Titration Curves
This video looks at Titration curves
A useful way to check that you can calculate the pH of a solution with confidence and accuracy is to construct a graph of the pH for different combinations of acid and alkali.
Strong acid / strong base
As an example, consider running 0.1 mol dm-3 sodium hydroxide solution into 25 ml of 0.1 mol dm-3 hydrochloric acid, 1 ml at a time. Calculate the pH at each stage.
It can be a good plan to construct a spreadsheet to plot the graph or else draw it by hand. If you can work with a friend to do alternate volumes, any irregularity in the graph will show that one of you has made an error. You can then check it out.
If you are good at spreadsheet calculations, you may like to try to work out the formula to convert each volume to its proper pH. This is a difficult task and although it will be satisfying to complete, don’t let it get in the way of the chemistry work!
Pure dilute acid solution:
0.1 mol dm-3 HCl has a [H+] = 0.1 pH = 1
Some alkali but not enough to neutralise the acid: eg 5 ml of alkali added.
5 ml of alkali contains 5 x 0.1 / 1000 moles of OH- ion = 0.0005
25 ml of acid contains 25 x 0.1 / 1000 moles of H+ ion = 0.0025
The total volume is 25 + 5 = 30 ml
The number of moles of H+ left unreacted is 0.002 [H+] = 0.002 x 1000/ 30 = 0.0667
pH = 1.176
Continue this process up to 24.99 ml of NaOH.
Continue at 25.01 ml of NaOH.
This time we have NaOH in excess and so we can find the number of moles of OH- and then the [OH-]. Since Kw = 1 x 10-14 = [H+] x [OH-] at 25 oC, we can calculate [H+] and then go on to find pH.
Keep going up to 50 ml of NaOH.
You might like to think about what is happening at exactly 25 ml of acid and 25 ml of alkali.
Try to calculate the pH for exact equivalence.
Try to explain why the calculation doesn’t work!
Doing the same thing for weak acid / strong base is a little more difficult.
Pure dilute acid solution:
pH = ½ pKa – ½ log10[HA]
Some alkali but not enough to neutralise the acid:
The [H+] must be calculated using the formula: Ka = [H+][A-] / [HA]
The moles of HA is the amount of acid that has not yet been neutralised by the alkali and can be found from
moles of acid at start - moles of alkali added
The moles of A- present is given by the moles of alkali added.
Total volume is calculated as before.
Go on to find the concentrations of [A-], [HA] and then use the formula to find [H+].
This system can continue up to 24.99 ml of NaOH.
Continue at 25.01 ml of NaOH. From here on, we don't need to worry about the weak or strong character of the acid since it is already neutralised.
A more detailed version of this is on the next page.
This video looks at strong acid titration and equivalence point
This video looks at the Equivalence point when titrating a weak acid
Category
