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Karnataka Board PUCPUC Science Class 11

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

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Question

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

Numerical
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Solution

c = 0.005

pH = 9.95

pOH = 4.05

pH = – log (4.105)

`4.05 = -log["OH"^-]`

`["OH"^-] = 8.91 xx 10^(-5)`

`"c"  alpha = 8.91 xx 10^(-5)`

`alpha = (8.91 xx 10^(-5))/(5xx10^(-3)) = 1.782 xx 10^(-2)`

Thus `"K"_"b" = "c"  alpha^2`

`= 0.005 xx (1.782)^2 xx 10^(-4)`

`= 0.005 xx 3.1755 xx 10^(-4)`

`= 0.0158 xx 10^(-4)`

`"K"_"b" = 1.58 xx 10^(-6)`

`"Pk"_"b" = - log "K"_"b"`

`= - log (1.58 xx 10^(-6))`

= 5.80

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Chapter 6: Equilibrium - EXERCISES [Page 237]

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NCERT Chemistry - Part 1 and 2 [English] Class 11
Chapter 6 Equilibrium
EXERCISES | Q 7.51 | Page 237
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