NCERT Class XI Chemistry Equilibrium Solutions

Question : 51

Total: 73

The pH of 0.005 M codeine (C18H21NO3) solution is 9.95. Calculate its ionization constant and pKb.

Solution:

[H+] = antilog (- 9.95) = 1.12 × 10−10
∴ [OH−] =

1.0×10−14

1.12×10−10

= 8.92 × 10−5 M
But Kb =

[M+][OH−]

[MOH]

(8.92×10−5)2

0.005

= 1.59 × 10−6
pKb = - log Kb = - log (1.59 × 10−6) = 5.80

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