Determine the degree of ionisation of `0.05 M NH_(3)` at `25^(0)C` in a solution `pH = 11 (K_(b) = 1.76 xx 10^(-5))`

Options

A. `0.0173`
B. `0.173`
C. `1.73`
D. `17.3`

Answer

A

Explanation

Let us consider the ionisation of
`NH_(4)OH`
`NH_(4)OH hArr NH_(4)^(+) + OH^(-)`
`{:("At time",t=0,C,O,O),("At equlibrium",C(1-alpha),,Calpha,Calpha):}`
`K_(b) = (CalphaCalpha)/(C(1-alpha)) = Calpha^(2){{:(if alpha lt lt 1),((1-alpha)~~1):}`
`alpha = sqrt((K_(b))/(C)) = sqrt((1.76 xx 10^(-5))/(0.05)) = 0.01876`
Under normal condition degree of dissociation be
`= 0.01876`
When `pH` is i.e. `pOH = 14-11=3`
The, `[OH^(-)] = 10^(-3)`
`:. K_(b) = (Calpha[OH^(-)))/(C(1-alpha)) = (0.05 xx alpha[10^(-3)])/(0.05(1-alpha))`
`1.76 xx 10^(-5)= (alpha10^(-3))/((1-alpha))`
`1.76 xx 10^(-2)(1-alpha) = alpha, 1.76 xx 10^(-2) = 1.0176alpha`
`alpha = (0.0176)/(1.0176) = 0.01729`
`= 0.0173`
(or)
`[OH^(-)] = sqrt(K_(b) xx C)`
`K_(b) = Calpha^(2)`
`[OH^(-)] = Calpha`