[suzy]

problem :

How do I have to calculate the pH of a  mixture of 10 mL 0,1M NH4H2PO4 and 10 mL 0,2M K3PO4
ammoniumion:
pKa = 9.25 ;
fosforic acid:
pKA1= 2.15 ;
pKA2 =7.20 ;
pKA3 = 12.38

solution: my trial

I first calculated the equivalents:
for NH4H2PO4 = 10*0.1 = 1mmol ;
for K3PO4 = 10*0.2 = 2mmol ;



reactions H3PO4 -> H ( +) + H2PO4(-) pKa1 = 2.15 thus pKb1 = 11.85   ;                                      
              H2PO4(-) -> H ( +) + HPO4(2-) pKa2 = 7.20 thus pKb2 = 6.80  ;
              HPO4(2-) -> H ( +) + PO4(3-) pKa3 = 12.38 thus pKb3 = 1.62  ;

              NH4H2PO4 -> NH4( +) + H2PO4(-) K3PO4 -> 3K( +) + PO4(3-) ;

Then I made react both reactants :
...NH4H2PO4 + PO4(3-) -> NH4 HPO4(2-) + HPO4(2-)
pKa....9,25....7,20..-..9,25....-....12,38
pKb.. -..11,85...1,62....-..11,85..-...  

I think that this is a buffer, but I don't know what to do with the 1mmol NH4?
So, I didn't take note of this concentration and I calculated the pH with the equation of Henderson-Hasselbach :  
pH = pKa + log (Base)/(Acid)
How can I find the concentrations of base and acid in this reaction?
I thought that it was 1mmol H2PO4(-)  and 2mmol PO4(3-) both as a base.   How can I find from those equivalents  the concentration of an acid and put them in the equation of Henderson-Hasselbach?  
Can someone help me, please?  
Thank you,
Suzy

[Demotivator]

I wouldn't use hendersson ...equation on a mixed system of various pKas. Well, here's my attempt.

The problem poses an interesting question of whether NH4+ equlibrium plays a role in the PH.  It turns out it does not if it is assumed that [H2PO4] and PO4 remain close to their original values, at equilibrium (a common assumption in buffer cases). First, I'll include NH4 in the calculations:

The following 3 equilibria are simultaneous with the same value for [H+]
[H+] = K[NH4]/[NH3]   (eq 1)

[H+]= K2[H2PO4]/[HPO4]   (eq 2)

[H+] = K3[HPO4]/[PO4]    (eq 3)

multiplying together yields:
[H+]3 = KK2K3[NH4][H2PO4]/[NH3][PO4]   (eq 4)

We know H2PO4 = .05M and PO4 = .1M and assume that the actual values are close to those.
Now, the problem is to determine a value for NH4/NH3.  Square eq 1  and multiply eq 2 by eq 3:
[H+]2 = K^2[NH4/NH3]^2
[H+]2 = K2K3[H2PO4]/[PO4]  (the goal of cancelling HPO4 out is achieved)
then divide the above two to yield:

1 = K^2[NH4/NH3]^2[PO4]/K2K3[H2PO4]
hence,
[NH4/NH3]^2 = K2K3[H2PO4]/K^2[PO4] = 6.3x10^-8* 4.2x10^-13* 0.05/(5.6x10^-10)^2 * 0.1
= .042
[NH4/NH3] = .205
Plugging this value and the others into the equation 4:
[H+]3 = 5.6x10^-10* 6.3x10^-8* 4.2x10^-13* .205* .05/.1
[H+]3 = 1.52x10^-30
[H+] = 1.15x10^-10
PH = 9.94

It turns out that whatever K is, it's effect is nullified by NH4/NH3, ie. larger K, smaller NH4/NH3, as long as it is in keeping with the assumption.

To illustrate, I exclude NH4+ and work only with eq 2 multiplied by eq 3. I wind up with the same result:
[H+]2 = K2K3[H2PO4]/[PO4]
[H+]2 = 6.3x10^-8* 4.2x10^-13* .05/.1
      = 1.32x10^-20
[H+] = 1.15x10^-10
pH = 9.94

[Borek]

if it is assumed that [H2PO4] and PO4 remain close to their original values, at equilibrium

It doesn't hold:

H3PO4        3.099e-013  (p =  12.51)
H2PO4^-    7.171e-005  (p =   4.14)
HPO4^2^-  1.479e-001  (p =   0.83)
PO4^3^-    2.016e-003  (p =   2.70)
H^+           3.059e-011  (p =  10.51)
OH^-         3.269e-004  (p =   3.49)
NH4+         2.581e-003  (p =   2.59)
NH3           4.742e-002  (p =   1.32)
K+             2.997e-001  (p =   0.52)
KOH           3.100e-004  (p =   3.51)

These are results of numerical attack. It seems hydrolisis is to strong.

pKb for KOH is 0.5.

[AWK]

There is an error of one order for concentration in Borek calculation (should be 0.05 and 0.1 approximately for concentration of H2PO4(-) and PO4(3-), respectively). Moreover, from the information on program he used, the activity, not concentrations, were used in his calculations, and this may give substantial differences of calculations with using multicharged ions.

[Borek]

There is an error of one order for concentration in Borek calculation (should be 0.05 and 0.1 approximately for concentration of H2PO4(-) and PO4(3-), respectively).

[PO4(3-)] + [HPO4(2-)] + [H2PO4(-)] + [H3PO4] = 0.15

I think it is consistent with the question posted. Correct me if I am wrong.

Moreover, from the information on program he used, the activity, not concentrations, were used in his calculations, and this may give substantial differences of calculations with using multicharged ions.

I haven't posted any information about the program I am using so you are just guessing. And you are wrong.

Results shown were for concentration calculations. Here are results for activities (substance formula, substance concentrations, activity coefficient, substance activity):

H3PO4         2.692e-013 * 1.000 =  2.692e-013 (p =  12.57)
H2PO4^-     3.614e-005 * 0.559 =  2.019e-005 (p =   4.69)
HPO4^2^-   1.386e-001 * 0.097 =  1.349e-002 (p =   1.87)
PO4^3^-     1.135e-002 * 0.005 =  6.005e-005 (p =   4.22)
H^+           1.690e-010 * 0.559 =  9.442e-011 (p =  10.02)
OH^-          1.896e-004 * 0.559 =  1.059e-004 (p =   3.98)
NH4+          1.156e-002 * 0.559 =  6.458e-003 (p =   2.19)
NH3            3.844e-002 * 1.000 =  3.844e-002 (p =   1.42)
K+              2.999e-001 * 0.559 =  1.675e-001 (p =   0.78)
KOH            5.615e-005 * 1.000 =  5.615e-005 (p =   4.25)

Ionic strength is 0.968 so it is too high for a Debye-Huckel theory anyway.

[AWK]

Sorry, fast thinking is worse than fast food. Borek is completely right.
But I think, it is quite sufficient to calculate hydrolysis of 0.15 M HPO4(2-)