Question

Write the reaction quotient expression for the ionization of NH₃ in water:

Short answer

The reaction quotient expression for the ionization of NH₃ in water:

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

Step-by-step solution

The reaction regarding the ionization of NH3 in water:

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).

The reaction quotient expression for the ionization of NH3 in water:

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:

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