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A buffer solution is prepared in which t...

A buffer solution is prepared in which the concentration of `NH_(3)` is 0.30 M and concentration of `NH_(4)^(+)` is 0.20 M. If equilibrium constant `K_(b)` for `NH_(3)` equals `1.8xx10^(-5)`, what is the pH of the solution ?

A

`9.08`

B

`9.43`

C

`11.72`

D

`8.72`

Text Solution

Verified by Experts

The correct Answer is:
B
`pOH=pK_(a)+(["Salt"])/(["Base"])`
`pK_(b)=-log K_(b)=-log(1.8xx10^(-5))=4.74`
`pOH=4.74+log((0.20)/(0.30))=4.56`
`pH=14-4.56=9.44`
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