My notes are actually with my teacher at the moment, so I'm working solely off memory - forgive me if my explanation is not as clear as it could be!
For question two:
Firstly set out what we know already. We're reacting HCl with NH3, and it makes NH4Cl. All of these are in the aqueous state. We have 20 cm3 of ammonia solution, which is 0.02 dm3. That reacts with a certain volume of hydrochloric acid solution, which has a concentration of 0.02 mol dm-3. The hydrochloric acid which is titrated into the ammonia has a volume of 37.35 cm3, which is 0.03735 dm3 (which we'll round to 0.037 dm3). We can set this out like this to help us:
HCl (aq) + NH3 (aq) -> NH4Cl (aq) HCl NH3Vol. 0.037 0.01Conc. 0.02 The unknown value which we now need to calculate is the concentration of NH3, and we can do that by working out the number of moles. The number of moles is equal to the concentration multiplied by the volume:
moles = 0.02 x 0.037 = 0.00074Because there is only one molecule of each reactant in the equation, we can simply directly transfer this across, so there are also 0.00074 moles of NH3. Going back to our table:
HCl (aq) + NH3 (aq) -> NH4Cl (aq) HCl NH3Vol. 0.037 0.01Conc. 0.02 Moles 0.00074 0.00074We can now use that same formula: moles = concentration x volume, and re-arrange it so that we find out concentration. Concentration therefore equals moles divided by volume, so:
conc. = 0.00074 / 0.01 = 0.074 mol dm-3This is our final answer. For the sake of completeness, we can fill it into our table:
HCl (aq) + NH3 (aq) -> NH4Cl (aq) HCl NH3Vol. 0.037 0.01Conc. 0.02 0.074Moles 0.00074 0.00074For question three, have you read the process detailed in
this document from AQA? It's pretty much exactly what you need.
For question four, you're basically doing the same thing as question two, but with different numbers and different molecules; hopefully my method detailed above will allow you to work this out for yourself!
EDIT: Unfortunately TSR's software removes excessive spaces, so my tables don't stick to their format. Sorry!