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Edexcel A2 Chemistry Unit 4 26/01/12 PM

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Can someone quickly explain an alkaline buffer solution to me please?
Reply 381
Original post by confused dot com
Can someone quickly explain an alkaline buffer solution to me please?


Buffer solution is above the pH of 7.
Consists of a weak base and its salt.
So, weak bases like Ammonia and a salt that is formed with ammona, like ammonium chloride etc.
They both form a buffer solution.
Original post by EffKayy

Original post by EffKayy
Buffer solution is above the pH of 7.
Consists of a weak base and its salt.
So, weak bases like Ammonia and a salt that is formed with ammona, like ammonium chloride etc.
They both form a buffer solution.


Sorry if I wasn't clear, but I meant what happens when an acid or alkaline is added :colondollar:
Reply 383
Original post by confused dot com
Sorry if I wasn't clear, but I meant what happens when an acid or alkaline is added :colondollar:


The same principle applies to both an acidic buffer or an alkaline buffer.

When acid is added (so the h+) which lies on the right hand side of the equillibrium, more H+ combines with the other product like NH3 to form NH4+. Also the equillibrium wants to restore itself so it shifts to the left.
When an alkaline is added (so OH), it reacts with the H+, causing it to be removed off the equillbrium as water is formed, so more of, say, NH4+ dissociates to replace the H+.

Therefore the H+ concentration doesnt change so the pH stays constant.
Original post by EffKayy

Original post by EffKayy
The same principle applies to both an acidic buffer or an alkaline buffer.

When acid is added (so the h+) which lies on the right hand side of the equillibrium, more H+ combines with the other product like NH3 to form NH4+. Also the equillibrium wants to restore itself so it shifts to the left.
When an alkaline is added (so OH), it reacts with the H+, causing it to be removed off the equillbrium as water is formed, so more of, say, NH4+ dissociates to replace the H+.

Therefore the H+ concentration doesnt change so the pH stays constant.



Okay, so correct me if I'm wrong here:

I guess you're talking about the equilibrium reaction: NH3 + H2O <=> NH4+ + OH-

So adding H+, in order to resist the change to pH that excess H+ would cause, some of the reservoir of NH3 reacts with H+, removing the excess and preventing change to pH

Adding an alkaline adds OH- to the reaction mixture, therefore some of the reservoir of NH4+ reacts with it, removing the excess OH- which minimises any change to pH.

Is this right?
Original post by areyousure?
How I remember is inert - total entropy is negative so they just don't react. Like noble gases :smile:

Stable is due to a high activation energy - so from the thermo you can see that if you increase temperature then more molecules have kinetic energy greater than EA so it becomes unstable..


That didn't really make much sense :P



Original post by oHellno
If the total entropy is negative then the reactants are thermodynamically stable and then reaction won't take place. But if the total entropy is positive then the reaction is feasible however if the activation energy is very high it's unlikely to go and so you would say the reactants are kinetically inert.


Alright guys, thanks!
Do we need to know the specifics of the Haber Process, Urea production or that section on catalytic converters?
Reply 387
Original post by confused dot com

Original post by confused dot com
Do we need to know the specifics of the Haber Process, Urea production or that section on catalytic converters?


are you talking about the HSW box on page 25-6 in the blue Ann Fulick book??
Reply 388
Original post by confused dot com

Original post by confused dot com
Do we need to know the specifics of the Haber Process, Urea production or that section on catalytic converters?


No, according to the specification we don't need to know the detail of industrial processes.
Original post by wam-bam

Original post by wam-bam
are you talking about the HSW box on page 25-6 in the blue Ann Fulick book??


Yep
Original post by oHellno
No, according to the specification we don't need to know the detail of industrial processes.


Okay, thanks for that. Also, how much do we have to know about the propanone and iodine reaction as my teacher never went through it :s-smilie:
Reply 391
Original post by confused dot com

Original post by confused dot com
Yep


the no, we don'y
Original post by wam-bam

Original post by wam-bam
the no, we don'y


Alright, thanks :biggrin:
Time to start solid revision for this....
btw, is there a specimen paper for this unit out there?
Original post by confused dot com

Original post by confused dot com
Okay, thanks for that. Also, how much do we have to know about the propanone and iodine reaction as my teacher never went through it :s-smilie:


Hi, does anyone know the answer to this please?
Original post by confused dot com
Hi, does anyone know the answer to this please?


yes-you need to learn it.
It isnt soely remembering but you would have to understand what is going on. As in how they are using this reaction to figure out the rate of reaction.
Original post by im so fresh

Original post by im so fresh
yes-you need to learn it.
It isnt soely remembering but you would have to understand what is going on. As in how they are using this reaction to figure out the rate of reaction.


Okay, so I know how they calculate rate and order through quenching the reaction and titration with sodium thiosulfate, but do we have to know the actual mechanism used to form iodopropanone?
Original post by confused dot com
Okay, so I know how they calculate rate and order through quenching the reaction and titration with sodium thiosulfate, but do we have to know the actual mechanism used to form iodopropanone?


if you mean the reaction, then yes most likely.
Original post by im so fresh

Original post by im so fresh
if you mean the reaction, then yes most likely.


Alright then, thanks for your help :smile:
Original post by Sandeep G
Yes you need to know Iodoform reaction


That is not the iodoform reaction, I believe this reaction is done to determine rates and rate mechanism. Iodoform reaction is using iodine and alkali.



How does everyone think tomorrow's paper will be? From what I see, a lot of papers in other subjects have been obscenely difficult. Anyone thinks we have a tough one set up for us? How much of IR and NMR do you think is going to come up?

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