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Chemistry: Post your doubts here!

L

LeoMessi

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Guys pls help in the following question regarding partition coefficients. I cant understand the mark scheme at all :cry:
The question Question.PNG
and the answer are attached Answer.PNG
 
AbbbbY

AbbbbY

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Kevin Su said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_22.pdf
For 3 part c (i) how do you find the pH? I got the equation Na2O + H2O -> 2NaOH but i have no idea how to get the pH of NaoH ? Would i have to guess based on whether the solution is a acid or a base?
Click to expand...

Umm. What is NaOH and can it even have a ph < 7. Think, my dear Watson, think. Even a first year OLevels student knows the pH of NaOH as I'm sure do you.
 
GCE As and a level

GCE As and a level

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AbbbbY said:
Easy question. Been asked here quite a lot recently.

Look, moles of NaOH used = 0.045
Moles of NaOH that react with HCl = 0.005

The remaining NaOH is what reacted with the acid so acid remaining = 0.040

Equilibrium concentrations = 0.04 0.04 0.06 0.06
If you don't get the abridged explanation let me know and I'll go step by step.

BTW, anyone A2 will tell you this is an incorrect question as the NaOH will hydrolyse the ester formed but let's just assume it magically doesn't.
Click to expand...
hey
could u explain this in depth?
why the "Equilibrium concentrations = 0.04 0.04 0.06 0.06" ??
thank you
 
AbbbbY

AbbbbY

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GCE As and a level said:
hey
could u explain this in depth?
why the "Equilibrium concentrations = 0.04 0.04 0.06 0.06" ??
thank you
Click to expand...

You've reacted Acid + Alcohol + HCl to give you an ester.
At equilibrium, some unreacted acid, unreacted alcohol HCl and the ester is present.

You react this with NaOH (Lets assume the ester is NOT hydrolysed here). The NaOH will react completely with HCl and the remainder will react with the unreacted acid.

0.005 mol of HCl is present so 0.005 mol will react with that. 0.045 mols were used up, so moles of acid in the titre = 0.045-0.005 = 0.040. Since it was taken at equilibrium, it's this no. of mols that's left behind.

Thus, 0.04 0.04 0.06 0.06
 
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