Buffer question??????

Question

As a technician in a large pharmaceutical research firm, you need to produce 250mL of 1.00M potassium phosphate solution of pH = 7.43. The pKa of H2PO4^- is 7.21. You have 2.00L of 1.00M KH2PO4 solution and 1.50L of 1.00M K2HPO4 solution, as well as a carboy of pure distilled H2O. How much 1.00M KH2PO4 will you need to make this solution?

Answer is in mL.

Any help would be appreciated.

Answer 1

there are a few  different ways to work this, I hope this isn't the messiest:

Henderson–Hasselbalch equation

http://en.wikipedia.org/wiki/Henderson_h...

pH = pKa + log [anion] / [acid]

7.43 = 7.21 + log [anion] / [acid]

0.22 =log [anion] / [acid]

10 to the x of both sides

1.6596 = [anion] / [acid]

--------------------------------------...

the ratio of concentrations has to be 1.6596 , (to less sig figs, I know)...

but the ratio of concentrations in a solution, (250 mls), is the same as the ratio of moles in that volume,..so I am going to convert the mls of each stuff added... into moles, where I  arbitrarially choose one to be where that I added "X" millilitres of it ,  & the other becomes that I added  "250-X" ml pf it by default:

1.6596 = [M anion] / [M acid]

1.6596 = [M K2HPO4] / [M KH2PO4]

now I am going to multiply the molarity times the volume of each to get the moles of each , (& yes it won't matter that I amusing mls instead of liters):

1.6596 = (250ml - X)(1.00 Molar K2HPO4) / (X ml)(1.00 Molar KH2PO4)

1.6596 = (250 - X) / (X)

1.6596 X = 250 - X

2.6596 X = 250 ml

X= 93.999 ml of 1.00 molar KH2PO4 "acid"

your answer is : mix 94.0 ml of 1.00 Molar KH2PO4 with 156 ml of 1.00 Molar K2HPO4