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`)

Options

A. 9.08
B. 9.43
C. 11.72
D. 8.73

Answer

B

Explanation

`pOH=pK_(b)+log.(|NH_(3)^(+)|)/(|NH_(3)|)`
`=-log (1.8 xx 10^(-5))+log. (0.2)/(0.3)`
`=4.74 +log .(0.2)/(0.2)=4.56`
`pH ==14-4.56=9.44`