Question

How would you prepare 10L of 0.045M potassium phosphate buffer, pH 7.5?

You can use the Henderson-Hasselbalch equation to calculate how much of each chemical species. This ratio will tell you how much of the 10L comes from the [A-} species, and how much from the [HA] species, giving you the volumes for your next calculation. Then you have to figure out how much of each to weigh out to make that calculated volume at 0.045M for each species.

The two chemical species of phosphate that I need are?

The ratio of the more basic to more acidic ( [A-]/[HA] ) from the Henderson Hasselbalch is?

Of the 10L total volume, what volume is [A-] and what volume is [HA]?

The amount of each solution to weigh out to yield 0.045 M is?

Answer 1

potassium phosphate KxPO4

for pH = 7.5

compare pKa vlaues for Phosphoric acid, the nearest pKa value is 2nd ionization of H3PO4, which is HPO4-2 and H2PO4-

pKa = 7.20

then, if pH goal = 7.5

the two chemical species phosphate required --> KH2PO4 and K2HPO4

apply Henderson-Hasselbalch equation

pH = pKa +log(HPO4-2/H2PO4-)

7.5 = 7.2 + log(HPO4-2/H2PO4-)

(HPO4-2/H2PO4-) = 10^(7.5-7.2)= 1.9952

(HPO4-2/H2PO4-) = 2 (Equation 1) and also the ratio asked

Total moles of acid:

mol = M*V = 0.045*10 = 0.45 mol of H2PO4- + HPO4-2

equation 2 --> H2PO4- + HPO4-2 = 0.45

from eqn 1 and 2:

(HPO4-2/H2PO4-) = 2 --> HPO4-2 = 2*H2PO4-

substitute in --> H2PO4- + HPO4-2 = 0.45

H2PO4- + 2*H2PO4-2 = 0.45

H2PO4- = 0.45/3 = 0.15 M

H2PO4- = 0.15 mol

and we know

HPO4-2 = 2*H2PO4-

HPO4-2 = 2*0.15 = 0.30 M

HPO4-2 = 0.30 mol

ratio is 1:2

that is, 3.333 L for HA; 6.667 L of A-

KH2PO4 and K2HPO4

mass of KH2PO4 = mol*MW = 0.15*136.085541 = 20.41 g

mass of K2HPO4 = mol*MW = 0.30*174.2 = 52.26 g