# Titration curve help!!

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Hi,

The curve is like this

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?

The curve is like this

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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#2

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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(Original post by

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.

**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.

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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#4

Just a more nice curve for a pKa 6.5 acid titrated with sodium hydroxide:

100% is the equivalence point (as defined by the stoichiometry).

100% is the equivalence point (as defined by the stoichiometry).

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(Original post by

Just a more nice curve for a pKa 6.5 acid titrated with sodium hydroxide:

100% is the equivalence point (as defined by the stoichiometry).

**Borek**)Just a more nice curve for a pKa 6.5 acid titrated with sodium hydroxide:

100% is the equivalence point (as defined by the stoichiometry).

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(Original post by

How much NaOH was added to reach the end point?

**Borek**)How much NaOH was added to reach the end point?

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#8

(Original post by

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.

**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.

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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(Original post by

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.

**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.

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#10

**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.

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(Original post by

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).

**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).

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#12

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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