coconut64
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Size:  64.7 KB I don't quite get this question because if the conc is halved then it is two times less effective, so surely you will need twice more of the base to reach the equivalence point. I don't get why two times less base is needed if you have a lower conc of base.

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Pigster
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The concentration of the acid is being halved, not the alkali.
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will'o'wisp
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(Original post by coconut64)
Name:  xxxxxxxxxx.png
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Size:  64.7 KB I don't quite get this question because if the conc is halved then it is two times less effective, so surely you will need twice more of the base to reach the equivalence point. I don't get why two times less base is needed if you have a lower conc of base.

ThanksAttachment 601288601290
Can't fulyl answer the question but doesn't this have to do with a balanced equation?
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Pigster
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(Original post by will'o'wisp)
Can't fulyl answer the question but doesn't this have to do with a balanced equation?
It is: CH3COOH + NaOH -> CH3COO-Na+ + H2O

But is doesn't matter what the stoichiometry is, the important thing is that if it requires X mol of acid to neutralise Y mol of alkali, it would require Y/2 mol of alkali to neutralise X/2 mol of acid.

ASSUMING that when the concentration of acid were halved, the volume wasn't also changed.
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coconut64
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(Original post by Pigster)
The concentration of the acid is being halved, not the alkali.
Right, so bascially what you are adding is 2 times stronger, so you need two times less of the base to neutralise the acid?

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will'o'wisp
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(Original post by Pigster)
It is: CH3COOH + NaOH -> CH3COO-Na+ + H2O

But is doesn't matter what the stoichiometry is, the important thing is that if it requires X mol of acid to neutralise Y mol of alkali, it would require Y/2 mol of alkali to neutralise X/2 mol of acid.

ASSUMING that when the concentration of acid were halved, the volume wasn't also changed.
Oh right, ok so same volume only conc changed got it ^-^ thanks!
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coconut64
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(Original post by Pigster)
The concentration of the acid is being halved, not the alkali.
Also, lower conc of acid always means that the pH will be higher?
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Pigster
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Initial pH will be (a bit) higher
AND
Vertical bit on the graph will happen when half the volume of NaOH is added (compared to original graph).

your homework = done

(or at least this bit)
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coconut64
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(Original post by Pigster)
Initial pH will be (a bit) higher
AND
Vertical bit on the graph will happen when half the volume of NaOH is added (compared to original graph).

your homework = done

(or at least this bit)
Thanks, this is not hw btw
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