Question

Write the expression of the reaction quotient for the ionization of HOCN in water.

Short answer

Therefore,the reaction quotient for the ionization of HOCN in water.

\({Q_c} = \frac{{c\left( {{H_3}{O^ + }} \right) \cdot c\left( {OC{N^ - }} \right)}}{{c(HOCN)}}\).

Step-by-step solution

Reaction regarding ionization of HOCN in water:

First, we will write reaction of dissociation:

\(HOCN(aq) + {H_2}O(l) \to {H_3}{O^ + }(aq) + OC{N^ - }(aq)\)

(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.

The Expression for the reaction quotient for the ionization of HOCN in water:

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

Note - water is not included into the equation because it is liquid and its activity is 1 , so it would be like this:

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