Problem 96 - Chemistry Matter and Change

Question

What is the \(\mathrm{pH}\) of a 0.200\(M\) solution of hypobromous acid \((\mathrm{HBrO}) ? K_{\mathrm{a}}=2.8 \times 10^{-9}\)

Step-by-Step Solution

Answer

The pH of the 0.200M hypobromous acid solution is approximately 4.63.

Step 1: Write the Ionization Equation

Hypobromous acid (\(\text{HBrO}\)) ionizes in water according to the equation: \(\text{HBrO} \rightleftharpoons \text{H}^+ + \text{BrO}^-\). This represents the dissociation of hypobromous acid into hydrogen ions (\(\text{H}^+\)) and hypobromite ions (\(\text{BrO}^-\)).

Step 2: Set Up the Equilibrium Expression

The equilibrium constant expression for the dissociation of \(\text{HBrO}\) can be written using the acid dissociation constant \(K_a\), which is \(K_a = \dfrac{[\text{H}^+][\text{BrO}^-]}{[\text{HBrO}]}\). Given that \(K_a = 2.8 \times 10^{-9}\), we will use this to find the hydrogen ion concentration.

Step 3: Define Initial Concentrations and Changes

Let the initial concentration of \(\text{HBrO}\) be \([\text{HBrO}]_0 = 0.200\,M\). Initially, \([\text{H}^+] = 0\) and \([\text{BrO}^-] = 0\). At equilibrium, let the change in concentration of \(\text{H}^+\) and \(\text{BrO}^-\) be \(x\), so \([\text{H}^+] = [\text{BrO}^-] = x\) and \([\text{HBrO}] = 0.200 - x\).

Step 4: Substitute into the Equilibrium Expression

Substitute the equilibrium concentrations into the equation: \(K_a = \dfrac{(x)(x)}{0.200-x} = 2.8 \times 10^{-9}\). Simplifying gives \(x^2 = (0.200-x)(2.8 \times 10^{-9})\).

Step 5: Make a Simplifying Assumption

Assume \(x\) is small compared to 0.200, so \(0.200-x \approx 0.200\). Substitute into the simplified equation: \(x^2 \approx 0.200 \times 2.8 \times 10^{-9}\).

Step 6: Solve for \(x\)

Calculate \(x^2 = 5.6 \times 10^{-10}\). Taking the square root, \(x \approx \sqrt{5.6 \times 10^{-10}} \approx 2.37 \times 10^{-5}\). Therefore, \([\text{H}^+] = 2.37 \times 10^{-5}\,M\).

Step 7: Calculate the pH

Use the formula \(\mathrm{pH} = -\log([\text{H}^+])\). Thus, \(\mathrm{pH} = -\log(2.37 \times 10^{-5}) \approx 4.63\).

Key Concepts

Equilibrium Expression

An equilibrium expression is a mathematical equation that relates the concentrations of reactants and products in a chemical reaction at equilibrium. In an aqueous solution, when an acid dissociates, it reaches a point where the rates of the forward and reverse reactions are equal. This balance is captured in the equilibrium expression. For hypobromous acid, the equilibrium expression is crucial for understanding how it ionizes in water. It is written using the acid dissociation constant \(K_a\), reflecting the relationship between the concentrations of hydrogen ions \([\text{H}^+]\), hypobromite ions \([\text{BrO}^-]\), and undissociated hypobromous acid \([\text{HBrO}]\). The expression is:

• \( K_a = \dfrac{[\text{H}^+][\text{BrO}^-]}{[\text{HBrO}]} \)

Understanding this balance helps predict how changes in conditions can shift the equilibrium and affect the pH of the solution.

Acid Dissociation Constant (Ka)

The acid dissociation constant, abbreviated as \(K_a\), is a measure of the strength of an acid in solution. It quantifies the extent of acid ionization by expressing the ratio of the concentration of the ionized form to the non-ionized form.
For hypobromous acid, a weak acid, the given \(K_a\) is \(2.8 \times 10^{-9}\). This small value indicates that only a tiny fraction of \(\text{HBrO}\) molecules dissociate into \(\text{H}^+\) and \(\text{BrO}^-\).

• High \(K_a\) values: Strong acids, more ionization
• Low \(K_a\) values: Weak acids, less ionization

Using \(K_a\), one can predict how an acid behaves in a solution and its resultant hydrogen ion concentration, which is pivotal for calculating the solution's pH.

Hypobromous Acid

Hypobromous acid (\(\text{HBrO}\)) is a weak acid formed when bromine dissolves in water. It is known for its relatively low dissociation in aqueous solutions, meaning it does not release many hydrogen ions compared to strong acids.
\(\text{HBrO}\) plays an important role in understanding weak acids' behavior and properties. In the case of finding pH, knowing how much of \(\text{HBrO}\) ionizes is crucial, as it forms both hydrogen ions \(\text{H}^+\) and hypobromite ions \(\text{BrO}^-\).
A generic equation for its ionization in water is:

• \(\text{HBrO} \rightleftharpoons \text{H}^+ + \text{BrO}^-\)

This reversible process indicates that hypobromous acid partially dissociates, establishing equilibrium that determines the concentrations of each component.

Equilibrium Concentration

Equilibrium concentration refers to the stable concentration of reactants and products present when a reversible reaction has reached equilibrium. For hypobromous acid in a solution, the concentrations of \(\text{H}^+\), \(\text{BrO}^-\), and un-ionized \(\text{HBrO}\) need to be determined.
Suppose you start with an initial concentration of hypobromous acid, \([\text{HBrO}]_0\), and no \(\text{H}^+\) or \(\text{BrO}^-\).

• Change in concentration: Let \(x\) be the concentration of \(\text{H}^+\) and \(\text{BrO}^-\) at equilibrium.
• Equilibrium: \([\text{HBrO}] = 0.200 - x\), \([\text{H}^+] = x\), \([\text{BrO}^-] = x\)

Substitute these values into the equilibrium expression to solve for \(x\), giving the required equilibrium concentrations.

Hydrogen Ion Concentration

Hydrogen ion concentration \([\text{H}^+]\) is a crucial factor in determining a solution's pH. In acid-base chemistry, the smaller the \(pH\), the higher the hydrogen ion concentration, indicating a more acidic solution.
For hypobromous acid, once the equilibrium concentrations are known, solving for \(x\) in the equilibrium expression gives \([\text{H}^+]\), which is the concentration of hydrogen ions at equilibrium.
In our case:

• Equilibrium \([\text{H}^+] = 2.37 \times 10^{-5}\,M\)

Once \([\text{H}^+]\) is calculated, the pH can be determined using the formula \(\text{pH} = -\log([\text{H}^+])\). A quick calculation gives us a pH of approximately 4.63, denoting a weakly acidic solution.