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### Topic: Problem of the week - 08/10/2012  (Read 13062 times)

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#### Borek ##### Problem of the week - 08/10/2012
« on: October 08, 2012, 05:21:07 AM »
0.01M solution of NaHA salt has a pH of 6.00. What is pH of the 0.01M H2A solution, if acid H2A is 10000 times stronger than HA-?
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#### Rutherford ##### Re: Problem of the week - 08/10/2012
« Reply #1 on: October 08, 2012, 11:58:20 AM »
Got the answer, but maybe it is too soon to write it (only Monday) so I can leave it open for other members.

#### CrazyAssasin

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• Mole Snacks: +4/-0 ##### Re: Problem of the week - 08/10/2012
« Reply #2 on: October 08, 2012, 01:02:32 PM »
HA- + H2O → H3O+ + A-2
[H+] = 10-6mol/L
Ka2=([H^+ ][A^(-2)])/([〖HA〗^-])=〖(〖10〗^(-6))〗^2/(0,01-〖10〗^(-6) )≈〖10〗^(-10)
Ka1= 10000Ka2=〖10〗^(-6)
Ka1=([H^+ ][HA^-])/([H_2 A])=x^2/(0,01-x)≈x^2/0,01
[H+]=√(〖10〗^(-2)∙〖10〗^(-6) )=〖10〗^(-4)mol/L
pH=-log_10⁡〖[H^+ ]=〗4

#### XGen

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• Mole Snacks: +9/-4 ##### Re: Problem of the week - 08/10/2012
« Reply #3 on: October 09, 2012, 01:34:46 PM »
I'm not sure if CrazyAssassin accounted for the fact that another reaction occurs. Along with the dissociation of HA- to H+ and A2-, it also "reassociates" to H2A and OH-. This increases the pH.

Here is my (most likely incorrect) attempt:

Using the formula pH = (1/2)(pKa1 + pKa2), we can write the following two equations:
pKa1 + 4 = pKa2
6 = (1/2)(pKa1 + pKa2)

Solving the system of equations, we get:
pKa1 = 4
pKa2 = 8

Using this to calculate pH, 0.01M of H2A is 3 (the second dissociation is negligible).

#### Rutherford ##### Re: Problem of the week - 08/10/2012
« Reply #4 on: October 09, 2012, 01:42:15 PM »
Why "most likely incorrect"? I got the same.

#### Borek ##### Re: Problem of the week - 08/10/2012
« Reply #5 on: October 09, 2012, 05:41:50 PM »
pKa1 + 4 = pKa2
6 = (1/2)(pKa1 + pKa2)

Perfect! With these two equations finding the solution is a breeze.

And the correct answer is 3.02 (but 3 is good enough ).
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#### mikyustanov

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• Mole Snacks: +0/-0 ##### Re: Problem of the week - 08/10/2012
« Reply #6 on: October 10, 2012, 07:26:39 PM »
I got a 9.9×10^(-11) Ka for HA-

And the pH is 3.00, neglecting HA- the second time.

#### Borek ##### Re: Problem of the week - 08/10/2012
« Reply #7 on: October 11, 2012, 04:08:55 AM »
I got a 9.9×10^(-11) Ka for HA-

No, correct values are

pKa1 = 4.00

and

pKa2 = 8.00
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#### Big-Daddy

• Sr. Member
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• Mole Snacks: +28/-94 ##### Re: Problem of the week - 08/10/2012
« Reply #8 on: November 09, 2012, 04:09:05 PM »
Using the formula pH = (1/2)(pKa1 + pKa2), we can write the following two equations:

I've never seen this formula - where did you get it/can you get other similar formulae and what does it apply to? For example I do not imagine it will solve directly for the pH of a weak diprotic acid - so does it refer only to amphiprotic salts like NaHA or to what? If you can explain the formula that would be great.

#### Borek ##### Re: Problem of the week - 08/10/2012
« Reply #9 on: November 09, 2012, 04:10:11 PM »
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#### Big-Daddy

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• Mole Snacks: +28/-94 ##### Re: Problem of the week - 08/10/2012
« Reply #10 on: November 10, 2012, 06:57:16 AM »
http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt

I see - would the pH Calculator not be capable of solving the same sum? (I assume that is how you found 3.02 since the pH=(1/2)(pKa1+pKa2) formula finds exactly 4 and 8 for the pKa values, but the page itself does not produce the final expression for H+ alone without approximations so I don't know if the calculator can do it. ... or did you stick with approximations for the amphiprotic (in finding pKa1 and pKa2) and run the exact calculation for diprotic acid with those two dissociation constants?)

#### Borek ##### Re: Problem of the week - 08/10/2012
« Reply #11 on: November 11, 2012, 04:25:46 PM »
BATE does the calculations using each time exactly the same full approach (each solution is treated as a mixture of acid and base of known Ca & Cb concentrations, we discussed it to the death in the past), so the answer is always accurate. There is no need (and it makes no sense) to use different models for each problem.

3.02 is an exact answer, found using exact numerical method. Not something you can do manually, but perfectly fit for computer model.

3 is an approximate answer, found using approximate methods - suitable for manual calculations. Apparently it is good enough.
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#### Big-Daddy

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• Mole Snacks: +28/-94 ##### Re: Problem of the week - 08/10/2012
« Reply #12 on: November 11, 2012, 04:51:09 PM »
BATE does the calculations using each time exactly the same full approach (each solution is treated as a mixture of acid and base of known Ca & Cb concentrations, we discussed it to the death in the past), so the answer is always accurate. There is no need (and it makes no sense) to use different models for each problem.

3.02 is an exact answer, found using exact numerical method. Not something you can do manually, but perfectly fit for computer model.

3 is an approximate answer, found using approximate methods - suitable for manual calculations. Apparently it is good enough.

Yes I know, we have discussed this at length. So the calculator carries forward the exact method for solving for the amphiprotic acid itself? After all there is no option to plug in "Ka1" and "10000Ka1" as your values, you actually have to put in numbers.

#### Borek ##### Re: Problem of the week - 08/10/2012
« Reply #13 on: November 11, 2012, 05:05:42 PM »
3.02 is an exact pH of the 0.01M H2A solution, after Ka1 and Ka2 were calculated using approximate method. However, it can be easily checked that these approximately found values fit pH 6.00 and Ka1=10000Ka2 as described in the question.
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#### Big-Daddy

• Sr. Member
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• Mole Snacks: +28/-94 ##### Re: Problem of the week - 08/10/2012
« Reply #14 on: November 12, 2012, 02:27:16 PM »
3.02 is an exact pH of the 0.01M H2A solution, after Ka1 and Ka2 were calculated using approximate method. However, it can be easily checked that these approximately found values fit pH 6.00 and Ka1=10000Ka2 as described in the question.

Ah I see - yes because for pH to be exactly 6 and Ka1 to be exactly 10000 Ka2 then these two values have to match exactly as well ... but had you used a different pH e.g. 5.9 then you would need to use the exact method to find Ka1 and Ka2 for the salt. I presume this could be done by simply finding the general formula for the salt with Ka1 and Ka2, replacing Ka1 with 10000Ka2 and then rearranging for Ka2, after which the rest is easy to find. I don't know if the pH calculator equation would help to find this general formula though? Or if I have to do the rearranging by hand ...