The pH of a 0.1 M solution of `NH_(4)OH` (having `K_(b)=1.0xx10^(-5))` is equal to

Options

A. 10
B. 6
C. 11
D. 12

Answer

C

Explanation

`[OH^(-)]=sqrt(K_(b)xxC)`
`=sqrt(1xx10^(-5)xx10^(-1))`
`sqrt(10^(-6))=10^(3)`
`K_(w)=[H^(+)][OH^(-)]`
`10^(-14)=[H^(+)][OH^(-3)]`
`[H^(+)]=10^(-11)`
Hence, `pH=- logH^(+)`
`=-log(1xx10^(11))=11`