Henderson-Hasselbalch equation:
pH = pKa + log([base] / [acid])
Kb for ammonia is given, but we need Ka for ammonium ion NH₄⁺ .
This can be found from Kw = Ka•Kb :
Ka = Kw/Kb = (1.0 x 10¹⁴) / (1.8 x 10⁻⁵) = 5.56 x 10⁻¹⁰
pKa = -log(Ka) = -log(5.56 x 10⁻¹⁰) = 9.25
= 9.25 + log(0.48 / 0.87)
= 8.99
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Henderson-Hasselbalch equation:
pH = pKa + log([base] / [acid])
Kb for ammonia is given, but we need Ka for ammonium ion NH₄⁺ .
This can be found from Kw = Ka•Kb :
Ka = Kw/Kb = (1.0 x 10¹⁴) / (1.8 x 10⁻⁵) = 5.56 x 10⁻¹⁰
pKa = -log(Ka) = -log(5.56 x 10⁻¹⁰) = 9.25
pH = pKa + log([base] / [acid])
= 9.25 + log(0.48 / 0.87)
= 8.99