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OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD

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Original post by Dinasaurus
Guys how do I calculate the pH of a buffer solution of a week acid, is it not Ka=[H]^2 / (HA)

So you do -log of the sqrt[HA]Ka


It's [H+] = Ka* [HA]/[A-] (I remember it by saying Kacid-over-salt because it sounds like a holiday destination lol)
Then pH=-log[H+]
Original post by lai812matthew
0.64-0.4 quick


how do I do this question please, I get that 0.24 represents the HCOOH.

But why do we need the 0.4? I Tthought the formula was Ka= Hsquared/ [HCOOH]
Time to rest. Best of luck tomorrow guys :biggrin:
Original post by Dinasaurus
Guys how do I calculate the pH of a buffer solution of a week acid, is it not Ka=[H]^2 / (HA)

So you do -log of the sqrt[HA]Ka


No..... Buffer is the normal equation of Ka=[H+][A-]/[HA]

[H+] doesn't equal [A-] so you can't make that assumption, thats only for weak acids.
Original post by HJS14
When we get those crazy equations to balance (usually near the end) how do you know whether to use H+ ions, OH-, H20, or electrons? EG last Q on 2014 paper


If you have oxygen on one side of the equation you need to add it to the other side with water.
Then since you've added water you need to balance the hydrogens by adding hydrogen ions to the other side.
You use OH- if it specifies an alkaline conditions or if it says something like "oxygen reacts with water" so obviously you can't balance the oxygens by adding water to the other side so you use OH- instead :smile:
Original post by HJS14
When we get those crazy equations to balance (usually near the end) how do you know whether to use H+ ions, OH-, H20, or electrons? EG last Q on 2014 paper

****ing hate these questions, I dont even know electrolytes are:
Untitled.png
Original post by Dinasaurus
how do I do this question please, I get that 0.24 represents the HCOOH.

But why do we need the 0.4? I Tthought the formula was Ka= Hsquared/ [HCOOH]


That's the equation for calculating pH of a weak acid :smile: But for a buffer solution the equation is different becuase you cant assume that [A-]=[H+] So instead of [H+]sqrd you use:
Ka=([A-][H+])/[HA]
So 0.4 represents the salt (A-)
Original post by BioStudentx
No..... Buffer is the normal equation of Ka=[H+][A-]/[HA]

[H+] doesn't equal [A-] so you can't make that assumption, thats only for weak acids.


So Ka is actually on the fraction?
So Ka/[HA] = [H][A]

So Ka/HA x [A] = H?

I am so lost
Original post by HJS14
When we get those crazy equations to balance (usually near the end) how do you know whether to use H+ ions, OH-, H20, or electrons? EG last Q on 2014 paper


I have no idea from the 2014 paper how someone would have known, but they accepted it without OH- but usually there is an indication ie last year talking about adding potassium hydroxide so we knew it was adding OH- ions
It would indicate if there is some sort of acid added, it's usually in the block of text that can easily be missed
If you know for sure it's not acidic or alkali then you must only add water
Electrons are always included in half equations
Original post by Dinasaurus
So Ka is actually on the fraction?
So Ka/[HA] = [H][A]

So Ka/HA x [A] = H?

I am so lost


Ka=([A][H])/[HA]
Which rearranges to [H]= Ka*[HA]/[A]
Original post by Dinasaurus
how do I do this question please, I get that 0.24 represents the HCOOH.

But why do we need the 0.4? I Tthought the formula was Ka= Hsquared/ [HCOOH]


hcooh+naoh-->hcoona+h2o, 0.4 moles of naoh getting 0.4 moles of hcoona.
Original post by Rust Cohle
****ing hate these questions, I dont even know electrolytes are:
Untitled.png


memorise the hydrogen fuel cell in acidic and alkali conditions - the two half equations then it's simply just recall
How do you use electrode potentials to determine whether a reaction is feasible or not? In the textbook it says E cell must be 0.4+ but in an exam paper there was a question with a feasible reaction that had an E cell of 0.37. How come its feasible?
Anybody got a summary on ligands and complex ions?
june 2015 4ciii) how do you get 0.12M in C2H5COOH concentration.
Original post by Dinasaurus
So Ka is actually on the fraction?
So Ka/[HA] = [H][A]

So Ka/HA x [A] = H?

I am so lost


It's just a rearrangement mate. You'll normally have 3 values and rearrange to find the unknown. Or 1 value and a ratio.
Original post by Dinasaurus
So Ka is actually on the fraction?
So Ka/[HA] = [H][A]

So Ka/HA x [A] = H?

I am so lost


finding pH of a buffer: use 'kacidoversalt' which is [H+] = Ka x [HA]/[A-]
then -log as normal
Original post by HasanAlam
How do you use electrode potentials to determine whether a reaction is feasible or not? In the textbook it says E cell must be 0.4+ but in an exam paper there was a question with a feasible reaction that had an E cell of 0.37. How come its feasible?


In theory if it's greater than 0 it's feasible. In practicality anything below 0.3 is unlikely to be feasible. You won't get docked marks for saying it's feasible if it's above 0.
Original post by HasanAlam
How do you use electrode potentials to determine whether a reaction is feasible or not? In the textbook it says E cell must be 0.4+ but in an exam paper there was a question with a feasible reaction that had an E cell of 0.37. How come its feasible?


it MAY not be feasible if it's below 0.4 V difference but doesn't mean it won't be
Original post by HasanAlam
How do you use electrode potentials to determine whether a reaction is feasible or not? In the textbook it says E cell must be 0.4+ but in an exam paper there was a question with a feasible reaction that had an E cell of 0.37. How come its feasible?


The 0.4V rule is just a general rule but it doesn't apply to everything so I'd just ignore it tbh.
There are two main answers they look for when they say 'why may this not be feasible?':
1) the conditions aren't standard
2) The activation energy is high

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