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The pH of a 0.1M solution of NH(4)Oh (ha...

The `pH` of a `0.1M` solution of `NH_(4)Oh` (having dissociation constant `K_(b) = 1.0 xx 10^(-5))` is equal to

A

`10`

B

`6`

C

`11`

D

`12`

Text Solution

Verified by Experts

The correct Answer is:
C
`K_(b) = 10^(-5), pK_(b) = 5`
`NH_(4)OH` is a weak base
`:. [overset(Theta)OH] = Calpha`
`[overset(Theta)OH] = sqrt(K_(b)xxC)`
`pOH = (1)/(2) (pK_(b) - logC)`
`= (1)/(2) (5-log 0.1) = (1)/(2)(5+1)=3`.
`pH = 14 - pOH = 14 -3 = 11`
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