If the acid dissociation constant, Ka, for an acid HA is 8 x 104 at 25°C, what percent of the acid is dissociated in a 0.50 M solution of HA at 25°C?

A

08%

Percent dissociated = 3.92%

Explanation:

The equilibrium of the weak acid, HA, is:

HA ⇄ H⁺ + A⁻

Where Ka = 8x10⁻⁴ is: [H⁺] [A⁻] / [HA]

As both H⁺ and A⁻ comes from the same equilibrium we can write the concentrations of the species as follows:

[H⁺] = X

[A⁻] = X

[HA] = 0.50M - X

Where X is reaction coordinate

Replacing:

8x10⁻⁴ = X² / 0.50-X

4x10⁻⁴ - 8x10⁻⁴X - X² = 0

Solving for X:

X = -0.02M. False solution. There is no negative concentrations

X = 0.0196M. Right solution.

Replacing:

[A⁻] = 0.0196M

Percent of the acid that is dissociated is:

[A⁻] / [HA]₀ * 100

[HA]₀ is its initial concentration = 0.50M

0.0196M / 0.50M * 100

b is the correct answer

mark me brainliest and hope this