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

`{:(,NH_(3),+,H_(2)O,hArr,NH_(4)^(+),+,OH^(-)),("Initial conc.",0.05M,,,,,,),("Eqm. conc. ",0.05(1 -alpha),,,,0.05 alpha,,0.05 alpha):}`
`K_(b)=([NH_(4)^(+)][OH^(-)])/([NH_(3)])`
`:. 1.77xx10^(-5)=((0.5 alpha)^(2))/(0.05(1-alpha))~=0.05 alpha^(2) or alpha^(2) = (1.77xx10^(-5))/(0.05) = 3.54xx10^(-4) or alpha = 1.88xx10^(-2)`
Alternatively, apply directly the formula : `alpha =sqrt(K_(b)//C)`
(ii) `[OH^(-)]=0.05 alpha = 0.05 xx 1.88xx10^(-2) = 9.4 xx 10^(-4) M`
`pOH = log (9.04xx10^(-4)) =4-0.9562=3.04 :. pH = 14-3.04 = 10.96`
(iii) For conjugate acid base pair, `K_(a) = (K_(w))/(K_(b))=(10^(-14))/(1.77xx10^(-5))=5.64xx10^(-10)`