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Topic: Finding Ka1 and Ka2 for Malonic Acid  (Read 22365 times)

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Offline Donaldson Tan

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Finding Ka1 and Ka2 for Malonic Acid
« on: May 03, 2005, 09:53:40 PM »
I am having difficulty with the question below. I listed what I have done below. I searched online for the literature values of Ka for malonic acid. They are Ka1 = 1.149E-3 amd Ka2 = 2.03E-6

Quote
At 25C, the pH of 19.5g of malonic acid (HOOC-CH2-COOH) in 0.250L water is 1.47. After adding 12.5g of NaOH to this solution, the pH was 6.03. What are the values of Ka1 and Ka2?

Mr of Malonic Acid = 104g/mol
19.5g of Malonic Acid = 19/104 = 0.1857moles

moles of NaOH added = 12.5/40 = 0.3175
moles of acidic hydrogen available = 0.1857X2 = 0.3715moles

interpretation: 12.5g of NaOH neutralised all H+ from malonic acid, leaving the solution with CH2(COO)22- (propan-dioate) ions, which hydrolyse to produce hydroxide in the solution. However, the pH is 6.03 (<7). Where did I go wrong?

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Offline AWK

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Re:Finding Ka1 and Ka2 for Malonic Acid
« Reply #1 on: May 04, 2005, 02:03:13 AM »
Quote
interpretation: 12.5g of NaOH neutralised all H+ from malonic acid, leaving the solution with CH2(COO)22- (propan-dioate) ions, which hydrolyse to produce hydroxide in the solution. However, the pH is 6.03 (<7). Where did I go wrong?
This interpretation is not true.
0.3175 is not equal to 0.3715, and after addition of NaOH you should calculate pH of a buffer solution based on K2 of malonic acid.
« Last Edit: May 04, 2005, 02:03:51 AM by AWK »
AWK

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Re:Finding Ka1 and Ka2 for Malonic Acid
« Reply #2 on: May 04, 2005, 10:27:54 AM »
I am having difficulty with the question below. I listed what I have done below. I searched online for the literature values of Ka for malonic acid. They are Ka1 = 1.149E-3 amd Ka2 = 2.03E-6Mr of Malonic Acid = 104g/mol
19.5g of Malonic Acid = 19/104 = 0.1857moles

moles of NaOH added = 12.5/40 = 0.3175
moles of acidic hydrogen available = 0.1857X2 = 0.3715moles

interpretation: 12.5g of NaOH neutralised all H+ from malonic acid, leaving the solution with CH2(COO)22- (propan-dioate) ions, which hydrolyse to produce hydroxide in the solution. However, the pH is 6.03 (<7). Where did I go wrong?

Did the original question actually ask for the Ka2 also?  Here goes nothing...

You'll first need to find the equilbrium concentrations

From the pH, find the hydronium concentration, from this you can solve for Ka

Ka=(x)(x)/(initial-x)

The hydroxide will react with the acid directly, y=concentration of hydroxide, from this you can find the final pH due to the first Ka alone

Ka=(new H30+)(x+concentration Na0H)/(initial-x-conc. Na0H)

after finding the new H30+, find the pH.  Subtract this from 6.03 and from here you can deduce the Ka2.


Offline Borek

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Re:Finding Ka1 and Ka2 for Malonic Acid
« Reply #3 on: May 04, 2005, 11:48:31 AM »
The hydroxide will react with the acid directly, y=concentration of hydroxide, from this you can find the final pH due to the first Ka alone

You are plain wrong.

Ka1 is not enough to calculate pH of the solution when the second proton is at least partially neutralized. In this case it is neutralized in 2/3 so you should use HH equation for calculations. AWK have already wrote it.
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Offline Donaldson Tan

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Re:Finding Ka1 and Ka2 for Malonic Acid
« Reply #4 on: May 05, 2005, 10:29:24 AM »
i am not sure how to solve this problem. i was thinking of tackling the problem from the point after NaOH was added.

let malonic acid be H2X

i assume that pH after NaOH was added is due to hydrolysis of HX- to yield X2-. In another words, Ka2 governs the pH of 6.03.

initial concentration of H2X before dissociation = (19.5/104)/0.250 = 0.75

at eqbm,
total protonic hydrogen available = 0.0625moles
[ HX- ] = 0.0625/0.250 - x = 0.25 - x
[ H+ ] = x
[ X2- ] = a + x
-lg x = 6.03 => x = 9.333E-7
mole balance: [ X2- ] + [ HX- ] = 0.75 => a = 0.50
Ka2 = (9.333E-7)(0.50+9.333E-7)/(0.25 - 9.333E-7) = 1.87E-6
pKa2 = 5.729

*i corrected my substraction error and find it arrives at the same answer given by borek
« Last Edit: May 06, 2005, 05:22:32 PM by geodome »
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

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Re:Finding Ka1 and Ka2 for Malonic Acid
« Reply #5 on: May 05, 2005, 10:30:45 AM »
now let's not get too hasty here.  I was performing an estimation, and I believe this would amount to an adequate approximation.  I don't believe that we can use the first part of the information given in the problem to calculate the second Ka, not enough information is given and the second aspect of the question certainly does not supplement the first.  If it can be done, it'll require some kind of mathematical measure.

Borek, borek, borek...you and HH....you seemed to be obssessed with it w/0 knowing its derivation.  It's for buffer solutions, and it is based on the assumption that the conjugate in the form of its salt is added upon the original acid.  This is why we can assuming that neither the conjugate or the acid, dissociate...or react with water...it's fundamental general chemistry and you should be familiar with it.  The problem isn't even buffer related.


I thought was that the final pH 6.03 would not have been adequately derived from the change in pH due to the first Ka, that is one could have subtracted the former acid concentration given by the first Ka, and the rest would have been due to the influence of the second acid.

Geodome, could you clarify the solution to the problem?  I would like to see it.

Offline Donaldson Tan

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Re:Finding Ka1 and Ka2 for Malonic Acid
« Reply #6 on: May 05, 2005, 10:54:43 AM »
*0.03 in the ICE table attached should be 0.03388 or 10-1.47
*changes in H+ in the ice table should A+2B

A+B = 0.03388
Ka2 = 2.37E-6
Ka2 = [H+]B/A => B/A = 5.509E-4 => B = (5.509E-4)A
A + (5.509E-4)A = 0.03388
A(1 + 5.509E-4) = 0.03388
A = 0.03386
B = 0.00002

Ka1 =(0.03388)(0.03386)/(0.75-0.03388) = 1.60E-3
pKa1 = 2.795

*i had corrected my workings. made a substraction error to find the amount of protonic hydrogen available.
« Last Edit: May 06, 2005, 05:29:19 PM by geodome »
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

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Re:Finding Ka1 and Ka2 for Malonic Acid
« Reply #7 on: May 05, 2005, 10:58:16 AM »
call me a dumb ass...allright, I've taken a wrong approach to this problem.  After the complete reaction with sodium hydroxide there will be .0539 moles of the secondary acid left.  Thus the secondary acid is the sole contributor of the pH, 6.03.

(2xmoles of acid)-moles of 0H-=.0539moles of secondary acid remaining

You should understand why the 2 is there.  I think this is where your problem may lie.

This should be perfectly correct.  geodome, where did you get those Ka values?

From here, you can go onto easily deduce the first Ka.


« Last Edit: May 05, 2005, 10:59:44 AM by GCT »

Offline Donaldson Tan

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Re:Finding Ka1 and Ka2 for Malonic Acid
« Reply #8 on: May 05, 2005, 11:02:30 AM »
This should be perfectly correct.  geodome, where did you get those Ka values?

I googled the Ka values and 'nick' them from http://people.ouc.bc.ca/kperry/125web/125inclanswers/125eqxprob.doc

I just realised i made another mistake. the changes for H+ in the ice table should be A + 2B, but it still arrive at the same answer for Ka1 because B is sooo.. small.
« Last Edit: May 05, 2005, 11:07:39 AM by geodome »
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

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Re:Finding Ka1 and Ka2 for Malonic Acid
« Reply #9 on: May 05, 2005, 11:20:45 AM »
I imagine that this problem is now pretty simple.  You'll need to work backwards in a sense.

You can find the second Ka2, since the 6.03 is entirely due to the second Ka2.

Since K is simply related to dissociation in this case and the due to the symmetry of the situation I think what you may be able to do is to solve for the overall Ka, and then dividing by Ka2.  Do you get what I'm saying?

Offline Borek

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Re:Finding Ka1 and Ka2 for Malonic Acid
« Reply #10 on: May 05, 2005, 08:02:48 PM »
the pH of 19.5g of malonic acid (HOOC-CH2-COOH) in 0.250L water is 1.47. After adding 12.5g of NaOH to this solution, the pH was 6.03. What are the values of Ka1 and Ka2?

I am not in a mood for looking for your errors, however here is a PROPER solution that gives PROPER results, regardles of GCT comments. H2Ma stands for malonic acid, I am assuming analytical concentrations 0.75M of H2Ma and 1.25M for NaOH:

1.
pH = 1.47, [H+] = 0.034M

Only first step of dissociation is important:
H2Ma <=> H+ + HMa-

thus
[H2Ma] = 0.75 - [H+]

and
[HMa-] = [H+]

Ka1 = [H+][HMa-]/[H2Ma]

substituting we get

Ka1 = [H+]^2/(0.75-[H+]) = 0.0016

and pKa1 = 2.79

2.
pH = 6.03

1.25M NaOH means that in the solution there are Ma(2-) and HMa- only. Simple stoichiometry gives
[HMa-] = 0.25
and
[Ma(2-)] = 0.5

HH equation:
pH = pKa2 + log([Ma(2-)]/[HMa-])

rearranging:
pKa2 = pH + log([HMa-]/[Ma(2-)])

using calculator:
pKa2 = 6.03 + log(0.25/0.5) = 5.73

To not look further pKa1/2 for malonic acid taken from BATE database are 2.847 and 5.696. That's almost perfect fit.
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Offline Borek

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Re:Finding Ka1 and Ka2 for Malonic Acid
« Reply #11 on: May 05, 2005, 08:06:04 PM »
Ka1 =(0.03388)(0.03387)/(0.75-0.03388) = 1.64E-3

Ka1 deviates from the literature value by 44%. Sigh.. where did my workings go wrong?

It doesn't deviate! 0.0016 is a right result.
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Re:Finding Ka1 and Ka2 for Malonic Acid
« Reply #12 on: May 06, 2005, 02:58:10 PM »
I googled the Ka values and 'nick' them from http://people.ouc.bc.ca/kperry/125web/125inclanswers/125eqxprob.doc

I just realised i made another mistake. the changes for H+ in the ice table should be A + 2B, but it still arrive at the same answer for Ka1 because B is sooo.. small.

actually, instead of the ice tables, have you considered solving the rest of the problem through using the percent dissociation of second acid proton?

where percent dissociation ratio of the second proton=

[H30+, second acid]/[H30+, first acid]

since [H30+, first acid]=initial concentration of acid in the second dissociation


Ka, general (for this problem)=[percent dissociation][percent dissociation]/[1-percent dissociation]

Offline Donaldson Tan

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Re:Finding Ka1 and Ka2 for Malonic Acid
« Reply #13 on: May 06, 2005, 05:35:37 PM »
i made a substraction error to find the protonic hydrogen available.

protonic hydrogen available
= 2 x intial moles of malonic acid - moles of NaOH added
= 2 x 19.5/104 - 12.5/40
= 0.0625moles

protonic hydrogen available per L of solution
= 0.0625/0.250
= 0.25 moles/L

after i corrected my substraction error, i realise i arrive at the same answer as Borek. You may make reference to the corrected posts above.

BTW I think Borek's method is elegant.
« Last Edit: May 06, 2005, 05:36:02 PM by geodome »
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Offline Borek

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Re:Finding Ka1 and Ka2 for Malonic Acid
« Reply #14 on: May 06, 2005, 06:47:12 PM »
BTW I think Borek's method is elegant.

You are welcome :)
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