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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
`[OH^(-)]=sqrt(K_(a)xx(NH_(4)OH))`
`=(10^(-5) xx 10^(-1))^(1//2) =1 xx 10^(-3)`
`[H^(+)]=(10^(-14))/(10^(-3))=10^(-11)`
Hence, `pH =-log 10^(-11)=11.0`
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