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

Answer

`K_(b) = 1.6 xx 10^(-6), pK_(b) = 5.8`

Explanation

`c = 0.005`
`pH = 9.95`
`pOH = 4.05`
`pH = – log (4.105)`
`4.05 = -"log" [OH^(-)]`
`[OH^(-)] = 8.91 xx 10^(-5)`
`calpha = 8.91 xx 10^(-5)`
`alpha = (8.91 xx 10^(-5))/(5 xx 10^(-3)) = 1.782 xx 10^(-2)`
Thus, `K_(b) = calpha^(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)` ltbr gt `Pk_(b) = - "log" K_(b)`
`= -"log"(1.58 xx 10^(-6))`
`= 5.80`