To prepare 800mL of 1M phosphate buffer of pH 7.2 (PKa=7.1), by using salt = K2HPO4 and acid = KH2PO4, the quantity of salt required is

78.04g of 0.448 moles must be added

Explanation:

Using H-H equation we can find the pH of the buffer:

pH = pKa + log [A⁻] / [HA]

Where pH is the pH of the buffer = 7.2

pKa = 7.1

[A⁻] = [K₂HPO₄]

[HA] = [KH₂PO₄]

Replacing:

7.2 = 7.1 + log [K₂HPO₄] / [KH₂PO₄]

0.1 = log [K₂HPO₄] / [KH₂PO₄]

1.2589 = [K₂HPO₄] / [KH₂PO₄] (1)

And as the  concentration of the buffer is:

1M = [K₂HPO₄] + [KH₂PO₄] (2)

Replacing (2) in (1):

1.2589 = 1M - [KH₂PO₄] / [KH₂PO₄]

1.2589 [KH₂PO₄] = 1M - [KH₂PO₄]

2.2589 [KH₂PO₄] = 1M

[KH₂PO₄] = 0.44M

And [K₂HPO₄] = 0.56M

In 800mL = 0.8L:

0.8L * (0.56mol / L) = 0.448 moles K₂HPO₄. The mass is -Molar mass K₂HPO₄: 174.2g/mol-:

0.448 moles * (174.2g / mol) =

the answer would be a