Answer
$$K_c = \frac{[NH{_4}^+][OH^-]}{[NH_3]}$$
Work Step by Step
$$K_c = \frac{[Products]}{[Reactants]} = \frac{[NH{_4}^+][OH^-]}{[NH_3]}$$
- The equilibrium expression for a chemical equation is equal to the multiplication of the concentration of the products divided by that of the reactants.
That expression does not include compounds in the solid or liquid sate. In this case, $H_2O(l)$ was not included for that specific reason.
Notice: The equilibrium coefficients of each molecule/ion will give the exponent. In that case, since all the coefficients are equal to 1, all the exponents are also equal to 1.
