The `pH` of `0.005 M` codenine `(C_(18)H_(21)NO_(3))` solution is `9.95`. Calculate its ionisation constant and `pK_(b)`.

Explanation

`C od + H_(2)O hArr CodH^(+)+OH^(-)`
`pH = 9.95 :. pOH = 14 - 9.95 = 4.05 , i.e., - log [OH^(-)]=4.05`
or `log [OH^(-)]=-4.05=bar(5).95 or [OH^(-)] = 8.913 xx 106(-5)M`
`K_(b) =([CodH^(+)][OH^(-)])/([Cod]) = ([OH^(-)]^(2))/([Cod])=(8.91xx10^(-5))^(2))/(5xx10^(-3))=1.588 xx 10^(-6)`
`pK_(b) = - log (1.588xx10^(-6))=6-0.1987=5.8`
`{:("(ii) Also",C_(6)H_(5)NH_(2),+,H_(2)O,hArr,C_(6)H_(5)NH_(3)^(+),+,OH^(-),,,),("Initial conc.",C,,,,,,,,,),("At. eqm.",C-C alpha,,C alpha,,C alpha,,,,,):}`
`K_(b)=(C alpha . C alpha)/(C (1-alpha))=(C alpha^(2))/(1-alpha) ~= C alpha^(2) :. alpha=sqrt(K_(b)//C)=sqrt(4.27xx10^(-10)//10^(-3))=6.53xx10^(-4)`
(iii) `pK_(a)+pK_(b)=14` (for a pair of conjugate acid and base )
`pK_(b)=-log (4.27xx10^(-10))=10-0.62=9.38 :. pK_(a) = 14-9.38=4.62`
i.e., `- log K_(a) = 4.62 or log K_(a) = - 4.62= bar(5).38 or K_(a) = ` Antilog `bar(5).38=2.399xx10^(-5)~=2.4xx10^(-5)`