0.1 M `NH_(3)` के जलीय विलयन की pH हैं: `(K_(b) = 1.8 xx 10^(-5))`

Options

A. `11.13`
B. `12.5`
C. `13.42`
D. `11.55`

Answer

a

Explanation

`NH_(4)OH ltimplies NH_(4)^(+) + OH^(-)`
`K_(b) = C.alpha^(2)`
`therefore alpha=sqrt((K_(b))/C)= sqrt((1.8 xx 10^(-5))/(0.1))=1.34 xx 10^(-2)`
`[OH^(-)]= C.alpha=1.34 xx 10^(-2) xx 0.1= 1.34 xx 10^(-3)`
pOH =`-log[OH^(-)]=-log(1.34 xx 10^(-3)) = 2.87`
`pH = 14-pOH=14-2.87=1.13`