Enzyme-catalyzed 0.03 M Find the pH at end of reaction.

There is an enzyme-catalyzed reaction in a solution. It is buffered with 0.04 M PHOSPHATE, pH 7.4.
0.005 mole/liter (M) of acid is formed.

What is the pH at the END?

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Apparently I am supposed to use two H-H equations.

pKa values for phosphoric acid are: 2.16, 7.21, and 12.32

I am supposed to use 7.21, but don't know why.


Any help would be so greatly appreciated!!!!!

http://chemistry.about.com/od/acidsbases…

a buffer has its greatest capacity to maintain pH against acids & bases if it is 50% cojugate base & 50% cojugate acid....
at that point the pH = the pKa of the acid

since your pH of 7.4, is closest to the Ka2 of 7.21 ...
it would be prudent to make the the buffer with..
that acid "HA" which is (H2PO4)-1 & its conjugate base "A-" which is (HPO4)-2

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we need to first find the concentration oof each coljugate,
all we know is that they total 0.04 Molar

H-H for the original buffer

pH = pKa + log [A-] / [HA]
7.4 = 7.21 + log [A-]/[HA]

0.19 = log [A-]/[HA]
doing a 10^x for both sides:

1.549 = [A-] / [HA]
so
1.549 HA = [A-]

since we have a 0.04 M PHOSPHATE bffer
we know that HA + A- = 0.04
substituting in:
HA + (1.549 HA) = 0.04
2.549 HA = 0.04

*******HA = 0.0157 Molar (H2PO4)-1
&
********A- must be the rest: 0.243 Molar (HPO4)-2

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producing 0.005 mole/liter (M) of acid
shifts the equilibrium
(H2PO4)-2 <== to the left <== (HPO4)-2 & H+
as the conjugate base, removes the excess H+ ions formed

this decreases the (HPO4)-2 :
[0.0243 Molar (HPO4)-2 ] - (0.005 mole/liter ) = 0.0238 Molar (HPO4)-2
& the shift <== to the left <==
increases the (H2PO4)-1:
(0.0157 Molar (H2PO4)-1) + (0.005 mole/liter ) = 0.0162 M (H2PO4)-1


now the second H-H

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

pH = 7.21 + log [0.0238]/[0.0162]

pH = 7.21 + log 1.469

pH = 7.21 + 0.167

pH = 7.38