OCR Salters Chemistry F335 12th June 2013 Exam revision thread

Original post by CameronLee

Anybody know of any exam style question resources bar past papers ?

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The Chemical Ideas textbook questions are pretty good :smile:

Reply 25

anyone else doing f334 aswell? is there a thread for it, I can't find anything :frown:

((

Reply 26

Willisme1

5

Original post by Tikara

anyone else doing f334 aswell? is there a thread for it, I can't find anything :frown:

((

Yeah, I'm doing F334 too :smile:

Revision is going to crank up in the next week or so for that!

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Reply 27

super121

5

Original post by Tikara

anyone else doing f334 aswell? is there a thread for it, I can't find anything :frown:

((

I'm doing it too :/
There is a thread for it, but it's a bit dead. Here's the link anyway: http://www.thestudentroom.co.uk/showthread.php?t=2307436

Reply 28

tsr1

can anyone explain why is H2SO3 named sulphuric (iv) acid???

Reply 29

super121

5

Original post by tsr1

can anyone explain why is H2SO3 named sulphuric (iv) acid???

It's sulfuric acid if its dissolved in water or hydrogen sulfate if its not. The number in the brackets tells us the oxidation state of sulfur.

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Reply 30

Harantony

10

Original post by tsr1

can anyone explain why is H2SO3 named sulphuric (iv) acid???

Expanding on the super121's point above:

In H₂SO₄ and H₂SO₃, sulphur has two different oxidation states, which can be worked out.

H₂SO₄:
Hydrogen nearly always has an oxidation state of +1. As there are two of these, the SO₄ anion must have a charge of 2-.
Oxygen nearly always has an oxidation state of -2. As there are 4 of these, the overall number will be -8 for oxygen. To balance this with sulphur and get an overall charge of the anion to be -2, we need an oxidation state of +6 on the sulphur, as 6-8=-2.

Therefore: hydrogen sulphate(VI) / Sulphuric(VI) acid.

H₂SO₃:
Hydrogen nearly always has an oxidation state of +1. As there are two of these, the SO₃ anion must have a charge of 2-.
Oxygen nearly always has an oxidation state of -2. As there are 3 of these, the overall number will be -6 for oxygen. To balance this with sulphur and get an overall charge of the anion to be -2, we need an oxidation state of +4 on the sulphur, as 4-6=-2.

Therefore: hydrogen sulphate(IV) / Sulphuric(IV) acid. :smile:

Reply 31

tsr1

Original post by Harantony

Expanding on the super121's point above:

In H₂SO₄ and H₂SO₃, sulphur has two different oxidation states, which can be worked out.

H₂SO₄:
Hydrogen nearly always has an oxidation state of +1. As there are two of these, the SO₄ anion must have a charge of 2-.
Oxygen nearly always has an oxidation state of -2. As there are 4 of these, the overall number will be -8 for oxygen. To balance this with sulphur and get an overall charge of the anion to be -2, we need an oxidation state of +6 on the sulphur, as 6-8=-2.

Therefore: hydrogen sulphate(VI) / Sulphuric(VI) acid.

H₂SO₃:
Hydrogen nearly always has an oxidation state of +1. As there are two of these, the SO₃ anion must have a charge of 2-.
Oxygen nearly always has an oxidation state of -2. As there are 3 of these, the overall number will be -6 for oxygen. To balance this with sulphur and get an overall charge of the anion to be -2, we need an oxidation state of +4 on the sulphur, as 4-6=-2.

Therefore: hydrogen sulphate(IV) / Sulphuric(IV) acid. :smile:

very well explained.. thank you :smile:

Reply 32

does anyone know if NaBH₄ reduces carboxylic acids?
I know it can reduce aldehydes and ketones to their respective alcohols but it doesn't really mention anything about reducing -COOH to a tertiary alcohol

Reply 33

also, if people don't have the A2 revision guide and want more exam style questions I can post some up :smile:

Reply 34

Original post by Tikara

does anyone know if NaBH₄ reduces carboxylic acids?
I know it can reduce aldehydes and ketones to their respective alcohols but it doesn't really mention anything about reducing -COOH to a tertiary alcohol

I really doubt that they will ever ask about the potential of NaBH₄ reducing carboxylic acids. It's nowhere in the data sheet, nor the textbooks, so I shan't worry.

Plus, remember carboxylic acids come from primary alcohols, as it initially oxidises into an aldehyde, then further oxidation gives you the carboxylic acid. :smile:

Reply 35

Original post by abzy1234

I really doubt that they will ever ask about the potential of NaBH₄ reducing carboxylic acids. It's nowhere in the data sheet, nor the textbooks, so I shan't worry.

Plus, remember carboxylic acids come from primary alcohols, as it initially oxidises into an aldehyde, then further oxidation gives you the carboxylic acid. :smile:

shiieeet - yeah I completely got that wrong didn't I :P tertiary alcohols don't oxidise silly me thanks vm :smile:

yeah they won't ask anything on this

(but searching google did suggest it doesn't and people usually use a slightly different reducing agent)

Reply 36

Original post by Tikara

shiieeet - yeah I completely got that wrong didn't I :P tertiary alcohols don't oxidise silly me thanks vm :smile:

yeah they won't ask anything on this

(but searching google did suggest it doesn't and people usually use a slightly different reducing agent)

Yeah once you actually go deeper and deeper into the crazy world of organic chemistry, you get all kinds of weird reagents!

Stick to our textbooks :tongue:

Reply 37