How Do You Prepare a 40 L Phosphate Buffer with pH 6.9?

[Eshi]

Homework Statement

Describe the preparation of 40 L of 0.0500M phosphate buffer, pH 6.9 starting with 1M solutions of KH2PO4 and K2HPO4. pKa= 7.2

Homework Equations

pH= pKa + log[A-]/[HA]3. The Attempt at a Solution [/b

6.9=7.2 + log[A-]/[HA]
-0.3= log[A-]/[HA]
[A-]/[HA]= 0.50

0.5[HA]= [A-]
0.5[HA]+1[A-] = 1.5

0.5/1.5 = 0.33 = 33% [HA]0.05 mol/L x 40L = 2 mol HA
2 mol x 1L/mol = 2L HA2 mol x 0.33= 0.66 mol A-
0.66 mol x 1L/mol = 0.66 L A-2L HA + 0.66L A- = 2.66 <---Edit: 2.66 L of acid/base

40L-2.66L= add 37.34 L water to make the 40L total.
I was told by someone that this solution is incorrect. Any suggestions why?

 

[Borek]

2L HA + 0.66L A- = 2.66

Earlier you wrote you need 2 moles, here you have 2.66.

Not that I am sure I understand precisely your working in other places.

 

[stat00]

In the equation pH=pKa+log[A]/[HA], the ratio of the log value must be 10^-3=.5. Therefore, [HA] must be double [A].