Answer:
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) =
