sapling week 2 #2

[dana hu 1B]

Hello,
This is the question I am stuck on:
The Ka of a monoprotic weak acid is 0.00743. What is the percent ionization of a 0.143 M solution of this acid?

I found that x=0.03258 and tried to solve for the % ionization= 0.03258/0.143= 22.8%, but it was wrong. Can anyone identify what I did wrong?
Thank you.

[Victor Li 2A]

Ka = 0.00743 = x^2/(0.143-x)
Use quadratic formula and x = 0.02909

Then x/0.143 * 100% = 20.3%

This is what I got based on the provided information, is your Ka expression correct?

[Melody Wu 2L]

I believe you used the approximation when you were finding x, which is causing the error. Since Ka is larger than 10^-4, you can't use the approximation and have to use the quadratic formula.
Thus you would do 0.00743 = x^2/(0.143-x) to get x and I think the rest of your steps are right :)

[Tracey Huynh 2F]

I agree with what Melody said. Since the Ka is not smaller than 10^-4, approximation won't work, so the quadratic equation does have to be used, in which you can then use the formula for determining % ionization after you get the x value that should correspond with [H3O+].

[Frankie Mele 3J]

In addition to the above statements, make sure you don't approximate x unless your K value is less than 10^-4. I think using the quadratic would be the only other way this problem could be solved for x.

[Anna Yang 1A]

I think it was mentioned that if the x value you find is less than 5% of the initial concentration, then the approximation can be used. But since its greater than 5%, you have to use the quadratic as K is not small enough to approximate it as not having an affect on the initial molarity (i.e. cannot say initial molarity -x is approximately the initial molarity).

[uclakat]

THank you for the help!!!