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Question

A 1.35 L buffer solution consists of 0.273 M propanoic acid and 0.104 M sodium propanoate. Calculate the pH of the solution following the addition of 0.068 mol HCl . Assume that any contribution of the HCl to the volume of the solution is negligible. The 𝐾a of propanoic acid is 1.34×10−5 . pH =

Answer 1

Let HA = propionic acid

A- = propanoate

H+ = HCl

A- + H+ ==> HA

0.104....0.068.......0.273.....Initial

-0.068..-0.068.....+0.068....Change

0.036.......0..........0.068......Equilibrium

Use the Henderson Hasselbalch equation:

pH = pKa + log [A-]/[HA]

pKa = -log Ka = -log 1.34x10-5 = 4.87

pH = 4.87 - log (0.036 / 0.068) = 4.87 + (-0,28)

pH = 4.59