Topic: ph calculation (Read 23862 times)

suzy · « **on:** March 26, 2005, 05:13:19 PM »

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 · « **Reply #1 on:** March 30, 2005, 10:43:29 PM »

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 · « **Reply #2 on:** March 31, 2005, 05:13:39 AM »

Quote

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 · « **Reply #3 on:** March 31, 2005, 07:22:54 AM »

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 · « **Reply #4 on:** March 31, 2005, 11:48:37 AM »

Quote from: AWK on March 31, 2005, 07:22:54 AM

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.

Quote

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 · « **Reply #5 on:** April 01, 2005, 05:06:41 AM »

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-)

Topic: ph calculation (Read 23862 times)

suzy

ph calculation

Demotivator

Re:ph calculation

Borek

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AWK

Re:ph calculation

Borek

Re:ph calculation

AWK

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