Problem 32 - Chemistry Matter and Change

Question

Calculate the \(K_{\mathrm{a}}\) for the following acids using the given information. \begin{equation} \text { a. }0.220 M \text { solution of } \mathrm{H}_{3} \mathrm{AsO}_{4}, \mathrm{pH}=1.50 \quad \text { b. } 0.0400 M \text { solution of } \mathrm{HClO}_{2}, \mathrm{pH}=1.80 \end{equation}

Step-by-Step Solution

Answer

For \(\mathrm{H}_3\mathrm{AsO}_4\), \(K_a \approx 5.3 \times 10^{-3}\); for \(\mathrm{HClO}_2\), \(K_a \approx 1.03 \times 10^{-2}\).

Step 1: Understanding the Problem

We are asked to calculate the acid dissociation constant, \(K_a\), for two acids based on concentration and pH values of their solutions. The \(K_a\) is an equilibrium constant that measures the strength of an acid by expressing the concentration of its ions in solution relative to the concentration of the undissociated acid.

Step 2: Finding \( [H^+] \) from pH

Use the pH value to calculate the concentration of hydrogen ions \([H^+]\). For an acid with pH \(1.50\) (part a), calculate: \([H^+] = 10^{-1.50}\). For an acid with pH \(1.80\) (part b), calculate: \([H^+] = 10^{-1.80}\).

Step 3: Setting Up the Equilibrium Expression

For a general acid dissociation reaction \(HA \rightleftharpoons H^+ + A^-\), the equation for \(K_a\) is: \(K_a = \frac{[H^+][A^-]}{[HA]}\). Since \([H^+] = [A^-]\) at equilibrium, substitute \([H^+]\) for \([A^-]\). The initial concentration of the acid \([HA]_0\) minus \([H^+]\) will give \([HA]\) at equilibrium.

Step 4: Calculating \(K_a\) for \(\mathrm{H}_3\mathrm{AsO}_4\) (Part a)

Given, \([H_3AsO_4]_0 = 0.220\, M\) and \(\text{pH} = 1.50\). Substitute:\[ [H^+] = 10^{-1.50} = 0.0316 \text{ M} \] where \([A^-] = [H^+] = 0.0316\, M\) and \([HA] = 0.220 - 0.0316 = 0.1884\, M\).\[ K_a = \frac{(0.0316)(0.0316)}{0.1884} = 5.30 \times 10^{-3} \].

Step 5: Calculating \(K_a\) for \(\mathrm{HClO}_2\) (Part b)

Given, \([HClO_2]_0 = 0.0400\, M\) and \(\text{pH} = 1.80\). Substitute:\[ [H^+] = 10^{-1.80} = 0.0158 \text{ M}\] where \([A^-] = [H^+] = 0.0158\, M\) and \([HA] = 0.0400 - 0.0158 = 0.0242\, M\).\[ K_a = \frac{(0.0158)(0.0158)}{0.0242} = 1.03 \times 10^{-2} \].

Key Concepts

Equilibrium Expressions

When dealing with acids and their dissociation in a solution, equilibrium expressions help us determine the acid dissociation constant, denoted as \( K_a \). This gives us a sense of the acid's strength by showing how much of the acid dissociates into ions in the solution.
The general equation for an acid \( HA \) dissociating in water can be written as:

• \( HA \rightleftharpoons H^+ + A^- \)

The equilibrium expression, or \( K_a \) value, can be derived as follows:

• \[ K_a = \frac{[H^+][A^-]}{[HA]} \]

Here, \([H^+]\) and \([A^-]\) are the concentrations of the ions formed, and \([HA]\) is the concentration of the undissociated acid at equilibrium.
Understanding this expression is crucial because it helps us quantify how strong an acid is by illustrating the degree of ionization in a solution.

pH Calculations

The pH of a solution is a measure of its acidity or basicity. For acids, pH calculations are essential to determine the concentration of hydrogen ions \([H^+]\) in the solution because this concentration directly impacts the \( K_a \) value.
To find this concentration, use the pH equation:

• \[ [H^+] = 10^{-\text{pH}} \]

In the original exercise, for the acid \( H_3AsO_4 \) with a pH of 1.50, we calculated \([H^+]\) as \( 0.0316 \text{ M} \). Similarly, for \( HClO_2 \) with a pH of 1.80, \([H^+]\) comes out to be \( 0.0158 \text{ M} \).
This process reiterates the importance of pH calculations in determining the degree of ionization of an acid, which in turn is a key factor in calculating \( K_a \). Accordingly, these calculations are integral in determining how acidic a solution is.

Acid Strength

Acid strength refers to how easily an acid can donate a proton (\( H^+ \)) in a solution. The acid dissociation constant, \( K_a \), is an indicator of this strength, with higher \( K_a \) values signifying stronger acids that dissociate more in solution.
Consider \( H_3AsO_4 \) and \( HClO_2 \) from our problem. By calculating the \( K_a \) values:

• \( K_a \) for \( H_3AsO_4 = 5.30 \times 10^{-3} \)
• \( K_a \) for \( HClO_2 = 1.03 \times 10^{-2} \)

We notice that \( HClO_2 \) has a higher \( K_a \) value compared to \( H_3AsO_4 \). This indicates that \( HClO_2 \) is a stronger acid than \( H_3AsO_4 \) as it dissociates more in solution.
Understanding the relationship between \( K_a \) and acid strength is crucial for predicting reactions in acids and balancing equations in chemistry. This indicates how much a given acid impacts its encompassing environment chemically, determining its practical applications.