Answer:
pH = 5.493
Explanation:
It is possible to find pH of a buffer using H-H equation:
pH = pKa + log₁₀ [A⁻] / [HA]
<em>Where [A⁻] is concentration of sodium butanoate and [HA] is concentration of butanoic acid.</em>
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pKa is -log Ka. As Ka of butanoic acid is 1.52×10⁻⁵, <em>pKa is 4.818</em>
Before reaction, moles of butanoic acid and sodium butanoate are:
butanoic acid: 1.50L × (0.118mol / L) = <em>0.177 moles butanoic acid</em>
sodium butanoate: 1.50L × (0.318mol / L) = <em>0.477 moles sodium butanoate</em>
NaOH reacts with butanoic acid thus:
NaOH + butanoic acid → sodium butanoate + water.
<em>Butanoic acid decreases concentration and sodium butanoate increases concentration</em>
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 = <em>0.360M</em>
[HA] [butanoic acid] = 0.114mol / 1.50L = <em>0.076M</em>
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Replacing in H-H equation:
pH = 4.818 + log₁₀ [0.360M] / [0.076M]
<em>pH = 5.493</em>
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