A1.36 l buffer solution consists of 0.109 m butanoic acid and 0.266 m sodium butanoate. calculate the ph of the solution following the addition of 0.067 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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Ответ:
the pH of the solution is 5.54
Explanation:
Butanoic acid = HA
n = mole
Sodium acetate = A
Ka = 1.52×10⁻⁵ thus pKa = -log of Ka = 4.82
n=C×V; n of HA=0.109×1.36=0.148 mole; n of A=0.266×1.36=0.362 mole
Adding 0.067 mole of NaOH, it reacts with HA and increase amount of A
After reaction, n of HA=0.148-0.067=0.081; n of A=0.362+0.067=0.429 mole
c=n/v and v=1.36L
using Hasselbach equation: pH = pKa+log[A]/[HA]
pH =4.82+log[(0.429÷1.36)/(0.081÷1.36)] = 5.54
Ответ:
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