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0.25M solution of pyridinium chloride was found to have a pH of 2.699. What is Kb for pyridine?
[Py-1] + [H20] ----> [PyOH] + [H+] C(1-a) Ca Ca [PyOH][H+] ------------------ = Kh [Py-1][H20] => Kh = Kw / Kb [H+] = Ca = (Kh.C) So we get Ph = – ½ [ log C + log Kw – log Kb ]So Pkb = – log C + Pkw – 2(Ph) Pkb = 9.2041Kb = 6.25 x 10-10 – Arnesh Nagdeve, IIT Bombay [Py-1] + [H20] ----> [PyOH] + [H+] C(1-a) Ca Ca [PyOH][H+] ------------------ = Kh [Py-1][H20] => Kh = Kw / Kb [H+] = Ca = (Kh.C) So we get Ph = – ½ [ log C + log Kw – log Kb ]So Pkb = – log C + Pkw – 2(Ph) Pkb = 9.2041Kb = 6.25 x 10-10 – Arnesh Nagdeve, IIT Bombay
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