Mo_maths
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Hi,

The curve is like this Name:  Screen Shot 2014-04-01 at 11.43.58.png
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Ph on the side, vol of 0.1M NaOH added to the weak acid 0.1M


I had to find the pka, so ph at equivalence, I got 6.5 is this wrong? where is half equivalence?
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EierVonSatan
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The half equivalence point occurs where half the volume of NaOH is added to get to equivalence. So if equivalence took 20 ml, then half equivalence is 10 ml.
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Mo_maths
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(Original post by EierVonSatan)
The half equivalence point occurs where half the volume of NaOH is added to get to equivalence. So if equivalence took 20 ml, then half equivalence is 10 ml.
The thing was, i was given data to plot, so
0.00 naoh had ph 5.1
1.00 naoh had ph 7.9

This was the biggest jump, so would the volume needed for half neutralisation be like 0.5cm^3. Is the rest of the curve there to trick me? after the ph is above 7 does this mean there is no acid left?
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Borek
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Just a more nice curve for a pKa 6.5 acid titrated with sodium hydroxide:

Image

100% is the equivalence point (as defined by the stoichiometry).
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Mo_maths
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(Original post by Borek)
Just a more nice curve for a pKa 6.5 acid titrated with sodium hydroxide:

Image

100% is the equivalence point (as defined by the stoichiometry).
At 0.00 NaOH the ph was 5.1, it then shot up to 7.9 when 1.00 NaOh was added. is the half equivalence between these two points or have i done it all wrong?
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Borek
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How much NaOH was added to reach the end point?
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Mo_maths
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(Original post by Borek)
How much NaOH was added to reach the end point?
That what I'm confused about, I though that the end point is between when the solution is acidic and basic i.e. between the jump from 5 to 8.
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EierVonSatan
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(Original post by Mo_maths)
That what I'm confused about, I though that the end point is between when the solution is acidic and basic i.e. between the jump from 5 to 8.
It's more likely that your end point is the part between 8 and 11 pH, where the graph is steepest. What does 28 refer to on your graph? Is it 28 ml?

Since both solutions are equal concentration, you would expect an equal volume of NaOH is needed to reach the equivalence point for the volume of weak acid (assuming it is monoprotic). So 1 ml isn't realistic if the volume you are titrating against is roughly 20-30 ml.
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Mo_maths
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(Original post by EierVonSatan)
It's more likely that your end point is the part between 8 and 11 pH, where the graph is steepest. What does 28 refer to on your graph? Is it 28 ml?

Since both solutions are equal concentration, you would expect an equal volume of NaOH is needed to reach the equivalence point for the volume of weak acid (assuming it is monoprotic). So 1 ml isn't realistic if the volume you are titrating against is roughly 20-30 ml.
Yeah I had a feeling id done it wrong, but that was only one mark ( finding volume of end point) would you say a pka of 6.5 sounds ok?
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Borek
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(Original post by Mo_maths)
That what I'm confused about, I though that the end point is between when the solution is acidic and basic i.e. between the jump from 5 to 8.
No, if the titrated acid is weak end point is at pH higher that that. For acid with pKa 6.5 pH of the endpoint is around 9.6 (as marked on the titration curve I posted).
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Mo_maths
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(Original post by Borek)
No, if the titrated acid is weak end point is at pH higher that that. For acid with pKa 6.5 pH of the endpoint is around 9.6 (as marked on the titration curve I posted).
Sorry I meant for half equivalence, the question was draw the graph and find half equivalence to work out ka.. or is that still wrong :/ we had to find pka from the graph
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Borek
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You can't know half-equivalence without knowing end point - half equivalence is when you added exactly half of the titrant. So you have to titrate till the end, find the end point, and then read from the curve what was pH when half of the titrant was added.
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