The `pH` of a `0.1M` solution of `NH_(4)Oh` (having dissociation constant `K_(b) = 1.0 xx 10^(-5))` is equal to

Options

A. 3
B. 10.5
C. 11
D. 7.5

Answer

C

Explanation

`{:(,NH_(4)OH ,hArr,NH_(4)^(+),+,OH^(-)),("Initial",0.1,,0,,0),("At eqm.",0.1(1- alpha),,0.1 alpha,,0.1alpha):}`
`:. K_(b)=(0.1 alpha xx 0.1 alpha)/(0.1 ) =0.1 alpha^(2)`
`0.1 alpha^(2)=1 xx 10^(-5) `
`alpha^(2)=10^(-4) implies alpha=10^(-2)`
`[OH^(-)]=c alpha=0.1 xx 10^(-2) =10^(-3)M`
`[H^(+)]=10^(-14)//10^(-3) =10^(-11):. pH =11`.