# How to Practice Titration Calculations

By John Brennan
Stockbyte/Stockbyte/Getty Images

Introductory chemistry classes and labs have you practice lots of titration-related calculations because these types of problems allow you to familiarize yourself with aqueous acid-base equilibria and develop skills you'll use throughout your chemistry career. It's a good idea to practice titration calculations because they're almost certain to show up on tests. You don't need to actually do a titration to practice -- you can just work problems on paper.

## Types of Problems

Add a titrant, whose precise concentration is known, to an analyte until all of the analyte is neutralized. In acid-base titrations (the most common kind), the titrant is always either a strong base or a strong acid. The point where the analyte is neutralized is called the stoichiometric point. If you track the pH of the analyte throughout the titration, you should get a nice S-shaped or sigmoid-shaped curve. The point on the graph with the steepest slope marks the stoichiometric point.

Try the following practice problems:

1) Use 50 mL of 4 M sodium hydroxide to reach the stoichiometric point when titrating a 250 mL sample of hydrochloric acid. What was the original concentration of the hydrochloric acid sample?

2) Use a 4 M sodium hydroxide solution to titrate 250 mL of analyte containing 0.5 M acetic acid. The Ka of acetic acid is 1.74 x 10^-5. What is the pH of the analyte a) before the titration starts, b) after adding 5 mL of NaOH and c) 1/2 of the way to the stoichiometric point?

The sections below will give you the skills you need to solve these problems, and the answers to both appear in the Tips section so you can double-check your work.

Divide titration problems into two broad classes: problems where the concentration of the analyte is unknown and you need to figure it out based on the volume of titrant added, and problems where you need to calculate the pH at various points during the titration.

## Analyte Concentration

Start with the first kind of problem -- the one where you need to figure out the concentration of analyte.

Multiply the volume of the titrant added by its concentration to get the number of moles of titrant. If you added 75 mL of the titrant and the concentration of the titrant was 0.5 M, for example, you would multiply 75 mL by 0.5 moles/liter after converting 75 mL to liters, as follows:

(75/1000) x 0.5 = 0.0375 moles.

Multiply this result by the number of molecules of titrant needed to neutralize one molecule of analyte. You now have the number of moles of analyte. To continue the example, if you need two molecules of titrant to neutralize one molecule of analyte, and you added 0.0375 moles of titrant, you would multiply 0.0375 x 2 = 0.075 moles.

Divide the number of moles of analyte by the original volume in liters of the analyte. If its original volume is 500 mL, for example, then 0.0375 / (500/1000) = 0.075 M or moles/liter for the analyte, and this would be the solution to your problem.

## pH During Titration

Remember that the pH at the beginning of the titration is just determined by the concentration of the analyte. These kinds of problems typically include the concentration of analyte and the Ka or Kb of the chemical it contains, so you know how much was originally present and you can calculate the pH using the Ka of the acid (or Kb of the base).

Example: Suppose that at the beginning of your titration, you have a 0.25 molar solution of acetic acid. The Ka for acetic acid is 1.74 x 10^-5. so you can calculate the pH using the equilibrium expression you learned earlier on in general chemistry. If you need a refresher on equilibrium constants, click on the third link under the Resources section for more info. Remember that acetic acid is a weak acid, so we can approximate the equation as follows:

1.74 x 10^-5 = x^2 / 0.25

so multiply both sides by 0.25 then take the square root to get the following:

x = (4.34 x 10^-6)^1/2

x = 2.08 x 10^-3

then take the negative log of this result to get pH, which yields the following:

pH = -log (2.08 x 10^-3) = 2.68

Recall that during the next part of the titration, the chemical in the analyte is neutralized to become its conjugate base (or conjugate acid), so the concentration of conjugate base/acid increases, while the concentration of the original species decreases. You can figure out the pH at any point during this phase with the following equation for an acid:

pH = log (B/A) + pKa

and the following equation for a base:

pOH = log (A/B) + pKb.

Note that the pKb is the negative log of the Kb, while the pKa is the negative log of the Ka. B represents the concentration of a conjugate base in these expressions, while A represents the concentration of a conjugate acid.

Example: If 10 mL of 5 M sodium hydroxide is added to 150 mL of 1 M acetic acid, what is the new pH?

Answer: 150 mL of acetic acid = (150 mL) / (1000 mL per L) = 0.15 liters.

Multiply this by the concentration to give 0.15 L x 1 M = 0.15 moles.

10 mL of NaOH = 10 mL / 1000 mL per L = 0.010 L.

Multiply this by the concentration to give 0.010 L x 5 M = 0.050 moles.

0.05 moles of NaOH will neutralize 0.05 moles of acetic acid, so the new number of moles of acetic acid is 0.1, while there are now 0.05 moles of conjugate base. Since you added titrant, however, the volume of the solution has now changed. Its total volume is equal to the volume of titrant added plus the volume of analyte originally present -- in other words, there are now 160 mL in all. You must divide the new amounts of acetic acid and conjugate base by the new volume to obtain the concentration of each, as follows:

0.05 moles / 0.16 L = 0.3125 M conjugate base.

0.1 moles / 0.16 L = 0.625 M conjugate acid.

Plug these numbers into the equation from above, as follows:

pH = pKa + log (0.3125 / 0.625)

pH = pKa + log 0.5

pH = pKa - 0.301

The pKa of acetic acid is 4.76, so the new pH would be 4.76 - 0.301.

Remember that the pH or pOH halfway to the stoichiometric point will be exactly equal to the pKa (for an acid) or the pKb (for a base). You can deduce this from the equations for pOH and pH during the second phase of the titration, because the log of anything over itself is 0.

Example: If a solution of acetic acid is titrated halfway to the stoichiometric point, what will the pH be? The Ka of acetic acid is 1.74 x 10^-5.

Answer: Since the Ka for acetic acid is 1.74 x 10^-5, the negative log of this number is 4.76, so the pH halfway to the stoichiometric point will be 4.76.