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Chapter 16: Problem 58

The acid-dissociation constant for chlorous acid \(\left(\mathrm{HClO}_{2}\right)\) is \(1.1 \times 10^{-2}\). Calculate the concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{ClO}_{2}^{-}\), and \(\mathrm{HClO}_{2}\) at equilibrium if the initial concentration of \(\mathrm{HClO}_{2}\) is \(0.0200 \mathrm{M}\)

Short Answer

Expert verified
At equilibrium, the concentrations of \(H_3O^+\), \(ClO_2^-\), and \(HClO_2\) are approximately 0.0149 M, 0.0149 M, and 0.0051 M, respectively.

Step by step solution

01

Write the balanced chemical equation

The dissociation of chlorous acid HClO₂ in water is represented by the following balanced chemical equation: \( HClO_2 (aq) \rightleftharpoons H_3O^+(aq) + ClO_2^-(aq) \)
02

Set up an ICE table

An ICE table (Initial, Change, Equilibrium) helps organize the initial concentrations and changes in concentrations as the reaction progresses to equilibrium. For HClO₂, H₃O⁺, and ClO₂⁻ set up the table: ══════╤═══════╤═══════╤═════════╗ │ Initial │ Change │ Equilibrium ══════╪═══════╪═══════╪═════════╣ HClO₂ │ 0.020 │ -x │ 0.020-x H₃O⁺ │ 0 │ +x │ x ClO₂⁻ │ 0 │ +x │ x ╚═════╝ Note: x represents the change in concentration as the reaction approaches equilibrium.
03

Write the expression of the acid-dissociation constant

Using the balanced chemical equation, we can write the acid-dissociation constant (Kₐ) expression: \[ K_a = \frac{[H_3O^+][ClO_2^-]}{[HClO_2]} \]
04

Plug in the values from the ICE table into the Kₐ expression

Substitute the equilibrium concentrations from the ICE table into the Kₐ expression: \[ 1.1 \times 10^{-2} = \frac{x \cdot x}{0.020 - x} \]
05

Solve for x

To find the value of x, we have a quadratic equation. In this case, because the acid-dissociation constant is much greater than the initial concentration, we can approximate the solution by assuming x << 0.020. This makes our equation simpler: \[ 1.1 \times 10^{-2} \approx \frac{x^2}{0.020} \] Solve for x: \[ x \approx \sqrt{(1.1 \times 10^{-2})(0.020)} \] \[ x \approx 0.0149 \, M \]
06

Calculate the equilibrium concentrations

Now that we have the value of x, we can find the equilibrium concentrations of the species: \[ [H_3O^+]_{eq} = x \approx 0.0149 \, M \] \[ [ClO_2^-]_{eq} = x \approx 0.0149 \, M \] \[ [HClO_2]_{eq} = 0.020 - x \approx 0.020 - 0.0149 \approx 0.0051 \, M \] At equilibrium, the concentrations of H₃O⁺, ClO₂⁻, and HClO₂ are approximately 0.0149 M, 0.0149 M, and 0.0051 M, respectively.

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Most popular questions from this chapter

The average \(\mathrm{pH}\) of normal arterial blood is 7.40 . At normal body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{w}=2.4 \times 10^{-14} .\) Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\), and \(\mathrm{pOH}\) for blood at this temperature.

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