Calcuate the degree of ionisation and pH of 0.05 M solution of a weak base having the ionization constant `(K_(b))` is `1.77xx10^(-5).` Also calculate the ionisation constnat of the conjugate acid of this base.

Explanation

If c is the initial conc. Of `NH_3` and `alpha` is its degree of ionisation ,
`{:(,NH_3+H_2O hArr, NH_4(+)+ ,OH^-),("Initial conc.",c,0,0),("At equi.",c(1-alpha), calpha,calpha):}`
`K_b=((calpha)(calpha))/(c(1-alpha))approx calpha^2`
`therefore calpha^2 =1.77xx10^(-5)`
`alpha =((1.77xx10^(-5))/0.05)^(1//2)=1.88xx10^(-2)`
Degree of ionization =`1.88xx10^(-2)`
`[OH^-]=calpha=0.05xx1.88xx10^(-2)`
`=9.4xx10^(-4)` M
`[H_3O^+]=(1xx10^(-14))/(9.4xx10^(-4))=1.06xx10^(-11)`
pH=-log `(1.06xx10^(-11))` =10.973
Now `K_a xx K_b =K_w` for acid-base pair
`K_a` (conjugate acid of `NH_3` ) =`K_w/K_b`
`=(1xx10^(-14))/(1.77xx10^(-5))=5.64xx10^(-10)`