Learning Goal 24
Describe the form of the titration curves for titration of a strong acid by a strong base, a weak acid by a strong base, or a weak base by a strong acid.

A titration is a procedure for carrying out a chemical reaction between two solutions by the controlled addition from a buret of one solution (the titrant) to the other, allowing measurements to be made throughout the reaction. For a reaction between an acid and a base, a titration is useful for measuring the pH at various points throughout the reaction.

A titration curve is a graph of the pH as a function of the amount of titrant (acid or base) added.


Strong Acid-Strong Base Titrations

Here is an example of a titration curve, produced when a strong base is added to a strong acid. This curve shows how pH varies as 0.100 M NaOH is added to 50.0 mL of 0.100 M HCl.

The equivalence point of the titration is the point at which exactly enough titrant has been added to react with all of the substance being titrated with no titrant left over. In other words, at the equivalence point, the number of moles of titrant added so far corresponds exactly to the number of moles of substance being titrated according to the reaction stoichiometry. (In an acid-base titration, there is a 1:1 acid:base stoichiometry, so the equivalence point is the point where the moles of titrant added equals the moles of substance initially in the solution being titrated.)

Notice that the pH increases slowly at first, then rapidly as it nears the equivalence point. Why?

Learning Goal 25
Calculate the pH at any point, including the equivalence point, in an acid-base titration.

At the equivalence point, the pH = 7.00 for strong acid-strong base titrations. However, in other types of titrations, this is not the case.

EXAMPLE:

What is the pH when 49.00 mL of 0.100 M NaOH solution have been added to 50.00 mL of 0.100 M HCl solution?

Because it is a strong acid-base reaction, the reaction simplifies to:

H+ (aq) + OH- (aq) H2O (l)

The original number of moles of H+ in the solution is:

50.00 x 10-3 L x 0.100 M HCl = 5.00 x 10-3 moles

The number of moles of OH- added is :

49.00 x 10-3 L x 0.100 M OH- = 4.90 x 10-3 moles

Thus there remains:

(5.00 x 10-3) - (4.90 x 10-3) = 1.0 x 10-4 moles H+ (aq)

The total volume of solution is 0.04900 L + 0.05000 L = 0.09900 L

[H+] = {1.0 x 10-4 moles / 0.09900 L } = 1.0 x 10-3 M

pH = 3.00


Titrations Involving a Weak Acid or Weak Base

Titration curve of a weak acid being titrated by a strong base:

Here, 0.100 M NaOH is being added to 50.0 mL of 0.100 M acetic acid.

There are three major differences between this curve (in blue) and the one we saw before (in black):

1. The weak-acid solution has a higher initial pH.

2. The pH rises more rapidly at the start, but less rapidly near the equivalence point.

3. The pH at the equivalence point does not equal 7.00.

POINT OF EMPHASIS : The equivalence point for a weak acid-strong base titration has a pH > 7.00.

For a strong acid-weak base or weak acid-strong base titration, the pH will change rapidly at the very beginning and then have a gradual slope until near the equivalence point. The gradual slope results from a buffer solution being produced by the addition of the strong acid or base, which resists rapid change in pH until the added acid or base exceeds the buffer's capacity and the rapid pH change occurs near the equivalence point.

EXAMPLE:

What is the pH when 30.0 mL of 0.100 M NaOH have been added to 50.0 mL of 0.100 M acetic acid?

STEP 1: Stochiometric calculation:

The original number of moles of HC2H3O2 in the solution is :

50.0 x 10-3 L x 0.100 M = 5.00 x 10-3 moles HC2H3O2

Similarly, there are 3.00 x 10-3 moles of OH- due to the NaOH solution.

The reaction goes to completion:

OH- (aq) + HC2H3O2 (aq) C2H3O2- (aq) + H2O (l)

OH- HC2H3O2 C2H3O2-
INITIAL
3.00 x 10-3 mol
5.00 x 10-3 mol
0
CHANGE
-3.00 x 10-3 mol
-3.00 x 10-3 mol
+3.00 x 10-3 mol
FINAL
0
2.00 x 10-3 mol
3.00 x 10-3 mol

The total volume is 80.0 mL.

We now calculate the resulting molarities :

[HC2H3O2] = { 2.00 x 10-3 mol HC2H3O2 / 0.0800 L } = 0.0250 M

[C2H3O2-] = { 3.00 x 10-3 mol C2H3O2- } / 0.0800 L } = 0.0375 M

STEP 2: Equilibrium calculation, using simplification:

Ka = { [H+][C2H3O2-] / [HC2H3O2] } = 1.8 x 10-5

[H+] = { KA [HC2H3O2] / [C2H3O2-] } = { (1.8 x 10-5)(0.0250) / (0.0375) } = 1.2 x 10-5 M

pH = -log(1.2 x 10-5) = 4.92

Titration curve of a weak base being titrated by a strong acid:

Here, 0.100 M HCl is being added to 50.0 mL of 0.100 M ammonia solution.

As in the weak acid-strong base titration, there are three major differences between this curve (in blue) and a strong base-strong acid one (in black): (Note that the strong base-strong acid titration curve is identical to the strong acid-strong base titration, but flipped vertically.)

1. The weak-acid solution has a lower initial pH.

2. The pH drops more rapidly at the start, but less rapidly near the equivalence point.

3. The pH at the equivalence point does not equal 7.00.

POINT OF EMPHASIS : The equivalence point for a weak base-strong acid titration has a pH < 7.00.


Titrations of Polyprotic Acids

An example of a polyprotic acid is H2CO3 which neutralizes in two steps:

H2CO3 (aq) + OH- (aq) H2O (l) + HCO3- (aq)

HCO3- (aq) + OH- (aq) H2O (l) + CO32- (aq)

The titration curve for these reactions will look like this, with two equivalence points.



Uses of Titrations

Learning Goal 26
Use titration data or a titration curve to calculate reaction quantities such as the concentration of the substance being titrated.

The most common use of titrations is for measuring unknown concentrations. This is done by titrating a known volume of the unknown solution with a solution of known concentration (where the two react in a predictable manner) and finding the volume of titrant needed to reach the equivalence point using some method appropriate to the particular reaction. Then, the volume and concentration of titrant can be used to calculate the moles of titrant added, which, when used with the reaction stoichiometry, gives the number of moles of substance being titrated. Finally, this quantity, along with the volume of substance being titrated, gives the unknown concentration.

For acid-base titrations, the equivalence point can be found very easily. A pH meter is simply placed in the solution being titrated and the pH is measured after various volumes of titrant have been added to produce a titration curve. The equivalence point can then be read off the curve.

EXAMPLE:

If 80.0 mL of 0.200 M NaOH are required to reach the equivalence point in a titration of 50.0 mL of hydrofluoric acid, what is the concentration of the hydrofluoric acid?

The neutralization reaction goes to completion:

HF + OH- F- + H2O

The number of moles of NaOH added was:

nNaOH = [NaOH] x VNaOH
nNaOH = 0.200 M x 80.0 mL
nNaOH = 16 mmol

Since each NaOH produces 1 OH-, nOH = nNaOH = 16 mmol

From the 1:1 stoichiometry between HF and OH-, nHF = nOH = 16 mmol

So, the concentration of the original hydrofluoric acid solution was:

[HF] = nHF / VHF
[HF] = 16 mmol / 50 mL
[HF] = 0.320 M

In the same way, knowing the equivalence point can also be used to calculate other unknown quantities of interest in acid base reactions, such as concentration of titrant or volume of solution being titrated, provided that enough other information is known to perform the calculations.