A2 pH Question Help. Watch

Delta, Δ
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Hello guys!

http://filestore.aqa.org.uk/subjects...W-MS-JAN11.PDF
http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF
for question 2dii, in the mark scheme, 5.0 ×10-4 mol is added to Y-, why is this so?
Since the NaOH is naturalized by HY, all of the NaOH is gone, so I don't understand why this affects Y- concentration. Can you please help? Thanks guys! =)
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Secret.
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(Original post by Delta, Δ)
Hello guys!

http://filestore.aqa.org.uk/subjects...W-MS-JAN11.PDF
http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF
for question 2dii, in the mark scheme, 5.0 ×10-4 mol is added to Y-, why is this so?
Since the NaOH is naturalized by HY, all of the NaOH is gone, so I don't understand why this affects Y- concentration. Can you please help? Thanks guys! =)

Y- is the salt that is formed, as the acid is still in excess and you added some base to it you would produce that much more salt (the amount of base added) so if you added 0.0005 of OH it would react with 0.0005 moles of weak acid and produce 0.005 more moles of the salt (Y-) [this will also reduce the moles of weak acid, HY, by 0.0005)

Not sure if that's explained clearly or not? If not, I'll try and clarify
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letsbehonest
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When NaOH is added, more of the conjugate base is being made, so you add the moles of NaOH to the moles of the existing conjugate base to find the total moles of conjugate

Then you subtract the moles of NaOH added from the moles of the acid because some of the acid is reacting with the NaOH to form the conjugate base. Remember that the acid is in excess so you will still have acid left over..

When you have your new moles of conjugate base, you divide it by the new total volume. You do the same for the acid, to find the concentration.

Then you use your Henderson–Hasselbalch equation to find [H+] then proceed to find pH
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letsbehonest
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(Original post by Delta, Δ)
Hello guys!

http://filestore.aqa.org.uk/subjects...W-MS-JAN11.PDF
http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF
for question 2dii, in the mark scheme, 5.0 ×10-4 mol is added to Y-, why is this so?
Since the NaOH is naturalized by HY, all of the NaOH is gone, so I don't understand why this affects Y- concentration. Can you please help? Thanks guys! =)
All of the NaOH is gone.. but gone where? It has gone to form the conjugate base Y-
You add them so you can find the total moles of Y- after the NaOH is added.

And it affects concentration because.. you will have a new total number of moles of Y- and also a new volume.
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