Question

As a technician in a large pharmaceutical research firm, you need to produce 400. mL of 1.00 M a phosphate buffer solution of pH = 7.36. The pKa of H2PO4− is 7.21. You have 2.00 L of 1.00 M KH2PO4 solution and 1.50 L of 1.00 M K2HPO4 solution, as well as a carboy of pure distilled H2O. How much 1.00 M KH2PO4 will you need to make this solution?

Answer 1

we know that

for a buffer

pH = pKa + log [conjugate base / acid ]

in this case

the conjugate base is HP042-

and

the acid is H2P04-

so

the equation becomes

pH = pKa + log [HP042- / H2P04-]

now

given

pH = 7.36

and

pKa = 7.21

so

using those values

we get

pH = pKa + log [HP042- / H2P04-]

7.36 = 7.21 + log [HP042- / H2P04-]

log [HP042- / H2P04-] = 0.15

[HP042- / H2P04-] = 1.41

so

[HP042- ] = 1.41 [H2P04-]

we know that

conc = moles / volume

as the final volume is same for both

ratio of conc = ratio of moles

so

moles of HP042- = 1.41 x moles of H2P04-

now

total moles of phosphate = conc x volume (L)

total moles of phosphate = 1 x 0.4

total moles of phosphate = 0.4

now

total moles of phosphate = moles of HP042- + H2P04-

total moles of phosphate = 1.41 moles of H2P04- + H2P04-

total moles of phosphate = 2.41 x moles of H2P04-

so

2.41 x moles of H2P04- = 0.4

moles of H2P04- = 0.166

now

we know that

moles = conc x volume (L)

so

moles of H2P04- = conc x volume (L)

0.166 = 1 x volume (L)

volume (L) = 0.166

volume = 166 ml

so

166 ml of KH2P04 is needed