A-level Chemistry Revision Squad!

I need help for these two questions. For the ∆H sol, I calculated it this way. -(-2526)-1890+641-801= +476 kJmol^-1. As for the second one, 476-2526=-2050kJmol^-1. However, the answers are (i) ∆Hosol = 641 – 801 = –160 kJ mol–1 [1] (ii) ∆Hohyd = (1890 – 2526 – 160)/2 = –398 kJ mol–1

Can anyone explain the mark scheme to me? I can't quite get it.

anyone doing edexcel ial ?

There's a REALLY LONG way, and there's a much shorter way.

Here's the short way, because **** the long way.
Lol I say short, but my post is long. That's only because I just really really want you to understand it sorry. The working out when you come to write it is short.

You agree that you have some moles of HA.
The question is saying that the student adds some NaOH. This amount is equal to a third of the amount needed to neutralise it.

In other words, a third of the HA has reacted with the NaOH. If it said, half the amount was added, then half of the HA would have been neutralised, and so on.

When we react the HA and NaOH, this forms our A^-.So as a result, the amount of A^- we formed is equal to one third of the HA we started with.

In other words...

$\mathrm{A}^- = \dfrac{1}{3}\times \mathrm{HA}.$

If you want to represent this as concentration, by all means divide it by the volume. But the volume will cancel out, which allows me to just use the moles.

In fact I don't need to do that because...

$\mathrm{H}^+ = \dfrac{\mathrm{K}_c \times \mathrm{HA}}{\mathrm{A}^-} \ \ \ \ \$ becomes $\ \ \ \mathrm{H}^+ = \dfrac{\mathrm{K}_c \times \mathrm{HA}}{\dfrac{1}{3} \times \mathrm{HA}}$

And now the HAs cancel out, which just leaves...

$\mathrm{H}^+ = \mathrm{K}_c \times 3$

This is now solvable for pH
It's just like that first question of where the HA is half the number of A

There's a REALLY LONG way, and there's a much shorter way.

Here's the short way, because **** the long way.
Lol I say short, but my post is long. That's only because I just really really want you to understand it sorry. The working out when you come to write it is short.

You agree that you have some moles of HA.
The question is saying that the student adds some NaOH. This amount is equal to a third of the amount needed to neutralise it.

In other words, a third of the HA has reacted with the NaOH. If it said, half the amount was added, then half of the HA would have been neutralised, and so on.

When we react the HA and NaOH, this forms our A^-.So as a result, the amount of A^- we formed is equal to one third of the HA we started with.

In other words...

$\mathrm{A}^- = \dfrac{1}{3}\times \mathrm{HA}.$

If you want to represent this as concentration, by all means divide it by the volume. But the volume will cancel out, which allows me to just use the moles.

In fact I don't need to do that because...

$\mathrm{H}^+ = \dfrac{\mathrm{K}_c \times \mathrm{HA}}{\mathrm{A}^-} \ \ \ \ \$ becomes $\ \ \ \mathrm{H}^+ = \dfrac{\mathrm{K}_c \times \mathrm{HA}}{\dfrac{1}{3} \times \mathrm{HA}}$

And now the HAs cancel out, which just leaves...

$\mathrm{H}^+ = \mathrm{K}_c \times 3$

This is now solvable for pH
It's just like that first question of where the HA is half the number of A

Has anyone got OCR A-level june 2015 past papers F321, F322 , F324 and F325??
Exams are almost here and I really need those past papers to revise as they are the only ones I haven't done yet!

So what would you call this reaction?

Yeah

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You made a mistake. Moles of HA after partial neutralisation would be 2/3 of original value.

@TeachChemistry

Hey, i'm stuck on Q6.2 on the new specimen paper:

The students collected a 20cm^3 sample of liquid and weighted it. The mass of the sample was 16g. The density of ethanol is 0.79gcm^-3 and that of water is 1.00gcm^-3. Use this data to work out the mass of ethanol in the sample collected. Assume volume of the sample = volume of water+ethanol.

I attempted by setting volume as x and for ethanol x-20 and rearranged for x to get 19.05g which is wrong... How would you come across doing this?

Which exam board?

Here you go.

@TeachChemistry

Hey, i'm stuck on Q6.2 on the new specimen paper:

The students collected a 20cm^3 sample of liquid and weighted it. The mass of the sample was 16g. The density of ethanol is 0.79gcm^-3 and that of water is 1.00gcm^-3. Use this data to work out the mass of ethanol in the sample collected. Assume volume of the sample = volume of water+ethanol.

I attempted by setting volume as x and for ethanol x-20 and rearranged for x to get 19.05g which is wrong... How would you come across doing this?

Hi guys Hope revision is going well for you all I have one question that I'm stuck on (from the AQA June 2013 paper)- I've managed to solve pretty much all the redox calculation questions except this one
I would appreciate any help- thank you!

Compound Z is a complex that contains only cobalt, nitrogen, hydrogen and chlorine.A solid sample of Z was prepared by reaction of 50 cm3 of 0.203 mol dm–3 aqueouscobalt(II) chloride with ammonia and an oxidising agent followed by hydrochloric acid.When this sample of Z was reacted with an excess of silver nitrate, 4.22 g ofsilver chloride were obtained.Use this information to calculate the mole ratio of chloride ions to cobalt ions in Z.Give the formula of the complex cobalt compound Z that you would expect to beformed in the preparation described above.Suggest one reason why the mole ratio of chloride ions to cobalt ions that you havecalculated is different from the expected value.

So what don't you follow in the mark scheme?

Moles of cobalt = (50 × 0.203)/1000 = 0.01015 mol
Moles of AgCl = 4.22/143.4 = 0.0294
Ratio = Cl- to Co = 2.9 : 1
[Co(NH3)6]Cl3 (square brackets not essential)
Difference due to incomplete oxidation in the preparation

Can anyone help me work out the pH of a contaminated buffer solution such as the one as follows:
Calculate the change in pH of 100cm3 of a buffer solution containing o.5moldm-3 of ethanoic acid anf 0.5moldm-3 of potassium ethanoate when it is contaminated by:
a. 1.0cm3 of 1.0moldm-3 of HCL
b. 2cm3 of 1.5moldm-3 potassium hydroxide

I sat there for 1 hour trying to get an answer but I have no clue how to work out all my class mates had different answers to each other..