You travel to a distant, cold planet where the ammonia flows like water. In fact, the inhabitants of this planet use ammonia (an abundant liquid on their planet) much as earthlings use water. Ammonia is also similar to water in that it is amphoteric and undergoes autoionization. The \(K\) value for the autoionization of ammonia is \(1.8 \times 10^{-12}\) at the standard temperature of the planet. What is the \(\mathrm{pH}\) of ammonia at this temperature?

The autoionization of ammonia can be represented by the balanced chemical equation: \[ 2NH_3 \rightleftharpoons NH_4^+ + NH_2^- \] From this equation, we can see that each mole of ammonia forms one mole of NH\(_4^+\) ion and one mole of NH\(_2^-\) ion as it autoionizes.

The \(K\) value for ammonia autoionization is given as \(1.8 \times 10^{-12}\). Define the change in the concentration of ammonia ([NH\(_3\)]) as x. Thus, when the system reaches equilibrium, the concentration of NH\(_4^+\) ion would be x and the concentration of the NH\(_2^-\) ion would also be x. Plugging these values into the equilibrium constant expression, we get: \[ K = \frac{[NH_4^+][NH_2^-]}{[NH_3]^2} = \frac{x^2}{x^2} = 1.8 \times 10^{-12} \]

Since the equilibrium constant expression is already in the form \(K = x^2/x^2\), simplifying and solving for x gives: \[ x = \sqrt{1.8 \times 10^{-12}} = 1.34 \times 10^{-6} \] The equilibrium concentration of NH\(_4^+\) ions, which is x, is found to be \(1.34 \times 10^{-6} \, \mathrm{M}\).

Now, we can use the definition of \(\mathrm{pH}\) to find the \(\mathrm{pH}\) of the ammonia solution: \[ \mathrm{pH} = -\log [H_3O^+] \] However, we have the concentration of NH\(_4^+\). We need to account for the ionization of NH\(_4^+\) and calculate the concentration of H\(_3\)O\(^+\) ions. Recall the ionization of NH\(_4^+\) can be written as: \[ NH_4^+ \rightleftharpoons NH_3 + H^+ \] Note that H\(^+\) ion concentration in aqueous solution is equivalent to the H\(_3\)O\(^+\) ion concentration. Since the ionization of NH\(_4^+\) is considered to be complete, we can assume that the concentration of NH\(_4^+\) ions is equal to the concentration of H\(^+\) ions. Thus, [H\(_3\)O\(^+\)] = \(1.34 \times 10^{-6} \, \mathrm{M}\). Now calculate the \(\mathrm{pH}\): \[ \mathrm{pH} = -\log (1.34 \times 10^{-6}) = 5.87 \] The \(\mathrm{pH}\) of ammonia at this temperature and equilibrium constant is approximately 5.87.