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Topic: Titration - pH at equivalence point  (Read 43865 times)

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Offline refid

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Re: Titration - pH at equivalence point
« Reply #15 on: June 27, 2006, 04:04:23 PM »


HA-

Hydrolizes

HA- + H2O <-> H2A +OH-

Offline Borek

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Re: Titration - pH at equivalence point
« Reply #16 on: June 27, 2006, 04:42:01 PM »
HA-

Right.

Quote
Hydrolizes

HA- + H2O <-> H2A +OH-

and dissociates!

pH of amphiprotic salt
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Offline refid

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Re: Titration - pH at equivalence point
« Reply #17 on: June 27, 2006, 05:18:00 PM »
still confused about this ... give the initial [] how can I figure out the [] at the equivalence point

Offline Borek

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Re: Titration - pH at equivalence point
« Reply #18 on: June 27, 2006, 06:02:02 PM »
still confused about this ... give the initial [] how can I figure out the [] at the equivalence point

Of the salt? M1V1 = M2V2 type calculation. Or find the final volume (sum of titrant and titrated solution), find moles of salt - and use concentration definition.
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Offline refid

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Re: Titration - pH at equivalence point
« Reply #19 on: June 27, 2006, 09:50:30 PM »
Ok got the pH and [Acid]

How to I find the volume required for the second equivalence poit?


pH second equivalence point is .5(ka1 + ka2) do this first?

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Re: Titration - pH at equivalence point
« Reply #20 on: June 28, 2006, 02:55:44 AM »
Ok got the pH and [Acid]

Show your calculations.

Quote
How to I find the volume required for the second equivalence poit?

Simple stoichiometry - answer is in the reaction equation.

Quote
pH second equivalence point is .5(ka1 + ka2) do this first?

???
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Offline refid

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Re: Titration - pH at equivalence point
« Reply #21 on: June 28, 2006, 01:36:59 PM »
10ml of .1M Diprotic Acid titrated with .1M NaOH  pka1=1.87 pka2=6.07

Vol. req. to reach eq pt 1 = 10ml

[HA-]=(.1M x .010L)/ 0.02L = 0.05M


HA- + H2O <-> H2A +OH-

pka1 = 10^(-1.87) =1.35E-2


ka = x^2 / ( 0.05-x )
6.75E-4 -1.35E-2x = x^2

x^2 + 1.35E-2x- 6.75E-4

x=0.0200
x=-0.0335

pH = -log (0.02) = 1.69

Quote
pH second equivalence point is .5(ka1 + ka2) do this first?

???
Quote

to solve the at the second equivalence point equatuion is this .5(ka1 + ka2) should i solve this first?
« Last Edit: June 28, 2006, 01:49:37 PM by refid »

Offline Borek

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Re: Titration - pH at equivalence point
« Reply #22 on: June 28, 2006, 03:19:57 PM »
10ml of .1M Diprotic Acid titrated with .1M NaOH  pka1=1.87 pka2=6.07

Looks like maleic ;)

Quote
Vol. req. to reach eq pt 1 = 10ml

[HA-]=(.1M x .010L)/ 0.02L = 0.05M

So far so good.

Quote
HA- + H2O <-> H2A +OH-

pka1 = 10^(-1.87) =1.35E-2


ka = x^2 / ( 0.05-x )
6.75E-4 -1.35E-2x = x^2

x^2 + 1.35E-2x- 6.75E-4

x=0.0200
x=-0.0335

pH = -log (0.02) = 1.69

No. HA- is amphiprotic, thus it is here that you should use (pKa1+pKa2)/2

Quote
pH second equivalence point is .5(ka1 + ka2) do this first?

Quote
to solve the at the second equivalence point equatuion is this .5(ka1 + ka2) should i solve this first?

Sorry, my English fails here.

At the second equivalence point you have just A2-. This is multiprotic base, but difference in stregth of both steps of basic dissociation is large enough - you may treat it as if only first dissociation step occurs.

Calculate ppH for both equivalence points using approach you should be aware off atm. Download my BATE and play with maleic acid/sodium hydroxide solution, to see how pH changes. Compare values displayed by the program with the results of your calculations, to see when simplified approach fails.
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Offline refid

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Re: Titration - pH at equivalence point
« Reply #23 on: June 28, 2006, 08:47:11 PM »
thanks Borek

i have another question lets say we have a monoprotic acid
and we know the pH of the equivalence point from the titration graph.
I know from the graph I can interpolate and the pH at half the volume of NaOH is the pKa, but
lets say :
pH at the equivalence point  4.89 and initial [ ] = 0.1M 10 ml mono. acid  and 0.1M NaOH

10ml of base req. to reach equivalence point     

5ml of base

how would I calculate the pKa with initial given [ ]? 
 
« Last Edit: June 29, 2006, 05:47:45 PM by refid »

Offline Borek

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Re: Titration - pH at equivalence point
« Reply #24 on: June 29, 2006, 03:36:47 AM »
but how would I calculate the pKa is initial C?

Sorry, I have no idea what you are asking about.

You are right about pH = pKa at 50% titration (although it doesn't work for very strong and very weak acids).
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Offline refid

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Re: Titration - pH at equivalence point
« Reply #25 on: June 29, 2006, 05:49:32 PM »

lets say :
pH at the equivalence point  4.89 and initial [ ] = 0.1M 10 ml mono. acid  and 0.1M NaOH

10ml of base req. to reach equivalence point     

5ml of base

how would I calculate the pKa with initial given [ ]? 
 



Offline Borek

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Re: Titration - pH at equivalence point
« Reply #26 on: June 29, 2006, 06:05:01 PM »
pKa doesn't depend on the initial concentration, if that's what you are asking about.

Take a look at the Henderson-Hasselbalch equation. Once 50% of the initial amount of acid was neutralized, ratio under log equals 1, thus log is 0, and pH = pKa,

Note this approach doesn't hold for very strong acids, very weak acids and diluted acids, as at 50% titration they are either dissociated, or hydrolysed.
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Offline refid

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Re: Titration - pH at equivalence point
« Reply #27 on: June 30, 2006, 01:18:45 AM »
no im saying given the pH at the equivalence point 1 ( monoprotic) how do I calculate the pKa (without using the graph).

Offline Borek

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Re: Titration - pH at equivalence point
« Reply #28 on: June 30, 2006, 03:03:43 AM »
no im saying given the pH at the equivalence point 1 ( monoprotic) how do I calculate the pKa (without using the graph).

Reverse procedure for the pH calculation at the equivalence point. Take equation for pH of a salt of weak acid (base) and solve it for Ka.
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Offline refid

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Re: Titration - pH at equivalence point
« Reply #29 on: June 30, 2006, 04:51:50 PM »
Oic.. also i been looking at your lecture note and I coudnt find any lecture about titration curve with  calculating pka/pkb. 

Im confused If an acid is titrated by a base then



the pka would around 2-3 . and pkb = 14 - pka

If an base is titrated by a acid then


pkb is 10.9  and  pka = 14 - pkb

An polyprotic Acid titrating with NaOH



pkb1= 14 - pka1
pkb2= 14 - pka2
pkb3= 14 - pka3

is this correct?
« Last Edit: June 30, 2006, 05:28:30 PM by refid »

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