In this lesson, we will learn:
- Why titrations are performed and their purpose as an experiment.
- How to use reaction stoichiometry to find moles and unknown concentrations from titrations.
- How to use titrations and mole calculations to find percentage purity of a sample.
- A titration is a lab experiment used to find the unknown concentration of an acid or base. We can use an acid or base sample with a known concentration (let’s call the ‘known’ chemical A) in a chemical reaction with a sample of a base or acid of unknown concentration that we’re investigating (let’s call the ‘unknown’ chemical B).
Chemicals A and B will react in proportional amounts (moles) to each other, so the amounts of A we used, in the end, tells us the amounts of B that’s present.
- Titration experiments work because volume AND concentration of chemical A are known, so we can calculate moles of A, which is related to moles of B (in their reaction), which is used with its volume to calculate A’s unknown concentration! Finding this works in a sequence as shown in the table below:
- The practical experiment goes like this:
- Known chemical A is gradually added to a precise volume of unknown B until the stoichiometric point is reached. The stoichiometric point is when the ratio of moles of A to B is the same as in the balanced chemical equation. This point is clear from a large change in pH (and color change with indicator added) because neutralization occurs.
The molar ratio in the reaction equation (the stoichiometry) and the volume of A that was needed to neutralize B is then used to find the number of moles of B. Now that moles and volume of ‘unknown’ B are known, the concentration of the unknown chemical B sample can be calculated.
- Titrations are excellent for finding acid/base concentrations because:
- Acids and bases are complementary in their chemistry so there are lots of ideal ‘partner’ chemicals for them.
- It is easy to measure pH and pH change of a mixture, and the pH can be used to find the concentration and moles of H3O+ or OH- in that same mixture. This is perfect for chemicals that produce these ions, like acids and bases!
- The most important factor in a successful titration is precision. The titration needs to stop precisely at the stoichiometric point (AKA equivalence point). To make sure this is accurate and precise, a few steps are taken in the practical method:
- A chemical called an indicator is added to the mixture. This chemical indicates a change in pH by a clear color change, and the equivalence point is exactly when the titration mixture changes color permanently. There are many different indicators for different pH ranges and combinations of weak/strong acids reacting with weak/strong bases. They are very sensitive to pH change – the addition of one extra drop of acid or base might cause a permanent color change.
- A trial run is performed on the very first titration experiment. This is a ‘rough run’ to establish roughly the volume of chemical A needed to reach the equivalence point. This is done to save time and still allow chemists to be precise in the approximate range where the color change – and equivalence point – is found.
- The final amounts of the known chemical A are added dropwise to get as precise a reading as possible. A single titration run with a volume reading obtained is called a ‘titer’.
- The precise runs – titers – are repeated until two readings fall within 0.05cm3 of each other. An average titer is then taken from these two readings.
- Once an average titer has been obtained, this is taken to be the volume where there is an equal number of moles of both acid and base.
Remember that your titer volumes will be in milliliters or cm3 but concentration is measured in moles per LITER, or dm3. You must convert the volume to liters! Dividing milliliters by 1000 will give you a value in liters.
The number of moles of chemical B can be calculated because its concentration and volume are both known:
Rearrange to give:
With this moles of B value obtained, recall the reaction stoichiometry:
aA + bB → cC + dD
The molar ratio is important for finding moles of B. The equivalence point occurs when b moles of A is equal to a moles of B. Use this to find the number of moles of unknown chemical B in the titration mixture at equivalence point.
- Once the number of moles of chemical B is found, you have moles and volume:
Now the unknown concentration can be calculated!
- Titrations involving partial neutralization can also be performed; this is where polyprotic acids (acids with more than one proton to donate) donate a proton one by one, so no one step completely neutralizes the acid. Take the example with phosphoric acid, H3PO4:
Reaction 1: H3PO4 + OH-
→ H2PO4- + H2O
Reaction 2: H3PO4 + 2OH-
→ HPO42- + 2H2O
Reaction 3: H3PO4 + 3OH-
→ PO43- + 3H2O
If you are asked to do a calculation with partial neutralization, you might need to find out the acid:base molar ratio in the reaction. For this, you’ll be given the volume and concentration of both acid and base so that moles of both can be found – then divide the larger number of moles by the smaller to find the ratio between them! To find the moles, the rearranged equation for concentration can still be used:
This is sometimes written as:
n = C * V Where n = moles, C = concentration (M), V = volume (L or dm3)
- Titration results can be used to find the percentage purity of a sample as well. This is done in the following method:
- Find the number of moles of the acid/base you added to the solvent to create the solution. For example, a 1.13g sample of NaOH was used. Find the number of moles of this by dividing the mass used by molar mass:
Find the expected concentration by dividing the volume of the solution made by this number of moles.
For example, the NaOH used was diluted to 250 mL (0.25 L):
Record your expected concentration.
- Perform the titration and find the concentration of your sample substance using your titer results with the method above. This is your actual concentration – if you have impurities in your sample, they won’t react but they still took up mass in your sample, so this actual concentration will be lower than the expected concentration.
For example, a 25.00 mL sample of the NaOH solution was titrated with 27.90 mL (or 0.0279 L) of 0.097 M HCl:
HCl and NaOH react in a 1:1 ratio, so 0.0027 mol HCl = mol NaOH:
- Use these two values to calculate percentage purity of your sample with the equation: