A 1.42 L buffer solution consists of 0.176 M butanoic acid and 0.280 M sodium bu

Question

A 1.42 L buffer solution consists of 0.176 M butanoic acid and 0.280 M sodium butanoate. Calculate the pH of the solution following the addition of 0.077 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.

Note: The buffer solution consists of the weak acid butanoic acid, C3H7COOH, and its conjugate base butanoate, C3H7COO–. The added NaOH, a strong base, will react with a stoichiometrically equivalent amount of butanoic acid to produce butanoate. Start by determining the initial amount, in moles, of butanoic acid and butanoate in the buffer solution before the addition of NaOH.

Explanation / Answer

   concentration of sodiumbutanoate = 0.28 M

concentration butanoicacid = 0.176 M

   pka = -logKa

        = -log(1.52*10^-5)

        = 4.82

concentration of NaOH added = 0.077/1.42 = 0.054 M

pH = pka + log(salt+NaOH/acid - NaOH)

    = 4.82 + log((0.28+0.054)/(0.176-0.054))

   = 5.26

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