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A 1.36 L buffer solution consists of 0.184 M butanoic acid and 0.331 M sodium butanoate....

A 1.36 L buffer solution consists of 0.184 M butanoic acid and 0.331 M sodium butanoate. Calculate the pH of the solution following the addition of 0.068 moles of NaOH. Assume that any contribution of the NaOH to the volume of the solution is negligible. The Ka of butanoic acid is 1.52 × 10^-5.

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Homework Answers

Answer #1

first find the pKa of the butanoic acid

pKa -logKa = -log1.52 x 10-5 = 4.8

moles of Butanoic acid = Molarity x volume = 0.184 M x 1.36 L = 0.25 mol

noles sodium btananoate = 0.331M x 1.36 L = 0.45 mol

when you add 0.068 moles of naOH that will react with 0.068 mol of butanoic acid and form the 0.068 mol of sodium butanoate

pH = pKa + log [moles of sodium butanoate +x / moles of butanoic acid + x]

where x = moles of NaOH

pH = 4.8 + log[0.45+ 0.068 / 0.25-0.068]

pH = 4.8 + log [0.518 / 0.182]

pH = 4.8 + 0.45

pH = 5.25

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