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

Options

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

Answer

D

Explanation

`[NH_(3)]=0.3M,[NH_(4)^(+)]=0.2M`
`k_(b)=1.8xx10^(-5)`
`NH_(3)+H_(2)OhArrNH_(4)^(+)+OH^(-)`
`K_(b)=([NH_(4)^(+)][OH^(-)])/([NH_(3)])rArr[OH^(-)]=(k_(b)xx[NH_(3)])/([NH_(4)^(+)])`
`[OH^(-)]=(1.8xx10^(-5)xx0.3)/(0.2)=2.7xx10^(-5)`
`pOH=-"log"[OH^(-)]=-"log"(2.7xx10^(-5))`
`=5-"log"2.7=5-0.43=4.57`
`therefore pH=14-pOH=14-4.57=9.43`