Chapter 14: Problem 76

Calculate the percent dissociation for a \(0.22-M\) solution of chlorous acid \(\left(\mathrm{HClO}_{2}, K_{\mathrm{a}}=0.012\right)\)

Short Answer

Expert verified

The percent dissociation for a 0.22 M solution of chlorous acid (HClO2) is approximately 23.36%.

Step by step solution

Write the acid dissociation equation for HClO2

The dissociation of chlorous acid in water can be represented by the following equation: \( HClO_2 \rightleftharpoons H^+ + ClO_2^- \)

Set up the equilibrium expression using the given Ka value

We can write the equilibrium expression for the above reaction as: \( K_a = \frac{[H^+][ClO_2^-]}{[HClO_2]} \) Since Ka is given as 0.012, we can plug it into the equation: \(0.012 = \frac{[H^+][ClO_2^-]}{[HClO_2]} \)

Set up an ICE table and calculate the concentrations at equilibrium

Let x represent the concentration of ions that dissociate from HClO2. ICE table: | | HClO2 | H+ | ClO2- | | --- | --- | --- | --- | | Initial | 0.22 | 0 | 0 | | Change | -x | +x | +x | | Equilibrium | 0.22-x | x | x | Now, we can plug these values into the equilibrium expression: \( 0.012 = \frac{x^2}{0.22-x} \)

Solve the equation for x, the concentration of dissociated ions

Make the assumption that x is much smaller than 0.22, so the equation can be simplified as: \( 0.012 = \frac{x^2}{0.22} \) Solve for x: \( x^2 = 0.012(0.22) \) \( x^2 = 0.00264 \) \( x = 0.0514 \) So, the concentration of the dissociated ions (H+ and ClO2-) is approximately 0.0514 M.