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Offline Anthasci

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Buffer system calculations
« on: June 11, 2012, 08:00:12 PM »
Greetings everyone,

I am currently struggling trying to understand buffer systems and more generally speaking, neutralising titrations of all varieties. Specifically, I came across this problem;

1.) Determine the pH of buffer solutions obtained  by mixing 50,00 mL of 0,200M NaH2PO4 with;
a) 50,00 mL of 0,120M HCl (1,97)
b) 50,00 mL of 1,120M NaOH (7,37)
K1=7,11*10-3
K2=6,34 *10-8
K3=4,2*10-13

for a), my idea was to write the following equation:
NaH2PO4 + HCl  :rarrow: H3PO4 + NaCl
which to my knowledge, can be regarded as a one-way?
Next, I wrote:
H3PO4 + H2O  ::equil:: H2PO4- + H3O+

So, NaH2PO4 and HCl react at a molar ratio of 1:1, which is 0,006 mol (some sodium dihydrogenphosphate is left out), resulting in a 0,06M solution of H3PO4 (total volume is now 100 mL)

Using the K1, I figured out that in the second equation, [H3O+] equals 0,017403, which in turn, results in the pH of 1,76. Did I go wrong somewhere or are the solutions wrong (written above)? Also, I'm guessing b) has pretty much the same idea? I got 6,68 for b) but I think this is wrong, since I have a strong feeling the system would have a basic pH.

Thanks!
« Last Edit: June 11, 2012, 08:10:15 PM by Anthasci »

Offline Borek

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Re: Buffer system calculations
« Reply #1 on: June 12, 2012, 04:00:43 AM »
It is not clear to me how you got your answers. You are right assuming H2PO4- reacts with added strong acid and strong base and trying to calculate concentrations of HPO42- and H3PO4 (but look below), but it is not clear to me what you do later. Common approach to the problem would be to put these concentrations into the dissociation constant definition - Ka is known, concentrations of acid and conjugate base are known, the only thing unknown is [H+]. Dissociation constant definition can be rearranged to the so called Henderson-Hasselbalch equation, which is very convenient for this type of calculations.

Unfortunately, this approach won't work for a, and answer given for a is wrong (pH is not 1.97, more like 2.11). H3PO4 is simply a little bit too strong acid for the assumption that H2PO4- gets quantitatively protonated by HCl, and finding the answer requires more complicated calculations (not THAT complicated - ICE table will do).

Answer given for b is correct, can't comment on your result as I have no idea how you got it.


Edit: I think I know what you did - you calculated pH using concentration of protonated H3PO4 only. You can't - solution already contained conjugate base, when you calculate pH starting from known concentration of an acid you assume initial concentration of the conjugate base to be zero.
« Last Edit: June 12, 2012, 04:31:57 AM by Borek »
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Offline AWK

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Re: Buffer system calculations
« Reply #2 on: June 12, 2012, 05:24:41 AM »
Quote
b) 50,00 mL of 1,120M NaOH (7,37)
? should be 0.120 instead 1.120 ?
Think about stoichiometry of neutralization in both cases. (both suggested answers are correct)
AWK

Offline Borek

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Re: Buffer system calculations
« Reply #3 on: June 12, 2012, 05:38:47 AM »
(both suggested answers are correct)

1.97 is incorrect. You can get it applying blindly HH equation, but it doesn't make it a correct answer.
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Offline Anthasci

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Re: Buffer system calculations
« Reply #4 on: June 12, 2012, 11:13:22 AM »
Borek - I don't think I understand. For a), you say that the concentrations of acid and conjugate base are known, but which numbers are they, here? Was I correct in saying that the H3PO4 is 0,06M? So, since 0,004 moles of NaH2PO4 are left over, does this mean it's concentration is now

[itex] \frac {0,004 mol}{0,1 L} [/itex]= 0,04M (A-)?

Then I'd go [H3O+]=Ka1[itex] \frac {0,06}{0,04} [/itex]

I understand this is a simplyfied solution, and probably what the prof. had in mind for us. Alternatively, what would be the correct solution?

Then I went over to b) and got lost - I see that from the molar ratio, some NaOH would be left unreacted - does this mean it dissolves in water and gives me a certain concentration of [OH-]? Calculating the [H3O+] from the dissociation of the acid, I can then find out how much [OH-] is left, since the reaction that takes place is:

OH- + H3O+  :rarrow: 2H2O






Offline Borek

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Re: Buffer system calculations
« Reply #5 on: June 12, 2012, 12:00:05 PM »
Then I'd go [H3O+]=Ka1[itex] \frac {0,06}{0,04} [/itex]

What you did to calculate concentrations of H3PO4 and H2PO4- was OK and would work for a slightly weaker acid with pKa of about 3 or higher.

Quote
I understand this is a simplyfied solution, and probably what the prof. had in mind for us. Alternatively, what would be the correct solution?

Note that assuming all phosphoric acid is in the form H3PO4 you are assuming there is no H+ in the solution, as it was all used to protonate the acid. But you just calculated that pH is 1.97, so you know phosphoric acid concentration is not 0.06 - which in turn means pH is not 1.97. For weaker acids this difference in insignificant, here is gives a relatively large error.

It can be relatively easy dealt with using ICE method.

     H3PO4   ::equil::   H+    +    H2PO4-

I     0.06            0            0.04
C      -x             +x           +x
E     0.06-x         x            0.04+x

[tex]K_{a1} = \frac{x(0.04+x)} {0.06-x}[/tex]

Solve for x - that's the H+ concentration, and it yields the correct pH.

Quote
Then I went over to b) and got lost - I see that from the molar ratio, some NaOH would be left unreacted - does this mean it dissolves in water and gives me a certain concentration of [OH-]?

I think you have a typo in your data and it should be 0.120, not 1.120 (otherwise you will not get 7.37 as the correct answer). But if 1.120 is correct, it means it is excess NaON that is responsible for the pH (especially when the excess is large).
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Offline Anthasci

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Re: Buffer system calculations
« Reply #6 on: June 12, 2012, 12:45:38 PM »
Thanks for the clarification!

Yeah, I suppose it makes more sense that cNaOH=0,120M

Though I am confused ... do I still keep working with the Ka in the b) problem, or Kb, since the pH > 7? I assume all of NaOH is used in the reaction

NaH2PO4 + NaOH :rarrow: Na2HPO4 + H2O

leaving me with a 0,06M solution of Na2HPO4 and a 0,04M NaH2PO4, since 0,004 moles out of 0,01 moles don't react and the total volume is 100 mL.

Using the same approach as before, based on the [H3O+], I get a pH of a little over 12.

Offline Borek

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Re: Buffer system calculations
« Reply #7 on: June 12, 2012, 12:53:56 PM »
Yes, you still work with pKa.

Show how you got 12, concentrations are OK but 12 is not.
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Offline Anthasci

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Re: Buffer system calculations
« Reply #8 on: June 12, 2012, 01:05:03 PM »
Ok, so this is the first equation:

NaH2PO4 + NaOH :rarrow: Na2HPO4 + H2O

Then:

HPO42- + H2O  ::equil:: PO43- + H3O+

The Ka3 is 4,2*10-13

cHPO42- = 0,06M
cPO43- = 0,04M

Actually, I'm not really sure whether to use Ka2 or Ka3 here

or does the leftover NaH2PO4 also dissolve in water or something like that?

EDIT: Solved it.

H2PO4-  ::equil:: H+ + HPO42-

knowing 0,04M H2PO4- and 0,06M HPO42-, I just solved it using ICE. Nice. Thank you very much for your aid.

Offline Borek

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Re: Buffer system calculations
« Reply #9 on: June 12, 2012, 01:56:29 PM »
Ka2, you have H2PO4- and HPO42- in the solution, this is the second dissociation equilibrium.

[tex]pH = pK_{a2} + log\frac{[HPO_4^{2-}]}{[H_2PO_4^-]}= pK_{a2} + log\frac{0.06}{0.04} = 7.37[/tex]
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Offline Anthasci

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Re: Buffer system calculations
« Reply #10 on: June 14, 2012, 07:58:32 PM »
Ok, so technically speaking, this following question is related to general equilibrium and solubility product, but I figured I'd just throw it in my existing topic.

5. Find the solubility of ZnC2O4 (g/L) at pH = 3
ZnC2O4  :rarrow: Zn2+ + C2O42-
Ksp=7,5*10-9
Kb1=1,8*10-10
Kb2=1,8*10-13

If I'm correct, s = [Zn2+] + [C2O42-]

Obviously the pH is acidic and I'm confused why I'm given the Kb. [itex] \frac {Kw}{Kb_1} = Ka_2 [/itex] and the same goes for Ka1 - these values seem to correspond with the pKa values for oxalic acid, so I figured it had something to do with it. Then I improvised the equations to get what I wanted (doubtedly correct):

a) C2O42- + H3O+  ::equil:: HC2O4- + H2O
b) HC2O4- + H3O+ ::equil:: H2C2O4 + H2O

Where for a), K corresponds with Ka1-1, same goes for b) and Ka2

I then arrived at the expression:

Ka1*Ka2*[H2C2O4] = [H3O+]2[C2O42-]

But I'm confused what [H2C2O4] equals? Since now I got 2 unknowns, including the [C2O42-], which I guess I'm trying to calculate.

Offline AWK

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Re: Buffer system calculations
« Reply #11 on: June 15, 2012, 01:27:44 AM »
Quote
If I'm correct, s = [Zn2+] + [C2O42-]
s = [Zn2+] = [C2O42-] + [HC2O4-] + [H2C2O4]
AWK

Offline Borek

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Re: Buffer system calculations
« Reply #12 on: June 15, 2012, 03:54:32 AM »
http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-acid-base

See equation 9.13 and think how it can be applied to the system.

And in future please don't hijack threads, start a new one.
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Offline Anthasci

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Re: Buffer system calculations
« Reply #13 on: June 15, 2012, 06:29:08 AM »
Ca will equal [itex] \frac {K_(sp)}{[C_2O_4^{2-}]} [/itex] and [A2-] = [C2O42-], right?

I'll keep that in mind in the future, sorry

Offline Borek

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Re: Buffer system calculations
« Reply #14 on: June 15, 2012, 06:44:04 AM »
Ca will equal [itex] \frac {K_(sp)}{[C_2O_4^{2-}]} [/itex]

No. Ca is the analytical concentration of the acid, so it is a sum of concentrations of all forms. See AWK post for a hint.

Quote
and [A2-] = [C2O42-], right?

Yes.

Beware: page I linked to uses overall dissociation constants, not stepwise ones.
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