The pH of 0.004M hydrazine solution...
Question

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}} )

JEE/Engineering Exams
Chemistry
Solution
105
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( 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 / / )