Achieve Week 2 Q2

[Teresa Dinh 3L]

The Ka of a monoprotic weak acid is 0.00567. What is the percent ionization of a 0.104 M solution of this acid?

I used a quadratic to solve this. I was wondering if anyone could show me how the ICE chart would look like for this situation?

[Devon Shao 1E]

HA(aq) + H2O(l) ⇌ A-(aq) + H3O+(l)
I 0.104, 0, 0
C -x, +x, +x
E 0.104-x, x, x

Thus, 0.00567 = (x*x)/(0.104-x).

Sorry, I'm not sure how I can better format this :\ but I hope this can make sense :)

[905715668]

You don't need to use the quadratic because it is a weak acid. You can just take out the X from the denominator of your Ka expression.

[905715668]

wait nevermind Achieve just said that assumption is not valid