# 17.3 pH Calculations Involving Titrations

##### Video Transcript

## pH Calculations for Strong Acid / Strong Base Titrations

Students are asked to perform pH calculations for 3 fundamental types of titrations:

Strong Acid / Strong Base

Weak Acid / Strong Base

Weak Base / Strong Acid

In this section we will deal with the first type: Strong Acid / Strong Base titrations. We will specifically look at titrating a strong acid with a strong base, though the calculations would be very similar for titrating a strong base with a strong acid. The calculations involved are generally easier than those involved in titrations involving a weak acid or base and can be broken up into the following 3 points in the titration:

1. Before the Equivalence Point

2. At the Equivalence Point

3. After the Equivalence Point

For all the titrations we will deal with you should note that either the acid, the base, or both are strong. Because of this the resulting neutralization reactions will go to completion, and because they go to completion we will treat them as limiting reagent problems. For all our titration calculations we will follow the same general procedure:

1. Calculate the number of moles of acid and base.

2. Have the neutralization reaction go to completion.

3. Based upon what is left after the neutralization reaction, calculate the pH.

There are 5 types of solutions you are required to be able to calculate the pH of (strong acid, strong base, weak acid, weak base, and buffer). These were presented in chapter 16 and earlier in this chapter in the case of buffers and are briefly summarized in the following table:

After neutralization is complete in any titration the pH of the resulting solution will be calculated using one of these 5 methods.

**For a strong acid / strong base titration specifically consider the titration of 150mL of 0.10M HCl with 0.125M NaOH for the following examples. **

### 1. Before the Equivalence Point

We should first note that the pH for the initial point before the start of the titration is straightforward to calculate as all that is present is 0.1M HCl, a strong acid.

pH = -log [H^{+}] = -log (0.1)

pH = 1

__Addition of 100mL 0.125M NaOH__

(0.10M)(0.150L) = 0.015moles HCl

(0.125M)(0.100L) = 0.0125moles NaOH

Now that the number of moles of acid and base have been determined we will have the neutralization reaction between HCl and NaOH go to completion.

HCl and NaOH react in a 1:1 ratio, and with fewer moles of NaOH it is the limiting reagent. After the neutralization reaction has run to completion we can see that there will be 0.0025 moles of excess HCl remaining, and that it is the only thing left in the solution that will affect the pH. The products of the neutralization are water and NaCl, a neutral salt, and won't affect the pH of the solution. To find the pH of the excess strong acid, we simply take the negative log of its concentration. At this point we know we have 0.0025 moles of HCl in the solution, but we have to divide this by the total liters of solution to convert it back to molarity. Having added 100mL NaOH to 150mL HCl the total volume is 250mL or 0.250L.

### 2. At the Equivalence Point

__Addition of 120mL 0.125M NaOH__

(0.10M)(0.150L) = 0.015moles HCl

(0.125M)(0.120L) = 0.015moles NaOH

Now that the number of moles of acid and base have been determined we will have the neutralization reaction between HCl and NaOH go to completion. Note that the number of moles are equal and that this is the equivalence point in the titration.

You can see that at the equivalence point there is no HCl or NaOH remaining in the solution. As stated above the products, water and NaCl, a neutral salt, will not affect the pH either, and therefore the pH at the equivalence point in a strong acid / strong base titration equal 7 as also explained in the previous lesson (17.2 Titrations and Titration Curves).

### 3. After the Equivalence Point

__Addition of 140mL 0.125M NaOH__

(0.10M)(0.150L) = 0.015moles HCl

(0.125M)(0.140L) = 0.0175moles NaOH

Now that the number of moles of acid and base have been determined we will have the neutralization reaction between HCl and NaOH go to completion.

We are past the equivalence point here and see that HCl is now the limiting reagent. After the neutralization reaction has run to completion we can see that there will be 0.0025 moles of excess NaOH remaining, and that it is the only thing left in the solution that will affect the pH. Again, the products of the neutralization (water and NaCl) won't affect the pH of the solution. To find the pOH of the excess strong base, we simply take the negative log of its concentration. At this point we know we have 0.0025 moles of NaOH in the solution, but we have to divide this by the total liters of solution to convert it back to molarity. Having added 140mL NaOH to 150mL HCl the total volume is 290mL or 0.290L.

For the titration of a strong base with a strong acid the titration curve would be inverted but the pH calculations would be very similar to the titration of a strong acid with a strong base.

## pH Calculations for Weak Acid / Strong Base Titrations

The pH calculations for the titration of weak acid with a strong base are generally more difficult than those for a strong acid with a strong base. One reason is that the salt formed in the neutralization reaction can't be ignored as it will be a basic salt rather than a neutral salt and will affect the pH of the solution. There are 4 areas on the curve requiring different methods of calculating the pH.

1. Initial Point

2. Before the Equivalence Point (excluding initial point)

3. At the Equivalence Point

4. After the Equivalence Point

Here, the base is strong. Because of this the resulting neutralization reaction will go to completion, and because it goes to completion we will also treat this as a limiting reagent problem. We will follow the same general procedure outlined above:

1. Calculate the number of moles of acid and base.

2. Have the neutralization reaction go to completion.

3. Based upon what is left after the neutralization reaction, calculate the pH.

**For these examples specifically consider the titration of 100mL of 0.10M HF with 0.10M NaOH. For these calculations I've provided you with the Ka for HF; Ka = 6.8x10 ^{-4}. **

### 1. Initial Point

Before the addition of any NaOH we just have a solution of a weak acid. In Lesson 16.4 pH Calculations for Weak Acids and Bases, we learned a shortcut for calculating the pH of a weak acid solution:

It turns out that this corresponds to a percent ionization of ~8%. Being more than 5% this probably should have been solved using the quadratic equation rather than the shortcut. But even had we used the quadratic the calculated pH still would have been 2.1 (2.10 vs 2.06) so the error here is minimal.

### 2. Before the Equivalence Point (excluding initial point)

__Addition of 25mL 0.10M NaOH__

(0.10M)(0.10L) = 0.01moles HF

(0.10M)(0.025L) = 0.0025moles NaOH

Now that the number of moles of acid and base have been determined we will have the neutralization reaction between HF and NaOH go to completion.

HF and NaOH react in a 1:1 ratio, and with fewer moles of NaOH it is the limiting reagent. After the neutralization reaction has run to completion we can see that there will be 0.0075 moles of excess HF remaining and 0.0025 moles of NaF formed. We now have a weak acid and its conjugate base present in solution and have a buffer solution. Technically, we could use the Ka or Kb expressions to solve for the pH, but it's much more common to use the Henderson-Hasselbalch equation instead. (Even if outside of the buffer region which requires within a 1:10 or 10:1 ratio of weak acid / conjugate base we could still use the Henderson-Hasselbalch equation to calculate the pH before the equivalence point.)

I show two forms of the Henderson-Hasselbalch equation. The ratio of the molarities of A^{-} to HA is equal to the ratio of the moles (n) of A^{-} to HA and either can be used. While the equation is normally presented as the first, it is much more convenient here to use the second so that we don't have to convert the moles of HF and NaF back into molarities.

#### 2. Half-Equivalence Point (still Before the Equivalence Point)

One special point that falls in the range of 'before the equivalence point is the half-equivalence point. At this point [A-] = [HA] and pH = pKa. If you can recognize you're at this point early on you can save a lot of time, but if you work it out you'll still arrive at the same answer, pH = pKa.

__Addition of 50mL 0.10M NaOH__

(0.10M)(0.10L) = 0.01moles HF

(0.10M)(0.025L) = 0.005moles NaOH

Now that the number of moles of acid and base have been determined we will have the neutralization reaction between HF and NaOH go to completion.

NaOH is still the limiting reagent. After the neutralization reaction has run to completion we can see that there will be 0.005 moles of excess HF remaining and 0.005 moles of NaF formed. We still have a buffer solution and will still use the Henderson-Hasselbalch equation to calculate the pH. With the moles of F^{-} and HF being equal, the ratio of moles F^{-} to moles HF equals 1 and log (1) = 0 and pH = pKa.

### 3. At the Equivalence Point

__Addition of 100mL 0.10M NaOH__

(0.10M)(0.10L) = 0.01moles HF

(0.10M)(0.10L) = 0.01moles NaOH

Now that the number of moles of acid and base have been determined we will have the neutralization reaction between HF and NaOH go to completion. Note that the number of moles are equal and that this is the equivalence point in the titration.

At this point the HF and NaOH have completely neutralized each other and the only thing remaining that will affect the pH is NaF, a weak base. In Lesson 16.4 pH Calculations for Weak Acids and Bases, we learned a shortcut for calculating the pH of a weak base solution:

But to use this equation we'll first have to calculate the Kb of F- from the Ka of HF. Since Ka x Kb = Kw, then Kb = Ka / Kw. This equation also requires us to plug in the molarity of the weak base. At this point we only know that we have 0.01 moles of weak base but we'll have to divide that by the total liters of solution to convert it into a molarity. As 100mL HF and 100mL NaOH have been combined, the total volume is 200mL, or 0.2L.

### 4. After the Equivalence Point

__Addition of 120mL 0.10M NaOH__

(0.10M)(0.10L) = 0.01moles HF

(0.10M)(0.12L) = 0.012moles NaOH

Now that the number of moles of acid and base have been determined we will have the neutralization reaction between HF and NaOH go to completion.

After the equivalence point we have excess NaOH, a strong base, remaining in the solution. We have also formed NaF, a weak base. As a strong base dissociates completely, whereas a weak base dissociates to a small extent (usually <5%). Therefore, we generally ignore the weak base and calculate the pOH as if it were a solution of a strong base alone. In this example we have 0.002 moles of NaOH now in a total volume of 220mL, or 0.22L.

## pH Calculations for Weak Base / Strong Acid Titrations

The pH calculations for the titration of weak base with a strong acid are very similar to those of a weak acid with a strong base. While not worked out explicitly here, the following curve shows the four areas involving a different way to calculate the pH and how to do so.

1. Initial Point

2. Before the Equivalence Point (excluding initial point)

3. At the Equivalence Point

4. After the Equivalence Point