The pH of 0.005 M codeine `(C_18 H_(21) NO_3) ` solution is 9.95 Calculate its ionisations contant and ` rhoK_b.`

Explanation

Codeine is a base. Therefore , it is expected to ionise in solutions as follows.
Codeine +`H_2O hArr " Codeine " H^(+) +OH^(-)`
Given ` " " pH = 9.95 `
` therefore " " pOH = 14 - 9.95 = 4.05`
i.e.,` " " - log_(10) [OH^(-)] = 4.05`
or ` " " [OH^(-)] = "antilog"_(10) ( - 4.05)`
` therefore ["Codeine " H^(+) ]= [OH^(-)] = 8.91 xx 10 ^(-5) `
and ` " " [" Codeine "]= 0.005`
` therefore " " K_b= ( [" Codeine "H^(+) ][OH^(-)])/(["Codeine "]) `
` " " = ( ( 8.91xx 10 ^(-5) ) xx (8.91 xx 10^(-5)))/([" Codeine"]) `
` " " = ((8.91xx 10 ^(-5))xx (8.91xx 10^(-5)))/(0.005)`
`" " = 1.587 xx 10 ^(-6)`
and ` " " rho K_b = - log_(10) K_b = - log_(10) 1.587 xx 10^(-6) = 5.8`