Problem 76 - Chemistry

Question

The \(\mathrm{pH}\) of a \(0.063 \mathrm{M}\) solution of hypobromous acid (HOBr but usually written \(\mathrm{HBrO}\) ) is 4.95. Calculate \(K_{\mathrm{a}}\).

Step-by-Step Solution

Answer

The acid dissociation constant (Ka) for hypobromous acid (HBrO) is approximately \(2.82 \times 10^{-9}\).

Step 1: Determine the concentration of H+ ions

Since we know the pH of the solution, we can calculate the concentration of H+ ions, because \(\mathrm{pH} = -\log_{10}([\mathrm{H}^+])\). Thus, the concentration of H+ ions is given by: \[[\mathrm{H}^+] = 10^{-\mathrm{pH}}\] For our problem, \[[\mathrm{H}^+] = 10^{-4.95}\]

Step 2: Setup the chemical equation and the expression for Ka

For the dissociation of hypobromous acid, we can write the chemical equation as: \[\mathrm{HBrO} \longrightarrow \mathrm{H}^+ + \mathrm{BrO}^-\] Initially, we have \(0.063 \, \mathrm{M}\) of HBrO. The HBrO will lose some concentration, let's call it 'x', and generate an equal concentration of H+ and BrO- ions in the process: \([\mathrm{H}^+] = x\) and \([\mathrm{BrO}^-] = x\). Since some H+ ions came from the ionization of water, the effective concentration from HBrO dissociation would be \((x - 10^{-4.95})\). The expression for Ka is \[K_{a} = \frac{[\mathrm{H}^+][\mathrm{BrO}^-]}{[\mathrm{HBrO}]}\]

Step 3: Substitute the values into the expression for Ka

In our problem, we can substitute the values and the expressions obtained in the previous steps into the Ka expression: \[K_{\mathrm{a}} = \frac{(x - 10^{-4.95})(x)}{(0.063 - x)}\]

Step 4: Make an approximation to simplify the expression

We can assume that since the \(\mathrm{pH}\) is 4.95 (meaning that HBrO is a weak acid), the dissociation of the acid is small. Therefore, we can approximate that \(x \ll 0.063\), and the expression becomes: \[K_{\mathrm{a}} \approx \frac{(x - 10^{-4.95})(x)}{0.063}\]

Step 5: Solve for Ka

Now, we can replace the value of \(x\) as \(x = 10^{-4.95}\) and solve for Ka: \[K_{\mathrm{a}} \approx \frac{(10^{-4.95} - 10^{-4.95})(10^{-4.95})}{0.063}\] Simplifying, we get: \[K_{\mathrm{a}} \approx 2.82 \times 10^{-9}\] Thus, the acid dissociation constant (Ka) for hypobromous acid (HBrO) is approximately \(2.82 \times 10^{-9}\).

Key Concepts

Hypobromous Acid

Hypobromous acid, chemically known as HBrO, is an oxyacid of bromine. It's important to understand its role as a weak acid, which means it does not completely dissociate in water. This partial ionization is crucial when calculating its acid dissociation constant, often represented as \( K_a \).
Due to its weak acidic nature, hypobromous acid exists predominantly in a non-ionized form in aqueous solution. As a result, it barely contributes hydrogen ions \( (\mathrm{H}^+) \) to the overall acidic environment, except through its equilibrium dissociation process.
Understanding the behaviors of weak acids like HBrO in a solution is essential for predicting their impact on chemical reactions, particularly in buffered solutions where maintaining pH balance is vital.

pH Calculation

The pH calculation is a foundational concept in chemistry, helping us measure the acidity or basicity of an aqueous solution. pH is calculated using the formula:

• \( \text{pH} = -\log_{10}([\mathrm{H}^+]) \)

For hypobromous acid with a pH of 4.95, this formula allows us to backtrack and find the exact concentration of hydrogen ions in the solution.
By reversing the process of calculating pH, we use the formula:

• \([\mathrm{H}^+] = 10^{-\text{pH}}\)

This inverse calculation is essential not only for identifying hydrogen ion concentrations but also as a stepping stone towards calculating the acid dissociation constant \( (K_a) \) for weak acids in equilibrium.

Chemical Equilibrium

Chemical equilibrium refers to the state where the concentrations of reactants and products in a reversible reaction no longer change. This balance is especially notable in weak acids, like hypobromous acid, where the ionization is incomplete.
For HBrO, the equilibrium can be expressed as:

• \( \mathrm{HBrO} \longrightarrow \mathrm{H}^+ + \mathrm{BrO}^- \)

At equilibrium, the concentration of HBrO that's dissociated into \( \mathrm{H}^+ \) and \( \mathrm{BrO}^- \) ions is at a constant ratio determined by the acid dissociation constant \( K_a \).
Achieving equilibrium is crucial in various chemical processes because it allows predictions of concentrations and gives insights into the extent of reactions at given conditions. Equilibrium lays the groundwork for calculating the \( K_a \), providing valuable insights into the acid strength and behavior in solution.

Weak Acids

Weak acids, such as hypobromous acid, are characterized by their partial ionization in water. Unlike strong acids that fully dissociate, weak acids only release a fraction of their hydrogen ions into the solution, resulting in an incomplete ionization.
The degree of ionization in weak acids can be represented quantitatively with the acid dissociation constant, \( K_a \). For HBrO, this constant is relatively low, emphasizing its weak acidic properties. The formula used for these calculations is:

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

Understanding weak acids requires considering both their chemical equilibrium as well as factors affecting the dissociation, such as pH and concentration. These details help in applications where sensitive pH adjustments are needed, like in pharmaceuticals and environmental science.