A weak acid (HA) is found to ionize 20.9%

when mixed at a concentration of 0.22 mol/L.
What is the value of Ka for this acid?

HA ----> A^- + H^+

Ka=[A^-][H^+]/[HA]

....HA........A^-....H^+

I ..0.22 ......0......0
C..-x..........x.......x
E.0.22-x....x.......x

Ka=[x][x]/[0.22-x]

Let x=0.22*(0.209)

Solve for x:

I believe the problem asks for Ka.

it is wrong

Typed something out of habit.

plug in the value for x and solve Ka:

I apologize about that.

I meant to say solve for Ka and not x.

To find the value of Ka (acid dissociation constant) for the weak acid, we can use the equation for percent ionization:

Percent ionization = (concentration of ions formed / initial concentration of acid) × 100

Given that the percent ionization is 20.9% and the initial concentration of acid is 0.22 mol/L, we can substitute these values into the equation:

20.9 = (concentration of ions formed / 0.22) × 100

To solve for the concentration of ions formed, we rearrange the equation:

concentration of ions formed = (20.9/100) × 0.22

Simplifying:

concentration of ions formed = 0.209 × 0.22

concentration of ions formed = 0.04598 mol/L

Now that we know the concentration of ions formed, we can use it to find the value of Ka. Since the weak acid has the generic formula HA, it can be represented as:

HA ⇌ H+ + A-

The Ka expression for this equilibrium is:

Ka = [H+][A-] / [HA]

Given that the concentration of H+ and A- (the ions formed) is 0.04598 mol/L, and the initial concentration of HA is 0.22 mol/L, we can substitute these values into the equation:

Ka = (0.04598)(0.04598) / 0.22

Simplifying:

Ka = 0.002108 / 0.22

Ka ≈ 0.00958

Therefore, the value of Ka for this weak acid is approximately 0.00958.