3.31 AS Inorganic Chemistry - Titration
Titration is a procedure of careful addition of one solution to another solution a little at a time until a specificend point is reached. Titrations can be carried out between many different pairs of reagents, although the most common titration is performed between an acid and a base.
Acid-base titrations
The reaction is carried out in the presence of an indicator, which shows the end-point of the titration as a colour change. The actual colour change depends on the type of indicator used.
The concentrations of acid and base used for titrations is important, as small additions must only change the pH level by small amounts for accuracy. Acceptable concentrations for the reagents are below 0.2 mol dm-3.
Theory
Mathematical calculation of the pH change as base is added to acid, shows that there is a large change in pH near the neutralisation point (end-point) of the titration. A graph of pH against volume of base added shows a curve with a rapid change near the end-point. This is called an inflexion.
Experimental
25 ml of dilute base are pipetted into a conical flask. A few drops of indicator solution are added. Dilute acid is added from burette a few drops at a time with swirling until the end-point is reached (colour change).
The titration is repeated until concordant results are obtained.
Typical results - titration of dilute hydrochloric acid and sodium hydroxide
- Molarity of sodium hydroxide = 0.098 mol dm-3
- Molarity of hydrochloric acid = approximately 0.05 - 0.2 mol dm-3
| expt | initial /ml ± 0.05 | final /ml ± 0.05 | titre /ml ± 0.1 |
| 1 | 0.00 | 18.60 | 18.60 |
| 2 | 0.00 | 18.45 | 18.45 |
| 3 | 0.00 | 18.45 | 18.45 |
| average of concordant results | 18.45 | ||
Calculation
Moles of sodium hydroxide in 25.0 ml = 0.025 x 0.098 = 2.45 x 10-3
Equation for the reaction:
NaOH + HCl 
NaCl + H2O
Hence, moles of acid = moles of base at the end-point.
Moles of acid = 2.45 x 10-3
Volume of acid from titration = 0.01845 dm3
Hence, concentration of acid = moles/volume = 2.45 x 10-3/ 0.01845 = 0.133 mol dm-3
Evaluation
The procedure of titration is very accurate when the apparatus is used correctly, but even so each step has a corresponding inaccuracy due to the manufactured tolerance of the glassware.
For example:
Grade B 25 ml pipette has an accuracy of ± 0.04 ml
This corresponds to a percentage inaccuracy of 100 x 0.04/25 = ± 0.16 %
The burette readings have an inaccuracy of one half of the minimum reading = ± 0.05 ml
Two readings per titre gives an inaccuracy of ± 0.1, corresponding to a percentage inaccuracy of 100 x 0.1/18.45 = ± 0.54% (this titration)
All of the percentage inaccuracies may be added in a procedure and converted back to an absolute inaccuracy of the final 'answer'.
Total percentage = 0.16 + 0.54 = 0.70 %
Hence, the calculated molarity of hydrochloric acid = 0.133 ± 0.001 mol dm-3
Note that only the two steps have been taken into account in this example. In a real experiment you would also have to take into account inaccuracy in any solutions prepared, masses weighed out etc.
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