Calculate `[OH^(-)]and %` dissociation of 0.01 M solution of ammonium hydroxide solution. The ionization constant for
`NH_(4)OH(K_(b))=1.8xx10^(-5)`

Explanation

`{:(NH_(4)OH,hArrNH_(4)^(+)+,OH^(-)),(C-Calpha,C alpha,Calpha):}`
`K_(b)=([NH_(4)^(+)][OH^(-)])/([NH_(4)OH])`
`K_(b)=(Calpha^(2))/(1-alpha)`
`1.8xx10^(-5)=(0.01xxalpha^(2))/(1-alpha)`
`alpha=sqrt(1.8xx10^(-3))=4.24xx10^(-2)[1-alpha=1]`
`%` dissociation `=4.24`
`[OH^(-)]=Calpha=0.01xx4.24xx10^(-2)`
`[OH^(-)]=4.24xx10^(-4)M`