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The pH of 0.004M hydrazine (NH(2).NH(2))...

The `pH` of `0.004M` hydrazine `(NH_(2).NH_(2))` solution is `9.7`. Calculate its ionisation constant `K_(b)` and `pK_(b)`.

Text Solution

Verified by Experts

`NH_2NH_2+H_2O hArr NH_2NH_3^(+)+OH^(-)`
`K_b=([NH_2NH_3^+][OH^-])/([NH_2NH_2])`
Now pH=9.7
-log `[H_3O^+]`=9.7
or log `[H_3O^+]` =-9.7
or =`bar(10).3`
Taking antilog on both sides
`[H_3O^+]=1.99xx10^(-10)`
`[OH^-]=K_w/([H_3O^+])=(1xx10^(-14))/(1.99xx10^(-10))`
`=5.025xx10^(-5)`
Now `[OH^-]=[NH_2NH_3^+]=5.025xx10^(-5)`
`K_b=(5.025xx10^(-5))^2/0.004=6.31xx10^(-7)`
`pK_b=-log k_b=-log 6.31xx10^(-7)`=6.0
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