Question

Calculate the \(\mathrm{pH}\) of a \(0.42 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) solution. \(\left(K_{\mathrm{b}}\right.\) for ammonia \(\left.=1.8 \times 10^{-5} .\right)\)

Short answer

The pH of the 0.42 M NH4Cl solution is approximately 5.32.

Step-by-step solution

Understand the Problem

We need to calculate the pH of a 0.42 M solution of \(\text{NH}_4\text{Cl}\). This salt dissociates in water to form \(\text{NH}_4^+\) and \(\text{Cl}^-\). The \(\text{Cl}^-\) ion is neutral in terms of acidity, while \(\text{NH}_4^+\) is the conjugate acid of \(\text{NH}_3\). Hence, we will focus on the hydrolysis of \(\text{NH}_4^+\) to find the pH.

Calculate \(K_a\) for \(\text{NH}_4^+\)

To find the \(\text{pH}\), we first need to calculate the acid dissociation constant \(K_a\) for \(\text{NH}_4^+\). Using the relation \(K_w = K_a \times K_b\) where \(K_w = 1.0 \times 10^{-14}\), the \(K_a\) can be calculated as: \(K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\).

Set Up the Equilibrium Expression

As \(\text{NH}_4^+\rightarrow \text{NH}_3 + \text{H}^+\) in water, we can use the ICE table (Initial, Change, Equilibrium) to find \([\text{H}^+]\). Initially, \([\text{NH}_4^+] = 0.42 \text{ M}\), \([\text{NH}_3] = 0\), and \([\text{H}^+] = 0\). At equilibrium, \([\text{NH}_4^+] = 0.42 - x\), \([\text{NH}_3] = x\), and \([\text{H}^+] = x\).

Solve for \(x\)

From \(K_a = 5.56 \times 10^{-10}\), we have \(K_a = \frac{x^2}{0.42 - x}\approx \frac{x^2}{0.42}\). Solving for \(x\), we get \(x = \sqrt{K_a \times 0.42} = \sqrt{5.56 \times 10^{-10} \times 0.42}\approx 4.84 \times 10^{-6}\text{ M}\).

Calculate \(\text{pH}\)

Now that we have \([\text{H}^+] = 4.84 \times 10^{-6}\text{ M}\), we can find the \(\text{pH}\) by using the formula \(\text{pH} = -\log([\text{H}^+]) = -\log(4.84 \times 10^{-6}) \approx 5.32\).

Key concepts

Acid Dissociation Constant

The acid dissociation constant, often represented as \(K_a\), is a crucial parameter in understanding the strength of an acid in solution. It measures how well an acid releases its hydrogen ions \((H^+)\) in water. The larger the \(K_a\), the stronger the acid, as it dissociates more in solution. To calculate \(K_a\) for a given compound, one needs to know the equilibrium concentrations of the species involved in the dissociation process.

In the case of ammonium ion \((\text{NH}_4^+)\), which forms from dissociation of ammonium chloride \((\text{NH}_4\text{Cl})\), we can use the relationship between \(K_a\), \(K_b\) (base dissociation constant), and \(K_w\) (ionic product of water):

• \(K_w = K_a \times K_b = 1.0 \times 10^{-14}\)

This formula allows the calculation of \(K_a\) when \(K_b\) is known. For ammonia \((\text{NH}_3)\), \(K_b = 1.8 \times 10^{-5}\), and thus \(K_a\) for \(\text{NH}_4^+\) can be found using:
\[ K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10} \]
This \(K_a\) value will be used to understand the degree of hydrolysis and the resulting \(\text{pH}\) of the solution.

Equilibrium Expression

The equilibrium expression is a mathematical representation of the concentrations of reactants and products in a chemical reaction at equilibrium.
This is essential in determining how a chemical system behaves when two opposing processes, such as forward and backward reactions, balance each other out.
For any general chemical equation:

• \(aA + bB \rightleftharpoons cC + dD\)

The equilibrium expression \(K\) is given by:
\[K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]
In the hydrolysis of ammonium ion \((\text{NH}_4^+)\), we start the process with an initial concentration of 0.42 M. As the equilibrium is reached, \(\text{NH}_4^+\) dissociates to form \(\text{NH}_3\) and \(H^+\):
\[\text{NH}_4^+ \rightleftharpoons \text{NH}_3 + \text{H}^+\]
An ICE (Initial, Change, Equilibrium) table is constructed to visualize the concentration changes. At initial stage, \(\text{NH}_3\) and \(H^+\) have 0 concentrations.
Then, at equilibrium, we have \([^+] - x\), \(x\), and \(x\) for the concentrations respectively. \(K_a\) is thus expressed as:
\[K_a = \frac{x^2}{0.42 - x}\]
Assuming \(x\) is very small compared to 0.42, we simplify to:
\[K_a = \frac{x^2}{0.42}\]
This simplification speeds up the calculation of equilibrium concentrations, critical in solving for the \(\text{pH}\).

Ammonium Chloride Solution

Ammonium chloride \((\text{NH}_4\text{Cl})\) dissolves in water to yield ammonium ions \((\text{NH}_4^+)\) and chloride ions \((\text{Cl}^-\)). While the chloride ion does not affect the acidity of the solution, the ammonium ion undergoes hydrolysis, significantly impacting the \(\text{pH}\).

Ammonium chloride solutions are often used in chemistry labs as controlled environments for studying acid-base equilibria due to their predictable behavior and the involved ionic reactions. They serve as excellent models for illustrating how salts of weak bases and strong acids can create acidic solutions because the ammonium ion releases hydrogen ions \((H^+)\) when it reacts with water.

• This reaction adds to the \(H^+\) concentration, making the solution acidic.
• The extent of this increase in \(H^+\) directly impacts the measured \(\text{pH}\) of the solution.

Understanding the properties of ammonium chloride helps in tackling broader concepts of salt hydrolysis and acidity.

Hydrolysis of Ammonium Ion

The hydrolysis of the ammonium ion \((\text{NH}_4^+)\) refers to its interaction with water, leading to a release of hydrogen ions \((H^+)\). This process occurs because \(\text{NH}_4^+\) is the conjugate acid of ammonia \((\text{NH}_3)\), a weak base. Due to this relationship, \(\text{NH}_4^+\) hydrolyzes according to the following reaction:
\[\text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_3 + \text{H}_3\text{O}^+\]
This transformation is key to understanding the nature of acidic solutions formed by salts of weak bases like ammonium chloride.

The equilibrium shift of this hydrolysis can be quantitatively analyzed using the previously mentioned equilibrium expression. This helps calculate the resulting concentration of \(H^+\) ions in solution, which is then used to determine the \(\text{pH}\) value.

• In essence, the degree of hydrolysis helps predict how much \(H^+\) is present.
• This prediction is crucial in various chemical applications and laboratory experiments.

Knowing this concept is particularly important in the fields of biochemistry and environmental science, where acid-base interactions play a significant role.