91Ó°ÊÓ

The reaction of dissociation of\({\rm{N}}{{\rm{H}}_3}\)is:

\({\rm{N}}{{\rm{H}}_3}(aq) + {{\rm{H}}_2}{\rm{O}}(l) \to {\rm{NH}}_4^ + (aq) + O{{\rm{H}}^ - }(aq)\)

(Note, there needs to be double arrow, because this is a reverse reaction).

Equation for reaction quotient is the same as for equilibrium constant, but we don't use equilibrium concentrations, but concentrations in any time before equilibrium.

\({Q_c} = \frac{{c\left( {NH_4^ + } \right) \cdot c\left( {O{H^ - }} \right)}}{{c\left( {N{H_3}} \right)}}\)

Water is not included into the equation because it is liquid and its activity is 1,

so, it would be like this:

\({Q_c} = \frac{{c\left( {NH_4^ + } \right) \cdot c\left( {O{H^ - }} \right)}}{{c\left( {N{H_3}} \right) \cdot 1}}\).