# The ( mathrm{pH} ) of ( 0.004 mathrm{M} ) hydrazine solution is ( 9.7 . ) Calculate its ionization constant ( K_{mathrm{b}} ) and ( mathrm{p} K_{mathrm{b}} )

( mathrm{NH} 2 mathrm{NH} 2+mathrm{H} 2 mathrm{O}-ldots->mathrm{NH} 2 mathrm{NH} 3++mathrm{OH}- )

From the given pH the Hion concentration can be measured.

So, we have

( [mathrm{H}+]=operatorname{antilog}(-mathrm{pH})=operatorname{antilog}(-9.7)=1.67 times 10^{wedge}-10 )

Now, ( [mathrm{OH}-]=mathrm{KW} /[mathrm{H}+]=1 times 10^{wedge}-14 / 1.67 times 10^{wedge}-10 )

( =5.98 times )

( 10^{wedge}-5 / / )

The concentration of the corresponding hydrazine ion is also the same as that of hydroxyl ion. The concentration of both these ions is very small so the concentration of the undissociated base can be

taken equal to ( 0.004 mathrm{M} )

Thus, ( mathrm{Kb}=[mathrm{NH} 2 mathrm{NH} 3+][mathrm{OH}-] /[mathrm{NH} 2 mathrm{NH} 2]=left(5.98 times 10^{wedge}-5right)^{wedge} 2 / 0.004= )

( 8.96 times 10^{wedge}-7 / / )

So, ( mathrm{pKb}=-log mathrm{Kb}=-log (8.96 times 10-7)=6.04 / / )