Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: refid on June 25, 2006, 06:53:09 PM
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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.
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Define an equivalence point.
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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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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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If its:
H2A + OH- <-> HA + H20
At equivance point:
HA + H20 <-> A- + H3O+
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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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yess sorry its diprotic
Acid + Base -> H20 and Salt?
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yess sorry its diprotic
That makes calculations a little bit harder.
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 (http://www.chembuddy.com/?left=pH-calculation&right=toc) - especially sections devoted to pH of salts calculation. Salts in general (http://www.chembuddy.com/?left=pH-calculation&right=pH-salt-solution) 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 (http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-simplified) section.
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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. ;)
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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
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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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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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First equivance pt H2A + 20H <-> A- + 2H20
That's already second equivalence point.
Charges are completely wrong.
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H2A + OH- <-> HA- + H2O
HA- <-> H+ + A^2-
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H2A + OH- <-> HA- + H2O
OK
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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HA-
Hydrolizes
HA- + H2O <-> H2A +OH-
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HA-
Right.
Hydrolizes
HA- + H2O <-> H2A +OH-
and dissociates!
pH of amphiprotic salt (http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt)
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still confused about this ... give the initial [] how can I figure out the [] at the equivalence point
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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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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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Ok got the pH and [Acid]
Show your calculations.
How to I find the volume required for the second equivalence poit?
Simple stoichiometry - answer is in the reaction equation.
pH second equivalence point is .5(ka1 + ka2) do this first?
???
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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
pH second equivalence point is .5(ka1 + ka2) do this first?
???
to solve the at the second equivalence point equatuion is this .5(ka1 + ka2) should i solve this first?
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10ml of .1M Diprotic Acid titrated with .1M NaOH pka1=1.87 pka2=6.07
Looks like maleic ;)
Vol. req. to reach eq pt 1 = 10ml
[HA-]=(.1M x .010L)/ 0.02L = 0.05M
So far so good.
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
pH second equivalence point is .5(ka1 + ka2) do this first?
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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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 [ ]?
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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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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 [ ]?
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pKa doesn't depend on the initial concentration, if that's what you are asking about.
Take a look at the Henderson-Hasselbalch equation (http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch#eq15.2). 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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no im saying given the pH at the equivalence point 1 ( monoprotic) how do I calculate the pKa (without using the graph).
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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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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
(http://img.sparknotes.com/figures/3/3a5994498f24d59f5d5d762b40844a2a/sasb.gif)
the pka would around 2-3 . and pkb = 14 - pka
If an base is titrated by a acid then
(http://www.chem.purdue.edu/gchelp/116exams/e3f005.gif)
pkb is 10.9 and pka = 14 - pkb
An polyprotic Acid titrating with NaOH
(http://www.bio.cmu.edu/courses/03231/LecF04/Lec03/PiTitrText.gif)
pkb1= 14 - pka1
pkb2= 14 - pka2
pkb3= 14 - pka3
is this correct?
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I am sorry but I am not able to understand what are you asking about, so I can't help you. Try to reword it.
Only thing I am sure about is that
pkb1= 14 - pka1
pkb2= 14 - pka2
pkb3= 14 - pka3
is wrong - check out these polyprotic dissociation constants definitions (http://www.chembuddy.com/?left=pH-calculation&right=polyprotic-dissociation-constants#eq3.10) (check out indexes).
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from the monoprotic acid titration curves is there relationship between the pka/pkb with pkw
like
14- pka = pkb
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This relathionship doesn't have anything to do with titration curve - it describes dissociation of acid and conjugated base. For example for acetic acid dissociation constant:
CH3COOH <-> H+ + CH3COO-
Ka = [H+][CH3COO-]/[CH3COOH]
conjugated base is acetate, and its basic dissociation looks like:
CH3COO- + H2O <-> CH3COOH + OH-
Kb = [CH3COOH][OH-]/[CH3COO-]
(water concentration assumed constant and omitted).
You may check that Ka*Kb = Kw, thus pKa + pKb = pKw (it is also described in the lecture on Bronsted-Lowry theory (http://www.chembuddy.com/?left=pH-calculation&right=bronsted-lowry-theory)).
So this is a general relationship - it holds for titration curve as well, but that's only a specific case, one of many.
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Oh I see, thanks for your help and clear up that matter for me :D