AS Chemistry Unit 1 Module 1 (Atoms and Reactions) from titrations

3) a formula

4) the no. of molecules of water of crystallisation

You carry out a titration as follows:-

~~Using a pipette, you add a measured volume of 1 solution to a conical flask.

~~The other solution is placed in a burette.

~~The solution in the burette is added to the solution in the conical flask until the reaction has JUST been completed. This is calle the end point of the titration. The volume of the solution added from the burette is measured.

You now know the volume of 1 solution that EXACTLY reacts with the volume of the other solution.

--------We identify the end point using an INDICATOR - the indicator must be a different colour in the acidic solution than in the basic solution.

Colours of some common acid-base indicators + the colour at the end point (Notice that this end point colour is in between the colours in the acidic and basic solutions.

 Indicator                            Colour in acid           Colour in base         End point colour    

methyl orange                            red                        yellow                    orange

bromothymol blue                    yellow                        blue                      green

phenolphthalein                     colourless                    pink                      pale pink

This assumes that the aqueous base has been added from the burette to the aqueous acid. If acid is added to base, the titration is complete when the solution goes colourless.

Calculating Unknowns from Titration Results

Analysis of titration results follow a set pattern, as shown in the worked examlpes below.

~The first 2 steps are always the same

~The 3rd step may be different, depending on the unknown that you need to work out.

[In AS Chemistry, any calculations that you carry out will be structured similarly to the examples below - for A2, you may have to work out these steps yourself].

E.g. 1) Calculating an unknown concentration

In a titration, 25.0cm^3 of 0.150 mol dm-3 sodium hydroxide NaOH(aq) reacted exactly with 23.40cm^3 of sulfuric acid, H2SO4(aq)

2NaOH(aq) + H2SO4(aq) --------> Na2SO4(aq) + 2H2O(l)

(a) Calculate the amount, in mol, of NaOH that reacted.

n(NaOH) = cV/1000 = (0.150x25)/1000 = 3.75x10-3 mol = 0.00375 mol

(b) Calculate the amount, in mol, of H2SO4 that was used.

equation:                                 2NaOH(aq) + H2SO4(aq)

moles from equation:          …