Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: sundrops 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! :)
-
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....
-
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
-
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?
-
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 (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.
-
Here is your curve, no endpoints as you have to do them by yourself, but at least you should be able to estimate values.
-
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! ;)
-
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
-
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
-
where did this magic equation come from?
It is derived twice on the page, with different approaches. Note that it is very specific case!
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 (http://www.chembuddy.com/?left=pH-calculation&right=polyprotic-dissociation-constants) - scroll down the page).
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
-
ok, so how do i go about determining how much NaOH I need?
-
to reach the equivalence points I mean
-
You may use MODIFY buttom to not post twice.
ok, so how do i go about determining how much NaOH I need to reach the equivalence points?
Simple stoichiometry. Read AWK post.
-
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
-
Last Question for the night:
Night? It is 2 p.m. ;)
is that alright?
Seems so.
-
yes night, lol, here's it is now morning actually.... around 9:15am lol
thanks for everything Borek. :D
-
In order to find the concentration you need to take the molarity of what you are titrating with:
.1M NaOH
set equal to
x mol/.02
Then mol to mol ratio
1/2
Then mols of Maleic over L
multiply them all and that will be your concentration