[Doctor.]

(Original post by masterhr1)
Does anyone have a link to a diagram/flowchart of all the organic synthesis route stuff?

being honest, it's best to just make one for yourself! You'll memorise it while you make it

[jcb123]

Are we supposed to know how to do the following?, as it is my work booklet but can't seem to find it anywhere on the spec...

Calculating pH of strong acids or bases after dilution

and

Calculating pH on mixing solutions of strong acids and bases

Thanks

[Doctor.]

(Original post by jcb123)
Are we supposed to know how to do the following?, as it is my work booklet but can't seem to find it anywhere on the spec...

Calculating pH of strong acids or bases after dilution

and

Calculating pH on mixing solutions of strong acids and bases

Thanks

Strong acids after dilution...I can't see it in spec either lol So no

PS How are ya finding the rest of the spec? Done the 2012 paper yet?

[jcb123]

(Original post by Doctor.)
Strong acids after dilution...I can't see it in spec either lol So no

PS How are ya finding the rest of the spec? Done the 2012 paper yet?

I only started revision today! done Kinetics, Equilibria and nearly finished Acids and Bases. I am finding Unit 4 very hard - don't want to look at questions just yet

I took Unit 5 in January (managed 91/120 - B... 96 the A) - retaking that in June.

How about you? - you completed all the topics and now onto past papers?!!!

[Doctor.]

(Original post by jcb123)
I only started revision today! done Kinetics, Equilibria and nearly finished Acids and Bases. I am finding Unit 4 very hard - don't want to look at questions just yet

I took Unit 5 in January (managed 91/120 - B... 96 the A) - retaking that in June.

How about you? - you completed all the topics and now onto past papers?!!!

Oh look at you nearly 3 topics in a day is actually awesome!!!! Well done

Generally I think most find CHEM4 the hardest . It's pretty tough, so much to learn .Well done on unit 5! I'm sure youll do great!

Me, hmmm well I did the content and all notes and stuff, did one past paper and was just...disheartened. LOL hence the '' smiley when I mentioned 2012 paper lol

Just need to keep at It I guess Buffers is not fun

[jcb123]

(Original post by Doctor.)
Oh look at you nearly 3 topics in a day is actually awesome!!!! Well done

Generally I think most find CHEM4 the hardest . It's pretty tough, so much to learn .Well done on unit 5! I'm sure youll do great!

Me, hmmm well I did the content and all notes and stuff, did one past paper and was just...disheartened. LOL hence the '' smiley when I mentioned 2012 paper lol

Just need to keep at It I guess Buffers is not fun

Buffers is what I need to do next! don't scare me

What grade do you need for Chem?

[Doctor.]

(Original post by jcb123)
Buffers is what I need to do next! don't scare me

What grade do you need for Chem?

need A*, I need full marks In CHEM4...lol.(already done Chem5)

[jcb123]

(Original post by Doctor.)
need A*, I need full marks In CHEM4...lol.(already done Chem5)

I take it you applied for Medicine and guessing you have an offer from Cambridge?!

[Doctor.]

(Original post by jcb123)
I take it you applied for Medicine and guessing you have an offer from Cambridge?!

1) Medicine yes, need an A* to meet grades (I'm a resitter)

2) Cambridge would never accept me

[jcb123]

(Original post by Doctor.)
1) Medicine yes, need an A* to meet grades (I'm a resitter)

2) Cambridge would never accept me

So where you got offers from?

[Glacier]

(Original post by jcb123)
Are we supposed to know how to do the following?, as it is my work booklet but can't seem to find it anywhere on the spec...

Calculating pH of strong acids or bases after dilution

and

Calculating pH on mixing solutions of strong acids and bases

Thanks

For strong acids and bases after dilution, you do: volume before dilution x concentration before dilution = moles, then moles / volume after dilution to get concentration of either H⁺ or OH^- after dilution. Then -log [H⁺] to get pH. If it's a base, then do [H⁺] = Kw/[OH^-] first to get [H⁺]. If it's a diprotic acid or base, such as H₂SO₄ or Ba(OH)₂, remember to multiply moles of acid or base by 2 to get the number of moles of H⁺ or OH^-.

When mixing acids and bases, first do concentration x volume to calculate the number of moles of acid and base to find which one is in excess (remember to check if either the acid or base is diprotic), then find the difference between the number of moles of H⁺ and OH^- and use -log [H⁺] to find the pH. If the excess is a weak acid HX then, you need to use K_a = [H⁺]²/[HX], and if it's a buffer solution, then you need to use K_a = [H⁺][X^-]/[HX], where [X^-] is the concentration of the salt, in order to find [H⁺].

I hope this makes sense.

[jcb123]

(Original post by Glacier)
For strong acids and bases after dilution, you do: volume before dilution x concentration before dilution = moles, then moles / volume after dilution to get concentration of either H⁺ or OH^- after dilution. Then -log [H⁺] to get pH. If it's a base, then do [H⁺] = Kw/[OH^-] first to get [H⁺]. If it's a diprotic acid or base, such as H₂SO₄ or Ba(OH)₂, remember to multiply moles of acid or base by 2 to get the number of moles of H⁺ or OH^-.

When mixing acids and bases, first do concentration x volume to calculate the number of moles of acid and base to find which one is in excess (remember to check if either the acid or base is diprotic), then find the difference between the number of moles of H⁺ and OH^- and use -log [H⁺] to find the pH. If the excess is a weak acid HX then, you need to use K_a = [H⁺]²/[HX], and if it's a buffer solution, then you need to use K_a = [H⁺][X^-]/[HX], where [X^-] is the concentration of the salt, in order to find [H⁺].

I hope this makes sense.

Thanks for that! kind of makes sense - can be my task tomorrow to sort these type of calculations out woo! what a joyful day I have ahead

[margaretob]

any one got the ms for aqa a2 chem 4 26th jan PLEASE

[Doctor.]

(Original post by Glacier)
For strong acids and bases after dilution, you do: volume before dilution x concentration before dilution = moles, then moles / volume after dilution to get concentration of either H⁺ or OH^- after dilution. Then -log [H⁺] to get pH. If it's a base, then do [H⁺] = Kw/[OH^-] first to get [H⁺]. If it's a diprotic acid or base, such as H₂SO₄ or Ba(OH)₂, remember to multiply moles of acid or base by 2 to get the number of moles of H⁺ or OH^-.

When mixing acids and bases, first do concentration x volume to calculate the number of moles of acid and base to find which one is in excess (remember to check if either the acid or base is diprotic), then find the difference between the number of moles of H⁺ and OH^- and use -log [H⁺] to find the pH. If the excess is a weak acid HX then, you need to use K_a = [H⁺]²/[HX], and if it's a buffer solution, then you need to use K_a = [H⁺][X^-]/[HX], where [X^-] is the concentration of the salt, in order to find [H⁺].

I hope this makes sense.

We have to know this?

[Glacier]

(Original post by Doctor.)
We have to know this?

I've seen it in quite a few papers, including question 4a on the Jan 12 paper.

[Doctor.]

(Original post by Glacier)
I've seen it in quite a few papers, including question 4a on the Jan 12 paper.

ahh I see it . Think I just got confused by your explaination lol -4a wasn't that hard, just uses basic concept of Aicd-Bases .

[MuffinMonster]

(Original post by masterhr1)
In the January 2010 paper, Q2, it asks to find the concentration of sulphuric acid when pH is 0.25.

After anti logging, [H+] is 0.56

But then in the mark scheme they do 0.56 x 0.5 to get the concentration of sulphuric acid. Can someone explain why we need to do this? I'm too dumb lol.

H2SO4 produce 2 H+ ions when it dissociates.

[thegodofgod]

(Original post by margaretob)
any one got the ms for aqa a2 chem 4 26th jan PLEASE

Is it Jan 12?

If so, see attachments

[thegodofgod]

(Original post by masterhr1)
In the January 2010 paper, Q2, it asks to find the concentration of sulphuric acid when pH is 0.25.

After anti logging, [H+] is 0.56

But then in the mark scheme they do 0.56 x 0.5 to get the concentration of sulphuric acid. Can someone explain why we need to do this? I'm too dumb lol.

Sulfuric acid is a diprotic acid, meaning it dissociates twice to release 2 protons when in solution:

H₂SO₄ --> 2 H⁺ + SO₄^2-

Looking at the stoichiometric ratios, for every mole of H⁺, you have 0.5 moles H₂SO₄.

[MuffinMonster]

The silver nitrate test wouldn't work on something like chloromethane, right? Only on halide ions?