Strategy: From the `pH`, calculate the hydrogen ion concentration. Knowing the `H^(+)` ion concentration and the ionic product of water, calculate the concentration of hydroxyl ions. Using the concentration of `OH^(-)` ions, calculate `K_b` and `pK_b`.
Solution:
`pH=-log a_(H^+)`
`9.7=-log a_(H^+)`
Taking the antilog of both sides, we get
`C_(H^+)=1.67xx10^(-10)`
The ionic product of water is given as
`K_w=C_(H^+)+C_(OH^(-))`
`:. C_(OH^(-))=(K_w)/(C_(H^+))=(1xx10^(-14))/(1.67xx10^(-10))`
`=5.98xx10^(-5)`
Write the reaction summary:
`{:(NH_(2)NH_(2)(aq.)+H_(2)O(l)hArrNH_(2)overset(+)NH_(3)(aq.)+OH^(-)),("Initial (M) 0.004 0.00 0.000"),("Change (M)"-5.98xx10^(-5) +5.98xx10^(-5)+5.98xx10^(-5)),(bar("Equilibrium (M)" (0.004-598xx10^(-5))5.98xx10^(-5)5.98xx10^(-5))):}`
Note that the concentration of the hydrazinium ion is also the same as that of the hydroxide ion. Since the concentration of both these ions is very small, the concentration of the unionized base can be taken (approximately) equal to `0.004M`.
Writing the basically constant
`K_b=([NH_2overset(+)NH_3][OH^(-)])/([NH_2NH_2])`
`=(5.98xx10^(-5))(5.98xx10^(-5))/(0.004)`
`=8.94xx10^(-7)`
`pK_b=-logK_b=-log(8.96xx10^(-7))`
`=6.1`
