Question

Please assist, as I am working against the clock! Determine the percent dissociation of a 0.18 M solution of hypochlorous acid, HClO. The acid has a Ka value of 3.5 x 10⁻⁸.

a. 7.9 x 10⁻³%
b. 4.4 x 10⁻²%
c. 6.3 x 10⁻⁹%
d. 3.5 x 10⁻⁶%"

Answer 1

The percent dissociation of a 0.18 M hypochlorous acid solution is calculated using its Ka value and is found to be approximately 0.044%.

Step-by-step explanation:

To calculate the percent dissociation of a 0.18 M hypochlorous acid, HClO, solution, we start by writing the dissociation equation for HClO in water:

HClO → H+ + ClO-

At equilibrium, the concentrations of H+ and ClO- are equal, and if x is the amount of HClO that dissociates, we have:

[H+] = [ClO-] = x

[HClO] = 0.18 - x ≈ 0.18 M (since x will be very small compared to 0.18)

Using the acid dissociation constant (Ka), we have:

Ka = (x)(x) / (0.18 - x) = 3.5 x 10⁻⁸

Assuming x<<0.18, this simplifies to:

Ka = x² / 0.18

We can calculate x by:

x = √(Ka × 0.18)

x = √(3.5 x 10⁻⁸ × 0.18) = √(6.3 x 10⁻⁸) ≈ 7.9 x 10⁻⁵

Then, the percent dissociation of HClO is calculated by:

Percent dissociation = (x / 0.18) × 100%

Percent dissociation = (7.9 x 10⁻⁵ / 0.18) × 100% ≈ 0.044%