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

Answer

A

Explanation

`pOH=pK_(b)+log.(["salt"])/(["base"])`
`=-log1.8xx10^(-5)+log.(0.2)/(0.30)`
`=5-0.25+(-0.176)`
`=4.75-0.176=4.57`
`therefore pH=14-4.57=9.43`