Question

Outline how you would prepare a propanoic acid/propanoate buffer with pH = 4.25, starting with 5.0 L of 0.050 M potassium propanoate solution and adding the acid component? Ka of propanoic acid = 1.x 10-5 .

Answer 1

pH = 4.25

pKa = 5.00

Moles of potassium propanoate = 0.050×5 = 0.25 mol

for acidic buffer

$pH = pKa + log(S/A)$

4.25 = 5.0 + log(0.250/A)

A = 1.41 mol

So, by adding [1.4 ] mol Acid component we can make buffer