Chemistry A Level Revision

I’m now at uni studying biomedicine, and i’ve made a study(ish) blog much like this one for my notes and other interesting science stuff. Give it a follow if you think it’ll interest you! I can help with all things biomedical science, and am also happy to give advice relating to studying, university and A levels :)

yeah I think they’re great, i’ve just finished sixth form and i used note pads throughout the two years. I had two separate pads for each subject one would be for notes/revision notes and I would have a second one solely for doing practice questions in

Practice question on the synthesis of aromatic compounds (F324)

Mark Scheme/Answer:

http://ocr.org.uk/Images/61007-mark-scheme-unit-f324-rings-polymers-and-analysis-january.pdf

In Kekulé's structure the carbons are arranged in a hexagon with alternating double and single bonds between them. However, this structure doesn't explain benzene's chemical and physical properties properly. 

(Bolded parts are the parts that are taken from the mark scheme for a 3 mark question based on this topic)

1) Low Reactivity: Benzene doesn't react like other alkenes despite the 3 double bonds. It doesn't take part in electrophilic addition reactions which are typical of alkenes, in other words it is resistant to electrophilic attack. 

2) Bond Length: Carbon single bonds and Carbon double bonds have different lengths, therefore if benzene's structure was as Kekule described it, it wouldn't be symmetrical, because of alternating bond lengths, however, his structure was commonly showed as a symmetrical molecule. In actual fact, the carbon to carbon bond lengths in benzene are all the same, 0.139nm, this bond length is the intermediate between carbon double bonds and carbon single bonds.

3) Hydrogenation: The enthalpy change of one carbon double bond is -120kJmol/-1 ( for hydrogenation of cyclohexene), therefore if benzene had 3 double bonds like kekule suggests the enthalpy change would be -360kj mol/-1. However this is not the case, the hydrogenation of benzene is less exothermic than expected. It is only -208kj mol/-1

A typical exam question would be something like this "Explain the evidence that led to scientists to doubt the model proposed by Kekulé"

This can be easy 3 marks if you structure your answer explaining the 3 reasons mentioned above clearly. As the mark scheme below shows.

You need to know:

the Ka of the acid

the concentration ratio between acid and base

Buffer consisting of a weak acid, HA, and its conjugate base, A-.

Ka = [H+][A-] / [HA]

therefore [H+] = Ka x [HA] / [A-]

Example: 0.05 moldm^3 of CH3COOH (weak acid - HA)

0.1 moldm^3 CH3COO- (A-)

Ka = 1.7x10^-5

So:

[H+] = 1.7x10^-5 x 0.05/0.1

[H+] = 8.5x10^-6

then -log10[H+] for pH. ( = 5.07)

pH

(similarly, the p in front is going to mean -log10 it - so:  if you're going to use the pKa methods)

With: 

Need to work out the pH of a weak acid?

Rearrange with what you know to get the [H+] so that you can put that into the pH equation

Example: you're given the equilibrium

the Ka expression would be:

You will most likely be given the Ka and asked to work from there. Ethanoic acid dissociation is 1.7 x 10-5

your solution is 0.2 molar

[CH3COO-][H+] / 0.2  =  1.7 x 10-5 

We assume that H+ = A-, as the acid will dissociate evenly

so your equation becomes:

[H+]^2 / 0.2 = 1.7 x 10-5 

[H+]^2 = 3.4 x 10-6

[H+] = sqrt 3.4 x 10-6

[H+] = 1.8  x 10-3

Then: pH = -log10(1.8 x 10-3  

= 2.74

Remember:

pH of a weak acid is going to be a figure close to 3. If you've got anything below 2 or above 5.5, you've probably gone wrong somewhere (especially if you're ending up alkaline)

measure pH to 2 decimal places unless otherwise instructed

Calculating pH of bases

Strong bases

There are two methods, i use the one without the P's. It's a little longer but easier to understand rather than just blindly accepting something as fact.

the formula you need to remember is:

[H+] = Kw / [OH-]

using these equations and usually some rearranging, you should be able to answer any pH question

to summarise:

Weak acids:

Strong bases:

[H+] = Kw / [OH-]

Strong acids - the concentration of the whole acid = [H+] because it fully dissociates