7.51. The pH of 0.005 M codeine (C18H21N03) solution is 9.95. Calculate the ionization constant and PKb.
7.51. The pH of 0.005 M codeine (C18H21N03) solution is 9.95. Calculate the ionization constant and PKb.
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1 Answer
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pH = 9.95,
pOH = 14 – pH = 14 − 9.95 = 4.05
[OH−] = 10−pOH = 10−4.05 = 8.913 × 10−5Codeine + H2? O? CodeineH+ + OH−
The ionization constant, Kb? = [CodeineH+] [OH−] / [codeine]?
= [ (8.913×10−5)× (8.913×10−5)] / 5×10−3
= 1.588×10−6.
pKb? = −log (1.588×10−6)
= 5.8
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