- Created by: chunks-42
- Created on: 05-05-15 17:51
How to perform a titration
1. Titrations allow you to ind out exactly how much acid is neededto neutralise a quantity of alkali.
2. You measure out some alkali using a pipette and put it in a flask, along with some indicator, e.g. phenolphthalein.
3. First of all, do a rough titration to get an idea where the end point is (the point where the alkali is exactly neutralised and the indicator changes colour). Add the acid to the alkali using a burette - giving the flask a regular swirl.
4. Now do a accurate titration. Run the acid in to within 2cm3 of the end point, then add the acid dropwise. If you don't notice exactly when the solution changed colour, you've overshot and your result won't be accurate.
5. Record the amount of acid used to neutralise the alkali. It's best to repeat this process a few times, making sure you get the same answer each time. This'll make your results are reliable.
Calculating Concentrations from titrations
E.g. 25cm3 of 0.5 M HCl was used to neutralise 35cm3 of NaOH solution. Calculate the concentration of the sodium hydroxide solution in mol dm-3.
First write a balanced equation and decide what you know and what you need to know:
HCl + NaOH --> NaCl + H2O (HCl= 25cm3, 0.5 M), (NaOH= 35cm3)
Now work out how many moles of HCl you have:
Number of moles HCl = conc x vol. / 1000 = 0.5 x 25/1000 = 0.0125 moles
From the equation, you know 1 moles of HCl neutralises 1 mole of NaOH. So 0.0125 moles of HCl must neutralise 0.0125 moles of NaOH.
Now it's a doddle to work out the concentration of NaOH.
Concentration of NaOH = moles of NaOH X 1000/volume (cm3) = 0.0125 x 1000/35 = 0.36mol dm3
Calculating Volumes from Titrations
Thisis usually used for planning experiments. You need to use the formula from previous again, but this tme rearrange it to find the volume.
E.g. 20.4cm3 of a 0.5 M solution of sodium carbonate with 1.5 M nitric acid. Calculate the volume of nitric acid required to neutralise the sodium carbonate.
Like before, first write a balanced equation for the reaction and decide what you know and what you want to know.
Na2CO3 + 2HNO3 --> 2NaNO3 + H2O + CO2
0.5 M 1.5 M
Now work out how many moles of N2CO3 you've got:
No. of moles of Na2CO3 = conc. x volume (cm3)/1000 = 0.5x20.4/1000 = 0.0102 moles
Calculating Volumes from Titrations continued
1 mole of Na2CO3 neutralises 2 moles of HNO3, so 0.0102 of Na2CO3 neutralises 0.0204 moles of HNO3.
Now you know the number of moles of HNO3 and the concentration, you can work out the volume:
Volume of HNO3 = number of moles x 1000/conc. = 0.0204 x 1000/1.5 = 13.6 cm3