Question

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 $\mathbf{\text{K}}$ value for the autoionization of ammonia is$\mathbf{1}\mathbf{.8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{12}}$ at the standard temperature of the planet. What is the$\mathbf{\text{pH}}$ of ammonia at this temperature?

Short answer

The$\text{pH}$ of ammonia at the standard temperature of the planet is $5.89$.

Step-by-step solution

Concept Introduction:

In autoionization, an excited atom becomes ionized and transitions to a lower energy state by releasing one of two or more excited electrons with combined energy greater than the atom's ionization energy. Such a process that produces an electron with the energy of a photon of optical wavelength.

Equation for Ammonia:

First, write the balanced equation for ammonia as below.

${\text{NH}}_{\text{3}}+{\text{NH}}_{\text{3}}\iff {\text{NH}}_{\text{4}}^{\text{+}}+{\text{NH}}_{\text{2}}^{-}$

Note that the equilibrium constant for the auto-ionization of ammonia is,

$K=1.8\times {10}^{-12}$

Since, ${\text{NH}}_{\text{3}}$is a solvent, it is not included in the equilibrium expression. Thus, in a neural solution of ammonia,

$\left[{\text{NH}}_4^{+}\right]=\left[{\text{NH}}_2^{-}\right]$

$$\begin{gathered} {\left[{\text{NH}}_4^{-}\right]}^{\text{2}}=1.8\times {10}^{-12} \\ \left[{\text{NH}}_4^{-}\right]=\sqrt{1.8\times {10}^{-12}} \end{gathered}$$

$\begin{array}{rcl}\left[{\text{NH}}_4^{+}\right]& =& \left[{\text{NH}}_2^{-}\right]\\ & =& 1.3\times {10}^{-6}\text{M}\end{array}$

Calculation for pH of Ammonia:

Since the theoretical situation allows${\text{NH}}_{\text{3}}$ to act like water, then it can follow the auto ionization of water,

${\text{H}}_{\text{2}}\text{O}\iff {\text{H}}^{+}+{\text{OH}}^{-}$

Next, write the$\left[{\text{NH}}_4^{+}\right]$ as $\left[{\text{H}}^{+}\right]$. To simplify the autoionization, the balanced equation is,

${\text{NH}}_{\text{3}}\iff {\text{H}}^{+}+{\text{NH}}_2^{-}$

Then, solve the $\text{pH}$of ammonia using the concentration of $\left[{\text{H}}^{+}\right]$which in this case, is ${\text{NH}}_4^{+}$,

$\begin{array}{rcl}\text{pH}& =& {\text{pNH}}_4^{+}\\ & =& -\text{log}(1.34\times {10}^{-6})\\ & =& 5.89\end{array}$

Hence, the$\text{pH}$ of ammonia is $5.89$.