Determine degree of dissociation of `0.05M NH_(3)` at `25^(@)C` in a solution of `pH = 11`.

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)OHhArrNH_(4)^(+)+OH^(-)`
At time `t=0" "COO`
At equilibrium `C(1-alpha)" "Calpha" "Calpha`
`K_(b)=(CalphaCalpha)/(C(1-alpha))=Calpha^(2){{:("if "alphaltlt1),((1-alpha)~~1):}`
`alpha=sqrt((K_(b))/(C))=sqrt(((1.76xx10^(-5))/(0.05)))=0.01876`
Under normal condition degree of dissociation will be =0.01876
when pH is i.e., `pOH=14-11=3`
Then, `[OH^(-)]=10^(-3)`
`:.K_(b)=(Calpha[OH^(-)])/(C(1-alpha))=(0.05xxalpha[10^(-3)])/(0.05(1-alpha))`
`1.76xx10^(-5)=(alpha10^(-3))/((1-alpha))`
`1.76xx10^(-2)(1-alpha)=alpha,1.76xx10^(-2)=1.0176alpha`
`alpha=(0.0176)/(1.0176)=0.01729" "=0.0173`
(or)
`[OH^(-)]=sqrt(K_(b)xxC)`
`K_(b)=Calpha^(2)`
`[OH^(-)]=Calpha`