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K(b) for NH(4)OH is 1.8 xx 10^(-5). The ...

`K_(b)` for `NH_(4)OH` is `1.8 xx 10^(-5)`. The `[overset(Theta)OH]` of `0.1 M NH_(4)OH` is

A

`5.0 xx 10^(-2)`

B

`4.20 xx 10^(-3)`

C

`1.34 xx 10^(-3)`

D

`1.8 xx 10^(-6)`

Text Solution

Verified by Experts

The correct Answer is:
C
`K_(b) = 1.8 xx 10^(-5)`
`pK_(b) = 4.7447`
`pOH_(W_(B)) = (1)/(2) (pK_(b) - log 0.1)`
`= (1)/(2) (4.7447 +1) = 2.87`.
`[overset(Theta)OH] = "Antilog" (-2.87)`
`= "Antilog" (-2-0.87 + 1-1)`
`= "Antilog" (bar(3).13)`
`= 1.34 xx 10^(-3)`
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