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24 May, 07:12

A 1.50 L buffer solution consists of 0.118 M butanoic acid and 0.318 M sodium butanoate. Calculate the pH of the solution following the addition of 0.063 moles of NaOH. Assume that any contribution of the NaOH to the volume of the solution is negligible. The K a of butanoic acid is 1.52 * 10 - 5.

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  1. 24 May, 07:27
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    pH = 5.493

    Explanation:

    It is possible to find pH of a buffer using H-H equation:

    pH = pKa + log₁₀ [A⁻] / [HA]

    Where [A⁻] is concentration of sodium butanoate and [HA] is concentration of butanoic acid.

    pKa is - log Ka. As Ka of butanoic acid is 1.52*10⁻⁵, pKa is 4.818

    Before reaction, moles of butanoic acid and sodium butanoate are:

    butanoic acid: 1.50L * (0.118mol / L) = 0.177 moles butanoic acid

    sodium butanoate: 1.50L * (0.318mol / L) = 0.477 moles sodium butanoate

    NaOH reacts with butanoic acid thus:

    NaOH + butanoic acid → sodium butanoate + water.

    Butanoic acid decreases concentration and sodium butanoate increases concentration

    Thus, after reaction, moles of butanoic acid and sodium butanoate are:

    butanoic acid: 0.177mol - 0.063mol = 0.114mol

    sodium butanoate: 0.477mol + 0.063mol = 0.540mol

    As total volume is 1.50L, concentrations are:

    [A⁻] [sodium butanoate] = 0.540mol / 1.50L = 0.360M

    [HA] [butanoic acid] = 0.114mol / 1.50L = 0.076M

    Replacing in H-H equation:

    pH = 4.818 + log₁₀ [0.360M] / [0.076M]

    pH = 5.493
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