Question

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

Answer 1

Hydrozine is a weak bease `pH = 0.07`
`:. pOH = 14 - 9.7 = 4.3`
a. `pH ("weak base") = 14 - [(1)/(2)(pK_(b) - logC)]`
or
`pOH("weak base") = pK_(b) - log 0.004`
`pK_(b) = 8.6 +log 4 xx 10^(-3)`
`= 8.6 +2log 2-3`
`= 8.6 +2 xx 0.3 - 3`
`= 6.2`
`:. pK_(b) = 6.2`
b. `pK_(b) =- log K_(b) = 6.2`
`log K_(b) = - 6.2 =- 6 - 0.2 +1 - 1 = bar(7).8`
`K_(b) - "Antilog" (bar(7).8)= 6.31 xx 10^(-7)`.