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Topic: Titration - pH at equivalence point  (Read 43407 times)

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Offline refid

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Titration - pH at equivalence point
« on: June 25, 2006, 06:53:09 PM »
How do you calculate the pH at the equivalence point of mole OH- and H+ are equal? Also the pKa1 and pka2 is also given.
« Last Edit: June 25, 2006, 08:50:47 PM by refid »

Offline Borek

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Re: Titration - pH at equivalence point
« Reply #1 on: June 25, 2006, 06:55:10 PM »
Define an equivalence point.
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Offline refid

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Re: Titration - pH at equivalence point
« Reply #2 on: June 25, 2006, 06:57:38 PM »
equal quantities of acid and base exist to cancel/neutralize each other out. For example 1 mol HCl and 1 mol NaOH

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Re: Titration - pH at equivalence point
« Reply #3 on: June 25, 2006, 07:21:00 PM »
OK. pKa is given so you deal with the weak acid. What you will have in solution at the equivalence point when titrating weak acid?
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Offline refid

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Re: Titration - pH at equivalence point
« Reply #4 on: June 25, 2006, 07:28:52 PM »
If its:

 H2A + OH- <-> HA + H20

At equivance point:

HA + H20 <-> A- + H3O+

Offline Borek

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Re: Titration - pH at equivalence point
« Reply #5 on: June 25, 2006, 07:34:31 PM »
Is it diprotic acid? You stated pKa is given, not pKa1 & pKa2.

Imagine you mix equimolar quantities of acetic acid and NaOH. What you will have in solution?

(hint: acid + base -> .)
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Offline refid

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Re: Titration - pH at equivalence point
« Reply #6 on: June 25, 2006, 08:52:10 PM »
yess sorry its diprotic

Acid + Base -> H20 and Salt?

Offline Borek

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Re: Titration - pH at equivalence point
« Reply #7 on: June 25, 2006, 09:10:12 PM »
yess sorry its diprotic

That makes calculations a little bit harder.

Quote
Acid + Base -> H20 and Salt?

And that's answer to your question - pH calculation for equivalence point is nothing else but pH calculation of salt solution.

It is getting late here (3 a.m.) Check out these pH calculation lectures - especially sections devoted to pH of salts calculation. Salts in general section will be probably a little bit too general, as the equation derived there is of no use for hand calculations, but three other sections - with simplified approaches - should be helpfull.

Also note that in the case of salt of diprotic weak acid and strong base you have (at equivalence point) solution of diprotic weak base (conjugated) - thus you may find pH using exactly the same approach you will use to calculate pH of diprotic weak base solution. Check polyprotic simplified section.
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Offline arnyk

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Re: Titration - pH at equivalence point
« Reply #8 on: June 25, 2006, 09:17:34 PM »
Diprotics are a bit annoying...I'm getting a bit tired too but I'll throw down some points.

Strong + strong should pretty much be straightforward (even if it's diprotic, you just gotta do a little stoichiometry).

Weak + strong, you'll get a salt.  That salt will have an effect on pH.  Remember that each the acid and base will form conjugate acid/bases, and strong bases form weak conjugate acids (and vice versa), weak acids form strong conjugate bases (and vice versa).  So it's these "strong" conjugates that will affect pH.  So write out the equation for the salt dissociating, identify the strong conjugate counterpart.

Hopefully Borek will come back with some diprotic stuff lol. ;)

Offline refid

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Re: Titration - pH at equivalence point
« Reply #9 on: June 25, 2006, 09:37:30 PM »

And that's answer to your question - pH calculation for equivalence point is nothing else but pH calculation of salt solution.


lets 10mL pf 0.200M diprotic acid requires 20mL 0.1000M NaOH to Equivalence point the mole Acid and mole Base neutralize eachother.. how can i find the [salt]?

First: H2A + OH <-> HA +H20 
then: HA +H20 <-> A- + H3O+  How can I Find HA concentration

mol( Acid or base) / Total Volume

I know how to find pH at second quivalence point pH = 1/2 (pKa1 + pka2)
still confuse about first pH at first equivalence point



               

Offline Borek

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Re: Titration - pH at equivalence point
« Reply #10 on: June 26, 2006, 03:39:17 AM »
First: H2A + OH <-> HA +H20 
then: HA +H20 <-> A- + H3O+  How can I Find HA concentration

First of all: balance these with regard to charges, I suppose they are confisuing as not balanced.

Second: think about stoichiometry - what kind of salt will you have present at first equivalence point? Hint: look at titles of my lectures ;)
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Offline refid

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Re: Titration - pH at equivalence point
« Reply #11 on: June 26, 2006, 09:37:49 PM »
First equivance pt H2A + 20H <-> A- + 2H20

moles of H2A = 2 x moles OH

Salt: moles of  A- = 1/2 OH

Molarity of A- = moles of OH / (2 x Total Volume)

A- + H2O <-> HA + Oh-
 
Kb = Kw/Ka1  = x^2 / ( [A-] - x)     x<< [A-]

kb = x^2 / [A-]

( kb[A-] )^(1/2) = x                                    where x equals [OH-]

Correct so far?

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Re: Titration - pH at equivalence point
« Reply #12 on: June 27, 2006, 03:19:26 AM »
First equivance pt H2A + 20H <-> A- + 2H20

That's already second equivalence point.

Charges are completely wrong.
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Offline refid

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Re: Titration - pH at equivalence point
« Reply #13 on: June 27, 2006, 01:16:51 PM »
H2A + OH- <-> HA- + H2O

HA- <-> H+ + A^2-


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Re: Titration - pH at equivalence point
« Reply #14 on: June 27, 2006, 03:03:48 PM »
H2A + OH- <-> HA- + H2O

OK

Quote
HA- <-> H+ + A^2-

It should be a neutralization reaction, not dissociation - we are talking about acid/base titration.

HA- + OH- <-> A2- + OH-

So, what salt is present in solution at first equivalence point?
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