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 pH of ammonia at this temperature?

The autoionization of ammonia is represented by the following equation: \[ 2NH_3 \rightleftharpoons NH_4^+ + NH_2^-\] We're given the equilibrium constant, \(K\), which can be expressed as: \[K = \frac{[NH_4^+][NH_2^-]}{[NH_3]^2} \]

Since the ammonia is in a pure state, let's assume that the initial concentration of ammonia is \(a\), and the initial concentrations of \(NH_4^+\) and \(NH_2^-\) are zero. As the autoionization proceeds, the concentrations at equilibrium can be represented as: \[ [NH_3] = a - 2x\] \[ [NH_4^+] = x\] \[ [NH_2^-] = x\] where x represents the change in concentration of ammonia during the autoionization process.

Plug the equilibrium concentrations into the equation for the equilibrium constant: \[K = \frac{[NH_4^+][NH_2^-]}{[NH_3]^2} = \frac{x^2}{(a - 2x)^2}\] We're given that \(K = 1.8 \times 10^{-12}\). So, \[1.8 \times 10^{-12} = \frac{x^2}{(a - 2x)^2}\]

Since the value of \(K\) is very small, we can assume that the change in concentration of ammonia (\(2x\)) is negligible compared to the initial concentration (\(a\)), so the equation becomes: \[1.8 \times 10^{-12} = \frac{x^2}{a^2}\] We can solve for x: \[x = a\sqrt{1.8 \times 10^{-12}}\]

The concentration of hydroxide ions (\(NH_2^-\)) is equal to x. Therefore, the pOH can be calculated using the following equation: \[pOH = -\log{[NH_2^-]} = -\log{x} = -\log{(a\sqrt{1.8 \times 10^{-12}})}\] Since ammonia has a similar amphoteric behavior to water, we can use the relationship between pH and pOH: \[pH + pOH = 14\] Therefore, we can find the pH by rearranging the equation above: \[pH = 14 - pOH = 14 + \log{(a\sqrt{1.8 \times 10^{-12}})}\] Since we don't have a particular value for the initial concentration of ammonia (\(a\)), we can't determine a numerical value for pH. However, this equation can be used to calculate the pH of ammonia at the standard temperature of the distant cold planet for any given initial concentration.