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. 8.73
B. 9.08
C. 9.43
D. 11.72

Answer

C

Explanation

`pOH=pK_(a)+"log"(["Conjugate acid"])/(["base"])`
`=-log+1.8x10^(-5)+"log"(0.2)/(0.3)`
`=4.744-0.176=4.567`
`:.pH=14-4.567=9.432`