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Topic: pH calculation of buffer NaH2PO4 + Na2HPO4  (Read 66183 times)

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Offline Matt17

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pH calculation of buffer NaH2PO4 + Na2HPO4
« on: October 06, 2013, 05:45:05 AM »
Hello!
The last problem with pH. Solutions of NaH2PO4 and Na2HPO4 (both are 0,1 mol/dm3) are mixed together in ratio 1 : 1. Values given:
Ka(H3PO4) = 6*10-3
Ka(H2PO4- = 6*10-8
Ka(HPO42-) = 5*10-13

I tried to solve it in the following way:
Dissociation:
NaH2PO4  ::equil:: Na+ + H2PO4-
Na2HPO4  ::equil:: 2Na+ + HPO42-
And then, according to Bronsted-Lowry's theory:
H2PO4- + H2::equil:: HPO42- + H3O+

And now I think that I can calculate the concentration of H2PO4- and HPO42- after dissociation using ICE table, or maybe dissociation constants (though I don't know how to use them in this case), and than use it to calculate the concentraltion of H3O+ from Ka (H2PO4-).
I don't know if this is correct so please check me.
My mother tongue is not English so please forgive me my mistakes. I try to write as clearly as I can.

Offline Borek

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Re: pH calculation of buffer NaH2PO4 + Na2HPO4
« Reply #1 on: October 06, 2013, 05:50:25 AM »
NaH2PO4 and Na2HPO4 (both are 0,1 mol/dm3) are mixed together in ratio 1 : 1

Plug and chug with a Henderson-Hasselbalch equation. First example on the buffer calculation page that I already linked to. It is just a matter of selecting the correct pKa value (the one related to the acid and its conjugate base dominating the solution).
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Offline Matt17

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Re: pH calculation of buffer NaH2PO4 + Na2HPO4
« Reply #2 on: October 06, 2013, 07:22:08 AM »
So it seems that I can assume that:
[tex]\frac{HPO_4^{2-}}{H_2PO_4^-} = 1[/tex] can I?
Which means that
[H+] = Ka(H2PO4-)
Isn't it to easy?

Well, it seems OK because I read on the Internet that this buffer has always pH = 7,22 and from these calculations we also have pH = 7,22.
My mother tongue is not English so please forgive me my mistakes. I try to write as clearly as I can.

Offline Borek

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Re: pH calculation of buffer NaH2PO4 + Na2HPO4
« Reply #3 on: October 06, 2013, 07:40:07 AM »
OK. Checked with the pH calculator built into Buffer Maker:

http://www.youtube.com/watch?v=ReKHRo7I9x0

By mistake I used 1M solutions instead of 0.1M, but it doesn't change anything. pH equals pKa2 (value of pKa2 is necessarily identical to the one you were given, as it was taken from the program database). At the beginning I had to change the defaults so that the program sums the volumes of solutions mixed and ignores ionic strength of the solution.

It is application of the same general rule as in the case of KF/HCl solution you asked earlier - if the concentrations of acid and conjugate base are identical, pH=pKa.

Note - it also means pKa=pH at half titration!
« Last Edit: October 15, 2019, 05:57:06 PM by Borek »
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline Matt17

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Re: pH calculation of buffer NaH2PO4 + Na2HPO4
« Reply #4 on: October 06, 2013, 08:11:09 AM »
Thank you very much.
My mother tongue is not English so please forgive me my mistakes. I try to write as clearly as I can.

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