Problem 66 - Chemistry An Atoms First Approach

Question

Calculate the percent dissociation for a \(0.22-M\) solution of chlorous acid (HClO_, \(K_{\mathrm{a}}=1.2 \times 10^{-2}\) ).

Step-by-Step Solution

Answer

The percent dissociation for the 0.22 M solution of chlorous acid is approximately 23.36%.

Step 1: Write the dissociation equation

First, we write the chemical equation for the dissociation of chlorous acid in water as: \(HClO \rightleftharpoons H^+ + ClO^-\)

Step 2: Define the equilibrium concentrations

Let the equilibrium concentrations of H^+ and ClO^- be x. The equilibrium concentration of HClO will then be (0.22 - x). The progression of the concentrations can be shown in the following table: | HClO | H^+ | ClO^- Initial | 0.22 | 0 | 0 Change | -x | x | x Equilibrium|0.22-x | x | x

Step 3: Write the Ka expression

We can write the Ka expression for the equilibrium as follows: \(K_a = \frac{[H^+][ClO^-]}{[HClO]}\)

Step 4: Substitute values into Ka expression and solve for x

Substitute the equilibrium concentrations from Step 2 into the Ka expression: \(1.2 \times 10^{-2} = \frac{x \cdot x}{0.22 - x}\) To simplify calculations, we can make an assumption that x will be much smaller than 0.22, so we can replace (0.22 - x) with 0.22: \(1.2 \times 10^{-2} = \frac{x^2}{0.22}\) Now, we can solve for x: \(x^2 = 0.22 \cdot 1.2 \times 10^{-2}\) \(x^2 = 2.64 \times 10^{-3}\) \(x = \sqrt{2.64 \times 10^{-3}} \approx 5.14 \times 10^{-2}\)

Step 5: Calculate the percent dissociation

Now, we can use the formula for percent dissociation: Percent Dissociation = \(\frac{[H^+]_{eq}}{[HClO]_{initial}} \cdot 100\%\) Substitute the values of x and the initial concentration of chlorous acid (0.22 M) into the formula: Percent Dissociation = \(\frac{5.14 \times 10^{-2}}{0.22} \cdot 100\% \approx 23.36\% \) The percent dissociation for the 0.22 M solution of chlorous acid is approximately 23.36%.

Key Concepts

Equilibrium Concentrations

In any chemical reaction involving dissociation, equilibrium concentrations refer to the amounts of reactants and products present at equilibrium. Equilibrium is a balanced state where the forward and reverse reactions occur at the same rate. For chlorous acid (HClO) dissociation, we write the equilibrium as: \[ HClO \rightleftharpoons H^+ + ClO^- \]To understand equilibrium concentrations:

• The initial concentration of chlorous acid is 0.22 M.
• We assume initially that no ions are formed, so the initial concentrations of \(H^+\) and \(ClO^-\) are zero.
• As the reaction moves towards equilibrium, concentrations change by certain amounts, often denoted by \(x\).

At equilibrium, we have:

• \( [HClO] = 0.22 - x \)
• \( [H^+] = x \)
• \( [ClO^-] = x \)

These concentrations allow us to calculate the equilibrium state of the solution.

Percent Dissociation

Percent dissociation is a measure of how much of an acid or base dissociates in solution. It shows what fraction of the original substance separates into ions at equilibrium. It is calculated using the formula:\[\text{Percent Dissociation} = \left( \frac{[H^+]_{eq}}{[HClO]_{initial}} \right) \times 100\%\]This calculation is very important:

• It provides insights into the strength of the acid; a higher percent dissociation indicates a stronger acid.
• For chlorous acid, after determining equilibrium concentrations, we found that \( x = 5.14 \times 10^{-2} \), which means \( [H^+]_{eq} = 5.14 \times 10^{-2} \).

Upon substitution, the percent dissociation is:\[\text{Percent Dissociation} = \left( \frac{5.14 \times 10^{-2}}{0.22} \right) \times 100\% \approx 23.36\%\]Thus, about 23.36% of the chlorous acid dissociates in the solution.

Acid Dissociation Constant

The acid dissociation constant, \( K_a \), is a vital component in understanding the strength of an acid. It represents the extent to which an acid can donate protons to the water, forming its conjugate base.
For chlorous acid, \( K_a \) is given as \( 1.2 \times 10^{-2} \). It is used in the equation:\[ K_a = \frac{[H^+][ClO^-]}{[HClO]} \]This equation is crucial for:

• Determining the equilibrium concentrations of the ions in solution.
• Understanding how much of the acid dissociates under given conditions.

In our example, using equilibrium concentrations derived under certain assumptions (\(x \ll 0.22\)), we substitute in to solve for \(x\):\[ 1.2 \times 10^{-2} = \frac{x^2}{0.22} \]Thus,\[ x^2 = 2.64 \times 10^{-3} \]And solving gives,\[ x \approx 5.14 \times 10^{-2} \]This calculation helps us understand the balance and the strength of chlorous acid in water. A larger \( K_a \) value means a stronger acid as more protons are released into the solution.