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

samuel de champlain. in 1608, king henry iv sponsored pierre dugua, sieur de mons and samuel de champlain as founders of the city of quebec with 28 men. this was the second permanent french settlement in the colony of canada.

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