Chapter 13: Problem 66

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

Short Answer

Expert verified

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

Step by step solution

Write the dissociation equation

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

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

Write the Ka expression

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

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: \(x^2 = 0.22 \cdot 1.2 \times 10^{-2}\) \(x^2 = 2.64 \times 10^{-3}\) \(x = \sqrt{2.64 \times 10^{-3}} \approx 5.14 \times 10^{-2}\)