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In a 1.55 m aqueous solution of a monoprotic acid, 4.54% of the acid is ionized. What is the value of its [tex]K_a[/tex]?

Answer :

To find the Ka of the monoprotic acid solution, the percent ionization is used to calculate the concentrations of the ionic and non-ionic forms. Those concentrations are then plugged into the equilibrium expression, resulting in the calculation of the Ka value to be approximately 3.35 × 10⁻³.

To calculate the acid dissociation constant (Ka) for a monoprotic acid in aqueous solution where 4.54% of the acid is ionized, we use the given concentration of the acid solution (1.55 M) and the degree of ionization. For a monoprotic acid, which we can denote as HA, the dissociation can be represented as:

HA ⇌ H⁺ + A⁻

Since 4.54% of the acid is ionized, we can calculate the concentrations of H⁺ and A⁻ at equilibrium. First, we determine the ionized concentration:

[H⁺] = [A⁻] = 1.55 M × 0.0454 = 0.0704 M

Also, the concentration of non-ionized acid would be:

[HA] remaining = 1.55 M × (1-0.0454) = 1.479 M

Then, we apply these values to the equilibrium expression for the dissociation of a weak acid:

Ka = [H⁺][A⁻] / [HA]

Substituting in the calculated concentrations:

Ka = 0.0704 M × 0.0704 M / 1.479 M

Ka = 0.00495776 / 1.479

Ka ≈ 3.35 × 10⁻³

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