Gen Chem Question

[purplepanda]

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A .015M solution of a weak acid is found to be 1.3% ionized what is its ka?
Answer: 2.6 x 10^-5


Some of my thoughts:
Ka = [H+][A-]/[HA] so... it will be [X][X]/[.15-X]
The ionized part is confusing? Does this mean that:

Kw * .013 = [H+][OH-]
SO... 1 x 10^-14 * .013 / [1 x 10^-7] = {[H+]?

I am confused? Thank you in advance!!

 

[OC1314]

Are you sure that is the answer?
I keep getting 2.6 x 10-6

 

[purplepanda]

It's pg. 78 in ACS gen chem. Yeah it's 2.6 x 10^-5, can you show me what your doing though. I am REALLY bad at Eq'm.

 

[iwannabadentist]

the trick to this question is that you can't really use any variables, because there is only one equation (the Ka) equation and you have the Ka as a variable already. So, as you might already know, a strong acid completely ionizes in water and a weak acid only partly ionizes in water. When it completely ionizes it would mean that ALL of the HA dissociates into H+ and A-. When it partially ionizes, would mean that only some of the HA would dissociate into H+ and A-. I guess you have to assume that it is a monoprotic acid because it doesnt state otherwise, but if it was something else like H2A or H3A, the equation would be different. But i guess for this one you have to assume its just HA. So the equation would be:
HA --> H+ + A-
HA --> 0.013(HA) + 0.013(HA)
1.3% dissociation means that 1.3% of HA becomes H+ and 1.3% of HA becomes A-.
so you would set this up a:

Ka = (0.013*0.015)(0.013*0.015)/0.015
= (0.013^2)*0.015
= 0.000169*0.015
=1.7*10^-4 * 1.5*10^-2
=2.6*10^-6

I think 2.6*10^-5 must be a typo.