AQA A2 CHEM4 & CHEM5 June 2015 [official discussion thread]

To the people discussing 8(c) on the June 2012 CHEM4 paper - was that the 5 marker where you had to draw isomers? Can't quite remember and school took our mock papers back in.

Hi can someone help me with this question.

It's Chem4, from June 2010, 5(a)(iii)

'Calculate the pH of pure water at 50degrees celsius'

Kw at 50 degrees is 5.47x10^-14


I've looked at the mark scheme but I'm not sure how they got the value for H+

Thank you for the help!


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Hi can someone help me with this question.

It's Chem4, from June 2010, 5(a)(iii)

'Calculate the pH of pure water at 50degrees celsius'

Kw at 50 degrees is 5.47x10^-14


I've looked at the mark scheme but I'm not sure how they got the value for H+

Thank you for the help!


Posted from TSR Mobile

Ok here goes.

We know that an equation for Kw is K_w = [H⁺][OH^-]

When it comes to pure water the concentration of OH^- and H⁺ are equal.

So this means that K_w for water is K_w = [H⁺]²


Now we ought to rearrange the equation about to make [H⁺] the subject.

[H⁺] = √K_w

[H⁺] =√5.48 x 10^-14

[H⁺] = 6.63


I hope that helps

Ok here goes.

We know that an equation for Kw is K_w = [H⁺][OH^-]

When it comes to pure water the concentration of OH^- and H⁺ are equal.

So this means that K_w for water is K_w = [H⁺]²


Now we ought to rearrange the equation about to make [H⁺] the subject.

[H⁺] = √K_w

[H⁺] =√5.48 x 10^-14

[H⁺] = 6.63


I hope that helps

Can definitely vouch for that. Also doing practice questions on n.m.r spectroscopy, specifically ones were you work out the molecule is very beneficial.

EDIT: What was question 8 a about, we missed that one out because when we did the mock last year we hadn't covered the area it was on yet, I don't know the specific area. I remember being really stumped with 8c, got it all right except P I think which was meant to be an optical carboxylic acid, it was such a silly mistake I made as well.

Hi there!

Calcium hydroxide dissociates in the following way:

Ca(OH)₂ ⇌ Ca²⁺ + 2OH^-

You're right Calcium hydroxide is a strong base so it will completely dissociate.

You'll notice from the equation above that the ratio between Ca(OH)₂ and the OH^- ions is 1:2.

This means that the concentration of the OH^- ions that have been dissociated will be double the concentration of Ca(OH)₂.

I believe all hydroxide molecules like Ca(OH)₂, KOH, NaOH, LiOH and etc are strong bases with exception of ammonium hydroxide, NH₄OH. So they're more easier to identify as strong bases.
But I don't know which are weak bases.

I found Q8c particularly difficult when I first did it in my MOCK but I believe the best way to master these questions is to do loads of practise questions on them because you'll find later on that they're actually not half bad and they'll become easy marks to grab.

I must say Q8a)i) was a beast! That got everyone in my year stumped!

Yes! Doing a lot NMR questions will boost your understanding and more importantly the time that you spend on those questions. I used to take a good 10 minutes answering those questions at the beginning of A2 but now its getting more easier with practise. And you're especially right with the forming structures questions.

Q8a was about naming and outlining a mechanism of how 5-hydroxy pentanoyl chloride will form a cyclic ester and have HCl as its by product.

Yh P was an optical carboxylic acid, with an alkene group attached.

Hey, sorry I keep on asking for help but would you mind going through part (c) aswell? I generally understand the buffer questions, but I got a little confused with this one. Thank you so much for the help! I really appreciate it


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not that person but thought i'd give it a go

So first things first, we need to work out the number of moles present of each solution.

Number of moles of H2s04 =25/1000x0.150
=3.75x10-3 moles

Number of moles of KOH = 30/1000x0.200
=6x10-3 moles

At this point its good to make the adjustment for the fact h2s04 is a diprotic acid. This means that for each mole of h2s04 that dissociates, 2H+ ions are "given off". This means that the moles of you H+ ions generated from the h2s04 (which is what will be used to neutralise the KOH) will actually double. From this point, think of the h2s04 moles as the moles of H+ (which has doubled)

For this reason, we can consider the H+ moles of the H2s04 to be 2x(3.75x10-3) = 7.5x10-3 moles.

From these two calculated values ( the moles of KOH and the "adjusted" moles of h+)we can spot that the moles of h+ is in excess because 7.5x10-3>6x10-3 moles. This means that we have an excess of H+ ions in the solution and we can plug this into the pH=-log(H+).

Remember for this stage that it is the concentration of H+ ions and therefore we need to take into account the volume of the solution. The volume of solution present has actually changed because you've added 25cm and 30cm of an acid and a base in. Therefore the total volume of the solution is now 55 cm3

So to account for this our final ph calculation=

pH=-log($\frac{1.5 \times 10^{-3} }{ \frac{55}{1000} }$)

=1.56

Hey, sorry I keep on asking for help but would you mind going through part (c) aswell? I generally understand the buffer questions, but I got a little confused with this one. Thank you so much for the help! I really appreciate it


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Thank you for the help! I understand it much better now
It was just weird in the mark scheme because it said moles of NaOH instead of moles oh KOH. But thank you anyway, I do understand it now



SIZE=1]Posted from TSR Mobile

You do not need to apologise for anything, I am always happy to help.

I must say when I first looked at this question I was a little confused at first as well.

So this is what your question is asking:

Calculate the pH of the solution formed when 25.0 cm³ of 0.150 mol dm^–3 aqueoussulfuric acid are added to 30.0 cm³ of 0.200 mol dm^–3 aqueous potassium hydroxide at25 °C. Assume that the sulfuric acid is fully dissociated.

The equation that forms here is:

H₂S0₄ + 2KOH → K₂SO₄ + 2H₂0

You'll have to work out the moles for KOH and H₂S0<sub>4.

Please note for the following calculations of moles I will write 10^-3 instead of ( ÷ 1000) because it is simpler for me.</sub>

The moles for KOH is:

n = 30 x 0.2 x 10^-3
n = 6 x 10^-3 mol

The moles for H₂SO₄:

n = 25 x 0.15 x 10^-3

n = 3.75 x 10^-3 mol

Now when you need to work out pH you need the value of [H⁺] which you'll get from the acid it comes from.

This is quite a special case as H₂SO₄ is a diprotic acid. It dissociates in the following way.

H₂SO₄⇌ 2H⁺ + SO₄^2-

This means that the molar ratio between H₂SO₄ and H⁺ is 1:2.

Hence to work out H⁺ you have to do 2 x H₂SO₄ which equals 7.5 x 10^-3 mol.

That is the initial molar value of H⁺ before it reacted with KOH.

The excess value of H⁺ is (7.5 x 10^-3) - (6 x 10^-3) = 1.5 x 10^-3 mol after it reacted with KOH.

That is the value present in the buffer solution.

Now you must find the concentration:

[H⁺] = (1.5 x 10^-3 x 10³⁾ ÷ (25 + 30) = 0.0273 mol dm^-3

pH = -log₁₀[H⁺]
pH = -log₁₀[0.0273]
pH = 1.56

I hope that helps, let me know if you need any further explanations.

I understand this question, but if you don't mind me asking, whenever we have a diprotic acid, will the equation react with 2 moles of everything?

I do not mind at all

Yes.

Whenever you have a diprotic acid it will react 2 moles worth of the acid in a buffer solution as H⁺ is what's causing the neutralisation and SO4^- is just a spectator ion.

And yes again.

You should multiply the moles by 2 before finding excess because as I said earlier it is the H⁺ that causes the neutralisation so you should work with H⁺ moles.

Does that make sense?

Yes, that makes sense and clears it up. Thanks

Do you know anywhere I can find lots of practice questions for them 8c kind of questions?

Ha! I just saw that mistake on the mark scheme too!

That is so weird, you'd think they'd be flawless innit.

Exactly!
They expect us to be flawless but they're making mistakes themselves.. ugh what can you say ...