Problem 103 - Chemistry An Atoms First Approach

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 \(K\) 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?

Step-by-Step Solution

Answer

The pH of ammonia at the standard temperature of the distant cold planet can be calculated using the equation: \[pH = 14 + \log{(a\sqrt{1.8 \times 10^{-12}})}\] where \(a\) is the initial concentration of ammonia. Without a specific value for \(a\), we cannot determine a numerical value for the pH. However, this equation can be used to calculate the pH for any given initial concentration of ammonia.

Step 1: Write down the autoionization of ammonia equation and its equilibrium constant expression

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: \[K = \frac{[NH_4^+][NH_2^-]}{[NH_3]^2} \]

Step 2: Analyze the initial and equilibrium concentrations of the autoionization of ammonia

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.

Step 3: Substitute the equilibrium concentrations in the equation for \(K\)

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, \[1.8 \times 10^{-12} = \frac{x^2}{(a - 2x)^2}\]

Step 4: Make an assumption for simplification and solve for x

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: \[x = a\sqrt{1.8 \times 10^{-12}}\]

Step 5: Find the pOH and pH

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.

Key Concepts

Equilibrium Constant

The equilibrium constant, often represented as \( K \), is a crucial concept in understanding chemical reactions at equilibrium. It measures the ratio of the concentrations of the products to the reactants at equilibrium for a reversible reaction. For the autoionization of ammonia, the equation is:

• \( 2NH_3 \rightleftharpoons NH_4^+ + NH_2^- \)

This equation shows that two molecules of ammonia \((NH_3)\) break down into one ammonium ion \((NH_4^+)\) and one amide ion \((NH_2^-)\). The equilibrium constant expression for this reaction is:

• \[ K = \frac{[NH_4^+][NH_2^-]}{[NH_3]^2} \]

This formula helps us evaluate how the concentrations of the ions relate when the reaction is at a balance. A small \( K \) value, such as \( 1.8 \times 10^{-12} \) (as given for ammonia), indicates that the reactants dominate, meaning very little product forms at equilibrium. Understanding \( K \) provides insight into the reaction's behavior and how it can be manipulated under various conditions.

pH Calculation

Calculating the pH of ammonia can be complex due to its amphoteric nature. The pH is a measure of the acidity or basicity of a solution. For ammonia, the pH computation involves calculating its pOH first, since ammonia in its autoionization produces \( NH_2^- \), akin to hydroxide ions:

• \[ pOH = -\log{[NH_2^-]} \]

Here, \( [NH_2^-] \) is the concentration of the amide ions, equated to \( x \). Knowing that \( pH + pOH = 14 \), we can rearrange this to find the pH:

• \[ pH = 14 - pOH \]

Since the initial concentration \( a \) of ammonia isn't provided, the exact pH can't be determined numerically. However, this method will yield the pH once \( a \) is known. It reflects how ammonia reacts similarly to water, requiring a balance between pH and pOH.

Amphoteric Substances

Amphoteric substances are chemicals that can act as both acids and bases. Ammonia is recognized as amphoteric, which means it can donate a proton (acting as an acid) or accept a proton (acting as a base). This dual capability is crucial on the theoretical planet where ammonia "flows like water."Ammonia's autoionization exhibits this behavior as it can simultaneously produce \( NH_4^+ \) and \( NH_2^- \). This ability to behave like water in terms of autoionization and pH balancing is fascinating. Being amphoteric provides flexibility in reactions, making ammonia essential in diverse chemical environments similar to water on Earth.

Chemical Equilibrium

Chemical equilibrium refers to the state in a closed reaction where the concentrations of reactants and products no longer change with time. For ammonia's autoionization:

• \[ 2NH_3 \rightleftharpoons NH_4^+ + NH_2^- \]

at equilibrium, the rates of the forward and reverse reactions are equal. This balance is carefully measured by the equilibrium constant \( K \).To predict and manipulate reactions, it's vital to understand that reaching equilibrium doesn't mean concentrations are equal, just stable over time. In the cold planet's environment, knowing how ammonia reaches equilibrium helps the inhabitants utilize it effectively in daily life, much like how we make use of water on Earth for various chemical processes. Understanding chemical equilibrium is fundamental to mastering this balance and ensuring efficient energy use and reaction management.