Question

In a 0.737 m solution, a weak acid is 12.5% dissociated. calculate ka of the acid.

Answer 1

The Ka of the acid is approximately 0.021.

Step-by-step explanation:

The dissociation of a weak acid can be represented by the equation:

HA ⇌ H⁺ + A⁻

In this case, the weak acid is 12.5% dissociated, which means that the concentration of the weak acid is 12.5% of the initial concentration. Let's assume the initial concentration of the weak acid is x M. So, the concentration of the weak acid is 0.125x M, and the concentration of the dissociated ions (H⁺ and A⁻) is also 0.125x M.

The ionization constant, Ka, can be calculated using the following equation:

Ka = [H⁺][A⁻] / [HA]

Substituting the values we obtained from the dissociation:

Ka = (0.125x)(0.125x) / (0.737x)

Simplifying the equation:

Ka = 0.015625x² / 0.737x

Cancelling out x:

Ka = 0.015625 / 0.737

Ka ≈ 0.021

Therefore, the ka of the acid is approximately 0.021.

Answer 2

Hello! Let me try to answer this :)

Thanks and please correct if there are any mistakes ^ ^