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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 the concentration of `NH_(4)^(+)` is `0.20 M`. If the equilibrium constant, `K_(b)` for `NH_(3)` equals `1.8xx10^(-5)`, what is the `pH` of this solution? (`log 2.7=0.43`)

A

8.73

B

9.08

C

9.43

D

11.72

Text Solution

Verified by Experts

The correct Answer is:
C
`p_(OH)=pK_(b)+log.(["Salt"])/(["Base"])=4.74+log.(0.20)/(0.30)`
`=4.74+(0.301-0.4.74)=4.74-0.176`
`=4.56`
`:. pH = 14 - 4.56 = 9.44`
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