Calculate the percentage dissociation of 0.5 M `NH_(3)` at `25^(@)C` in a solution of pH = 12.

Explanation

`{:(,NH_(4)OH,hArr,NH_(4)^(+),+,OH^(-),,),("Initial conc.",C "mol " L^(-1),,0,,0,,),("After disso.",C-Calpha ,,C alpha,,C alpha,,):}`
pH = 12 means `[H^(+)]=10^(-12) or [OH^(-)]=10^(-2)`
`:. [OH^(+)]=C alpha = 10^(-2) or alpha = (10^(-2))/(C) = (10^(-2))/(0.5) = 2 xx 10^(-2) or 2 % `.