`pH=9.95`
`therefore` कोडीन एक क्षार (base) है|
कोडीन `+ H_(2)O ltimplies कोडीन H^(+) + OH^(-)`
`{:(1,0,0),(1-alpha,alpha,alpha):}`
`therefore [OH^(-)]=C.alpha`
या `(10^(-14))/([H^(+)]) = C.alpha (therefore [H^(+)][OH^(-)]=10^(-14))`
pH=9.95
`therefore -log[H^(+)] 1.12 xx 10^(-10)`
`therefore (10^(-14))/(1.12 xx 10^(-10)) = 0.005 xx alpha`
`therefore alpha = 0.0179` या `1.79%`
`K_(b) = ([कोडीन^(+)][ओह^(-)])/([कोडीन]) = (Calpha xx Calpha)/(C(1-alpha)) = (Calpha^(2))/(1-alpha)`
`=(0.005 xx (0.0179)^(2))/(1-0.0179)`
`=1.63 xx 10^(-6)`
`pK_(b) = -log_(10)(1.63 xx 10^(-6)) = 5.78`
