Buffer solution with H2PO4- HPO4--

[alberto]

Hi, everybody! I'm an Italian student of pharmaceutical chemistry, so I apologize for my English.
However, here is my problem:
Calculate the concentration (mol/L) of HPO4-- in a buffer solution (pH: 7.30), which is 0.2 mol/L in H3P04.
I know it comes out a solution of H2PO4- , but I cannot understand how to manipulate the H&H expression in order to solve the problem. Maybe there's a formula to get the pH of such a strange buffer solution, but, INCREDIBLE AS IT MAY APPEAR, it seems to be impossible to find (neither on the web nor on my book).

[H+]=6.32E-8*([H2PO4-]/[HPO4--])

7.11E-3=[H+][H2PO4-]/[H3PO4]=x^2/(0.2-x)

6.32E-8/7.11E-3=[H3PO4][HPO4--]/[H2PO4-]^2

This is everything I worked out.

PLEASE HELP ME 'COZ THIS PROBLEM IS MAKING ME GOING CRAZY!

[charco]

Bongiorno
(scusi il italiano)

Strikes me that to do a buffer calculation you need to know the concentration of the sodium phosphate (or whatever the negative ion is in the buffer) - you cannot solve for two unknowns in one equation

pKa = pH + log[A-]/[HA]

or does phosphoric acid self-buffer?

H3A = H2A- + H+
H2A- = HA(2-) + H+
HA(2-) = A(3-) + H+


No, it can't be a self buffer with a pH of 7.3 that's just silly - how could it (phosphoric acid) be basic?


Unless this is the value that the question is asking for - the value of the Na2HPO4 concentration... hmmm

ok then taking [H3A] as 0.2M
pH = 7.3 = -log(10)[H+]
therefore H+ ion concentration = 5.011exp-8

ka = [H+] x [H2A-]/[H3A]

but as ka is unknown and [H2A-] also then it can't be done without the pKa values

[alberto]

I found the solution this morning at lunch (glucose is always the best!). When posted my message, I made a mistake because I didn't read the text of the problem carefully.
In fact, I have just a solution (which is not a buffer one), that is [H3PO4] = 0.2 M.
Then I've to add HPO4-- in order to make the pH=7.30.

Here we are:

[H3PO4]=0.2 M

We have two different kind of reactions. The former is a neutralisation
H3PO4 + HPO4-- = 2H2PO4- => [H2PO4-]=0.4M and

[HPO4--]'=0.2M

The latter is the buffer solution

[H+] = Ka2 * [H2PO4-]/[HPO4--]

5.012E-8 = 6.32E-8 * 0.4/x => [HPO4--]'' = 0.504M

So, in the end, the total amount of HPO4-- required is

[HPO4--] = [HPO4--]' + [HPO4--]'' = 0.2 + 0.504 = 0.704M

[Borek]

Originally Posted by alberto

I found the solution this morning at lunch (glucose is always the best!).

[HPO4--] = [HPO4--]' + [HPO4--]'' = 0.2 + 0.504 = 0.704M

I am impressed

Best,
Borek