You need to prepare 250 mL of a 50 mM potassium phosphate buffer at pH=6. In the

ID: 884443 • Letter: Y

Question

You need to prepare 250 mL of a 50 mM potassium phosphate buffer at pH=6. In the lab you have a 1 M stock solution of H3PO4, and solid salts of NaH2PO4, (FW=119.96 g/mole) Na2HPO4*7H2O (FW=268.07 g/mole), Na3PO4*12H2O (380.12 g/mole). Given that the three pKa values of phosphate are 2, 7, and 12, choose the appropriate reagents to prepare this buffer and calculate how much of each one you will need to add. At what pH values will phosphate buffers have the maximum bufer capacity? How could you double the buffering capacity?

Explanation / Answer

pKa's of phosphoric acid are 2,7 and 12 If required pH is 6,

then, 7 will be used. This means, monopotassium dihydrogen

phosphate and dipotassium monohydrogen phosphate (diprotic (H2PO4-)

and monoprotic (HPO4--) potassium salts) will be used.

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

Now, for A- you put K2HPO4 concentration, and for HA you put KH2PO4 concentration:

6 = 7 + log(A/HA)

log(A/HA)= -1

A/HA = 0.1

HA+A= 0.0125

A/HA=0.1

HA=0.01136 M

A = 0.00114 M

This means, you need to put 0.01136 moles of KH2PO4

and 0.00114 moles of K2HPO4 salts in 250 ml of solution.

mass of KH2PO4 = 0.01136*119.96 = 1.36 grams

mass of K2HPO4 = 0.00114*268.07 = 0.3055 grams

dissolve 1.36 grams of NaH2PO4 , 0.3055 grams of Na2HPO4*7H2O in 250 ml

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You need to prepare 250 mL of a 50 mM potassium phosphate buffer at pH=6. In the

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