A 0.200 M solution of a week acid, HA, is 9.4% ionized. The molar concentration of H+ is 0.0188 M. the Acid-dissociation constant, Ka, for HA is...?​

We are given:

Initial concentration of HA: 0.200 Molar

The acid is 9.4% ionized

Dissociation constant (α) = (Percent Ionized) / 100 = 0.094

Molar concentration of H+ = 0.0188

Let's Chill! (making the ICE box):

Reaction:       HA         ⇄       H⁺      +      A⁻

Initial:           0.200M               -                -

Equilibrium: 0.200(1-α)      0.200α       0.200α

while we're here, let's confirm the given equilibrium concentration of H⁺ ions

from the table here, we can see that the equilibrium concentration of H⁺ ions is 0.200α, we know that α = 0.094

[H⁺] = 0.200α = 0.200 * 0.094 = 0.0188 M

which means that we're on the right track

We're basically scientists at this point (finding the dissociation constant):

Acid dissociation is nothing but the equilibrium constant, but for the dissociation of Acids

From the reaction above, we can write the equation of the acid dissociation constant:

Ka = [H⁺][A⁻] / [HA]

now, let's take the values from the 'equilibrium' row of the ice box the plug those in this equation

Ka = (0.200α)(0.200α) / [0.200(1-α)]

Ka = (0.200α)²/[0.200(1-α)]

plugging the value of α

Ka = (0.200*0.094)² / [0.200(0.906)]

Ka = (0.0188)² / 0.1812

Ka = 1.95 * 10⁻³

the answer is b & c i believe : )