Calculate `pH` of `0.002 N NH_(4)OH` having `2%` dissociation

Options

A. `7.6`
B. `8.6`
C. `9.6`
D. `10.6`

Answer

C

Explanation

`NH_(4)OH` is a weak base and partially dissociated
`{:(,NH_(4)OH,hArr,NH_(4)^(+),+,OH^(-)),("Concentration",1,,0,,0),("before dissociation",,,,,),("Concentration",1-alpha,,alpha,,alpha),("after dissociation",,,,,):}`
`:. [OH^(-)]=C alpha=2xx10^(-3)xx(2)/(100)=4xx10^(-5)M`
`pOH= -log[OH^(-)]`
`= -log4xx10-5=4.4`
`pH=14-4.4=9.6`