Question

Determine the percent dissociation of a 0.18 M solution of hypochlorous acid, HClO. The kₐ for the acid is 3.5 times 10⁻⁸.

A) 3.5 times 10⁻⁶%
B) 6.3 times 10⁻⁹%
C) 7.9 times 10⁻³ %
D) 4.4 times 10⁻² %
E) 1.1 times 10⁻²%

Answer 1

The percent dissociation of a 0.18 M solution of hypochlorous acid, given Kₐ of 3.5 x 10⁻⁸, is calculated to be approximately 4.4 times 10⁻² %, which corresponds to answer choice D.

Step-by-step explanation:

To determine the percent dissociation of a 0.18 M solution of hypochlorous acid (HClO), we first need to write the dissociation equation:

HClO ⇌ H+ + ClO-

For a weak acid like HClO, which partially dissociates in water, the equilibrium expression for its dissociation is given by:

Kₐ = [H+][ClO-] / [HClO]

When the concentration of a weak acid and its dissociation constant (Kₐ) are known, the degree of dissociation (α) can be determined. Assuming x is the concentration of H+ that dissociates, the initial concentration becomes (0.18 - x) for HClO.

Because Kₐ for HClO is very small (3.5 x 10⁻⁸), x can be considered negligible in comparison to the initial concentration, hence the concentration of undissociated HClO is approximately 0.18 M. The Kₐ expression becomes:

Kₐ = x²/0.18

Solving for x:

x = √(Kₐ * 0.18) = √(3.5 x 10⁻⁸ * 0.18) = √(6.3 x 10⁻⁹) = 7.9 x 10⁻⁵ M

The percent dissociation is then calculated using the formula:

% dissociation = (x / initial concentration) * 100

Substituting the values we have:

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

Therefore, the correct answer is D) 4.4 times 10⁻² %.