Volumetric Analysis (2) Acid Base Titration Curves

In my previous blog I discussed the method of titration used in volumetric analysis.

I discussed one type of acid base titration: a weak acid titrated with a strong base.

To say a weak acid is titrated with a strong base means that the strong base is in the burette and the weak acid solution is in the conical flask. 

Why are there only three types of acid base titration?

The reason has to do with the way pH changes in the flask as the titration proceeds.

It is possible to monitor the pH of the solution in the flask using a pH probe linked to a data logger.

When this experiment is carried out the results look like this:

These are what are called titration curves: 

1. Titration curve for a strong acid titrated with a strong base

Here is the titration curve for the titration of 50ml of 0.100M HCl solution (strong acid) with a 0.100M solution of NaOH (a strongly alkaline solution):

We can see that there is barely any change in pH until close to the end point at 50ml NaOH solution added. 

Then there is sudden increase in pH which is why it is a very sharp end point: addition of a very small volume of the alkaline solution causes a rapid and large increase in pH of the order of 8 pH units. 

At the end point the pH is 7.

The pH change affects both methyl red and phenolphthalein dyes, methyl red is red below pH 5 and yellow above pH 5 so as the pH changes rapidly so does the colour of the indicator therefore it is very important at the end point we add very small drops of the alkali or we “overshoot” it.

Either phenolphthalein or methyl red can be used to detect the end point of a strong acid strong base titration.

So what happens if instead we have a weak acid in the flask instead of a strong acid?

2. Titration curve for a weak acid titrated with a strong base.

Here is the titration curve (green line) for the titration of 50ml of 0.100M CH3COOH acetic acid solution (weak acid) with a 0.100M solution of NaOH (a strongly alkaline solution):

As the alkali is added to the weak acetic acid the pH rises and levels off at about pH 5 this region is called the buffer region because the pH hardly changes as the alkali is added.

In this region there is no sharp change in pH so the methyl red indicator cannot work. 

But above pH 7 there is again a sharp change in pH as the buffer breaks down. 

This region covers about 4 pH units and phenolphthalein changes colour between pH 8 and 10 so it can be used to find the end point of a weak acid strong base titration. 

Finally what happens to the pH in a weak base strong acid titration?

3. Titration curve for a weak base titrated with a strong acid.

Here is the titration curve for the titration of 50ml of 0.100M solution of ammonia NH3 (a weakly alkaline solution) with a 0.100M HCl hydrochloric acid solution (strong acid). 

This time we start with the alkali in the titration flask so the pH is high at around 10 (the blue line).

As the strong acid is added the pH does not fall dramatically but levels off in an alkaline buffer solution almost to the end point.

As the end point approaches the buffer solution breaks down and the pH falls rapidly by about 4 pH units.

Here phenolphthalein cannot be used as an indicator but methyl red changes colour at these pHs so it can be used to detect the end point. 

Titration calculations and how to be successful at them.

Here is a set results for the titration of a strong base with a strong acid: 25ml of approx. 0.1M sodium hydroxide solution titrated with 0.1M hydrochloric acid solution using methyl red indicator.

Pipette solution	Sodium hydroxide			approx. 0.1 mol/dm3		25ml
Burette solution	Hydrochloric acid			0.1mol/dm3
Indicator	Methyl red
Burette rdgs	Rangefinder	1	2	3	(4)
Final rdg (ml)	25.20	25.10	25.00	24.90
First rdg (ml)	0.00	0.00	0.20	0.00
Volume used (ml)	25.20	25.10	24.80	24.90
Mean titre (ml)	24.85

Calculate the concentration of the sodium hydroxide solution.

Here is a first but not the only method.

1. Calculate the number of moles of hydrochloric acid used in the titration.

Use the equation n=cV

n = number of moles of the compound,

c = the concentration of the solution (in mol/dm³) and

V= the volume of the solution used in the titration in Litres (dm3)

Therefore:  n = 0.100 * 24.85/1000  =  0.002485 moles. 

2. Calculate the number of moles of sodium hydroxide that react with 0.002485 moles hydrochloric acid

We go to the equation for the reaction:

NaOH    +    HCl    =   NaCl   +   H2O

And this shows 1 mole of the acid reacts with 1 mole of the alkali.

Therefore the number of moles of sodium hydroxide in the 25ml of its solution were 0.002485 mol.

3. Next we use the answers to 1 and 2 to determine the concentration of the sodium hydroxide like this:

Use the equation:  c  =  n/V

Therefore the concentration c of the sodium hydroxide in the flask is: 

0.002485moles

0.025 dm3

or  0.0994 mol/dm³

Notes: 

a)  To calculate the mean titre we ignored the 25.10 value because it was too far away from the other two values.  We took the two closest values.

b)  We add zeros to the measured values of the readings since a burette can be read to 4 significant figures.  That’s why we also quote the mean titre to 4 significant figures.

c)   The final answer cannot now be quoted to 4 significant figures rather three sig figs are appropriate since we have used values to four sig figs and only 2 sig figs (25ml) in the calculation.

d)  We can give the final answer then to 3 sig figs i.e. 0.099M at best.