugate acid produced: 0.002 mol conjugate acid
```

Now that we have determined how much weak acid and base is in solution, all that’s left to do is to plug in these numbers, along with the pKa that we can acquire from the given information in the question to get our final answer.

$$\mathrm{pK}_{\mathrm{a}}=-\log \left(2.32 \times 10^{-11}\right)=10.63$$

$$\mathrm{pH}=10.63+\log (0.018 / 0.002)=11.6$$

### Part B

If you looked at the textbook definition of the midpoint, you might see that it would say something along the lines of “where the pH is equal to the pKa.” In which case we could easily say:

$$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}=10.6$$

Easiest question ever, right! But your next question might be, (or might not be, depending upon how interested you are) *why* is this the case? The answer has to do with the fact that the midpoint is where you halfway through neutralizing a solution. Again, if we were to look at the titration curve, that would be roughly….here, where the curve levels out:

Again, in this region the pH is relatively stable, and we have acidic and basic components in the solution, which means that it is again acting much like a buffer solution. However, since we are halfway to neutralization, that means that half of the weak base has been converted to its conjugate acid, and half of it is still weak base. In other words, they are present in a 1:1 ratio. If we were to plug this in to our Henderson-Hasselbach equation, we would get:

$$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log (1)=\mathrm{pK}_{\mathrm{a}}+0=\mathrm{pK}_{\mathrm{a}}$$

It’s worth noting that midpoints only occur if you have either a weak acid or base present in your titration. Reactions between strong acids and bases change the pH too quickly to have those present in their curves. When in doubt, draw out or imagine the rough sketch of the curve to help you envision what is happening around that point.

### Part C

Here, we are being asked the pH of the equivalence point (also known as the stoichiometric point), or the point where all of the titrant and the analyte have reacted, and neither is present in excess. This point corresponds to the steepest portion of the curve, where pH changes rapidly. In this case, buffer equations will not help us, since the solution is not acting in this manner at this point. Instead, we need to use another definition of pH, one that we are perhaps more familiar with:

pH = -log [H+]

However, in order to calculate how much [H+] we have in solution at this point, we first need to calculate what volume of solution we have, and therefore, how much titrant we added to reach this point. Initially, we had 0.02 mol weak base, so we need 0.02 mol \(\mathrm{HNO}_{3}\) to react in a 1:1 ratio without excess.

0.02 mol \(\mathrm{HNO}_{3}\)/0.20 M \(\mathrm{HNO}_{3}\) = 0.1 L \(\mathrm{HNO}_{3}\) added

0.1 L \(\mathrm{HNO}_{3}\) + 0.2 L weak base solution = 0.3 L solution at equivalence point

0.02 mol conjugate acid (BH+) produced

At equivalence point [BH+] = 0.02 mol/0.3 L of solution = 0.66 M BH+

Why do we need to know the concentration of conjugate acid? As you might recall, ions are constantly dissociating in solution, and our conjugate acid is no better. At the equivalence point, it is in equilibrium, and it is this dissociation that is the source of our H+. Because this is an equilibrium problem, we use an ICE table that we set up as follows:

```
[BH+ ] ⇆ [B] + [H+]
I 0.66 0 0
C -x +x +x
E 0.66-x x x
```

From here, we can solve for x by using the equation that defines the relationship between Kₐ and [H+], and then plug our subsequent result into our original pH equation. Had we been titrating a weak acid, we would have solved for [OH-] using Kb, and would have had the additional step of calculating pH from pOH.

$$K_a = \frac{[\mathrm{H+}][\mathrm{B}]} {[\mathrm{BH+}]} = \frac{x^2} {0.66 - x}$$

$$\mathrm{x}=1.24 \times 10^{-6}$$

$$\mathrm{pH}=-\log \left(1.24 \times 10^{-6}\right)=5.9$$

### Part D

The only tricky part of this problem compared to the other parts, which had us find out how much weak base had reacted with the strong acid, is that in this case we have more moles of strong acid in solution than our initial base, as you can see calculated below. In other words, in this far right portion of the titration curve, any changes in pH is due to the contribution of the strong acid, since it is the primary source of H+ in the weak base-depleted solution. The rest of the problem works the same way as part c, with the only exception being the lack of an ICE table, since the solution is not at equilibrium.

120.0 mL HNO₃ = 0.120 L HNO₃

(0.120 L HNO₃)(0.20 M HNO₃ ) = 0.024 mol HNO₃

Initially, we had 0.02 mol weak base, B.

0.024-0.02 = 0.004 mol unreacted HNO₃ = 0.004 mol H+ (assuming 100% dissociation)

(0.200 L weak base solution + 0.120 L HNO₃ added) = 0.320 L total solution

[H+] = 0.004 mol H+/0.320 L solution = 0.0125 M

pH = -log [H+] = -log 0.0125 = **1.9**

### What Have We Gathered

So you made it! After all this long and hard work, you have finally managed to figure out how to solve this one titration problem, but not just that: you have learned how to navigate a titration curve, and to use the knowledge of its basic structure as a template for figuring out how to find specific values. When we are in areas of the curve where the pH is increasing slowly, and we have small amounts of both acidic and basic components, we can use buffer-related equations since the solution is acting as a buffer solution. Likewise, when we are aware that we are in an area of the curve where pH is rapidly changing or we do not have either an acidic or basic component, we know when to abandon that particular tool of acid-base chemistry in favor of other ones. In the end, though there are different forms of titration scenarios, and titration curves, and it can get confusing trying to determine when to know which one, by slowly working your way through the problem, and using your existing knowledge of acid-base chemistry, you can manage to tackle acid/base titration.