A 1.49 L buffer solution consists of 0.264 M propanoic acid and 0.149 M sodium propanoate. Calculate the pH of the solution following the addition of 0.073 mol HCl . Assume that any contribution of the HCl to the volume of the solution is negligible. The K a of propanoic acid is 1.34 × 10 − 5 .

pH = 4.37

Explanation:

In 1,49L of buffer solution you will contain:

1.49L × (0.264mol Propanoic acid / 1L) = 0.393mol propanoic acid

1.49L × (0.149mol Sodium propanoate / 1L) = 0.222mol Sodium propanoate

Reaction of HCl with sodium propanoate is:

HCl + sodium propanoate → Propanoic acid + water

That means in the reaction you consume sodium propanoate producing propanoic acid doing moles of each compound are:

0.393mol propanoic acid + 0.073mol = 0.466 mol propanoic acid

0.222mol Sodium propanoate - 0.073mol = 0.149 mol Sodium propanoate

Using Henderson-Hasselbalch formula for propanoic acid / sodium propanoate buffer:

pH = pka + log₁₀ [Sodium propanoate] / [ Propanoic acid]

pH = pka + log₁₀0.149 mol /0.466 mol

pKa for this buffer is -log1.34x10⁻⁵ = 4.87

Replacing:

pH = 4.87 + log₁₀0.149 mol /0.466 mol

pH = 4.37

I hope it helps!

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