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Topic: Weak Diprotic Acid: maleic acid  (Read 42339 times)

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sundrops

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Weak Diprotic Acid: maleic acid
« on: October 27, 2005, 02:55:56 AM »
Maleic acid is a weak diprotic acid with a pKa1 = 1.87 and pKa2 = 6.07.

10mL 0.1M maleic acid is titrated with 0.1M NaOH.

Calculate the volume of NaOH required to reach each equivalence point, and calculate the pH at each equivalence point.


ok here's as far as I've got:

I've converted the pKa values to Ka values using Ka=10^-pKa
thus, Ka1 = 1.35E-2
        Ka2 = 8.51E-7

maleic acid which I will represent as H2A ionizes in a 2 step process.

H2A <-> H+ +HA-
HA- <-> H+ + A2-

then I did this:

Ka1 = 1.35E-2 = [H+][HA-] / [H2A]
Ka2 = 8.51E-7 = [H+][A2-] / [HA-]

and thats as far as I've got at the moment, a push in the right direction would be appreciated!  :)

« Last Edit: October 27, 2005, 02:58:01 AM by Mitch »

sundrops

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Re:Weak Diprotic Acid
« Reply #1 on: October 27, 2005, 02:58:41 AM »
oops - there's more!

i used M1V1=M2V2 to determine the volume of NaOH used in total,not i just need to determine the volume to reach the equivalence points, and then somehow the pH.  ::)

oh boy....

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Re:Weak Diprotic Acid: maleic acid
« Reply #2 on: October 27, 2005, 03:48:12 AM »
Write down stepwise neutralization reactions
H2A + NaOH =
NaHA + NaOH =
From above reactions calculate volume of NaOH, then recalculate concentrations

At the firts equivalence point you havs 0.05 M NaHA-.
At the second equivalence point you hav 0.03333 M Na2A
Calculate pH for hydrolysis of salts
AWK

sundrops

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Re:Weak Diprotic Acid: maleic acid
« Reply #3 on: October 27, 2005, 04:12:56 AM »
ok, so I have:

H2A + NaOH = NaHA + H2O
NaHA + NaOH = Na2A + H2O

net reaction is: H2A +2NaOH = Na2A +H2O


how did you figure out my concentrations at the equivalence points?

and how do I calculate the volume of NaOH using those equations?
just by M1V1=M2V2 again?

well I did that for Ka1 and found that I used 0.1L NaOH, I then used my ice table, and found that the pH is 1.98. Does this look ok so far? and these numbers are all for my first equivalence point right?

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Re:Weak Diprotic Acid: maleic acid
« Reply #4 on: October 27, 2005, 04:15:22 AM »
Once I clicked on the reply I have found that AWK already answered... So only one more link, for the calculation of first endpoint (pH of sodium hydrogen maleate):

http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt#12.11

Equation 12.11 allows calculation of pH of amphiprotic substance solution. Beware, it is simplified and not always can be used, but in this case the difference between correct result and simplified is only about 0.06 unit.
« Last Edit: October 27, 2005, 04:24:22 AM by Borek »
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Re:Weak Diprotic Acid: maleic acid
« Reply #5 on: October 27, 2005, 04:22:50 AM »
Here is your curve, no endpoints as you have to do them by yourself, but at least you should be able to estimate values.
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sundrops

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Re:Weak Diprotic Acid: maleic acid
« Reply #6 on: October 27, 2005, 04:39:02 AM »
wow, incredible. I read all the sections in my text on acids and bases and there weren't any equations like that in it - thats like MAGIC, makes this chem stuff simpler  ;D
I'm slowly destressing lol.

You guys are fantastic,

I'm going to try to work this out, but don't go anywhere - I may need you yet!  ;)

sundrops

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Re:Weak Diprotic Acid: maleic acid
« Reply #7 on: October 27, 2005, 04:57:35 AM »
for that equation, 12.11 that gives the overall pH - does it not?

how would i determine the pH at each equivalence point?





sorry for all the questions

sundrops

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Re:Weak Diprotic Acid: maleic acid
« Reply #8 on: October 27, 2005, 05:04:48 AM »
ok nvmd i read the post saying that that was the equation for equivalence point number one.  but how do u determine that? where did this magic equation come from? lol

and for equivalence point 2 would it just be pH = pKa1 + pKa2? because at equivalence point 1, all the H+ of the first dissociation are reacting but in the second equivalence point its ALL the H+ ions from both reactions reacting.

and then to determine the volume used to reach these equivalence points would it just be:

2*pKa1 = volume NaOH

2pKa2 / vol NaOH1 = vol NaOH2


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Re:Weak Diprotic Acid: maleic acid
« Reply #9 on: October 27, 2005, 05:41:52 AM »
where did this magic equation come from?

It is derived twice on the page, with different approaches. Note that it is very specific case!

Quote
and for equivalence point 2 would it just be pH = pKa1 + pKa2?

No. At the second end point you have solution of(CH2)2(COO)22- of known concentration - and this is a conjugated base. Use the same approach you used for calculation of pH of acetate, just be carefull when calculating pKb (pKa and pKb for polyprotic, conjugated acids and bases - scroll down the page).

Quote
and then to determine the volume used to reach these equivalence points would it just be:

2*pKa1 = volume NaOH

2pKa2 / vol NaOH1 = vol NaOH2

Ka values will never get into the calculation of volumes needed for titration. Volumes are about stoichiometry, Ka is about equilibrium
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sundrops

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Re:Weak Diprotic Acid: maleic acid
« Reply #10 on: October 27, 2005, 06:28:54 AM »
ok, so how do i go about determining how much NaOH I need?

sundrops

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Re:Weak Diprotic Acid: maleic acid
« Reply #11 on: October 27, 2005, 06:38:38 AM »
to reach the equivalence points I mean

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Re:Weak Diprotic Acid: maleic acid
« Reply #12 on: October 27, 2005, 07:01:24 AM »
You may use MODIFY buttom to not post twice.

Quote
ok, so how do i go about determining how much NaOH I need to reach the equivalence points?

Simple stoichiometry. Read AWK post.
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sundrops

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Re:Weak Diprotic Acid: maleic acid
« Reply #13 on: October 27, 2005, 07:52:03 AM »
i'm sorry, I didn't know that there was a modify button... :-\

I figured it out with your guys' help, thanks.


Last Question for the night:

you weigh out 0.0010 moles of a solid acid that has a molecular weight of 130g/mole.

Determine the volume of NaOH needed to titrate it if the acid is monoprotic and if it is diprotic.

Now I did it for the monoprotic one (easy peasy),
0.130g / 130g/mol = 0.001mol
o.oo1mol / o.1M = 0.01L NaOH

Now for the diprotic one. (I've decided that I hate diprotic acids  :P)
do I do the same but just multiply it by 2?
in this case;  2*(0.01L) = 0.02L NaOH

is that alright?



Thanks again everyone! ;D

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Re:Weak Diprotic Acid: maleic acid
« Reply #14 on: October 27, 2005, 08:08:35 AM »
Last Question for the night:

Night? It is 2 p.m. ;)

Quote
is that alright?

Seems so.
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