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

Calculate the degree of dissociation of 0.5 M NH_(3) at 25^(@)C in a s

1.	Calculate the degree of dissociation of 0.5 M NH_(3) at 25^(@)C in a solution of pH = 12.
Answer» Solution :`{:(,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 % `.