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percent ionization for butanoic acid using Ka value

(a) Calculate the percent ionization of 0.0075 M butanoic acid (Ka = 1.5 x 10-5).

I did this problem, but answer didn't match the book.

I set up ice chart and everything, but I don't how to use the table feature in here to show it to you. I used the equation:

HC4H7O2 H+ + C4H7O2-

For my equilibrium terms, I got (0.0075 - x), x, x, respectively.

For my Ka expression, I had: x2 / (0.0075 - x)

Finally my quadratic equation: -x2 - 1.4E-14x + 1.05E-6 = 0

x = 0.001024695 M, and percent ionization = (0.001024695 M /0.0075 M) = 13.66%

Also, I have a question about the following problem too:

"Explain why a mixture of HC2H3O2 and NaC2H3O2 can act as buffer while a mixture of HCl and NaCl cannot."

My answer was that a buffered solution contains a weak conjugate acid-base pair, not strong acid like HCl. I have a feeling that it wont be enough on a AP exam. Is it correct?

Based on the [HA] and the Ka value I'm coming up with a [H+] of 3.28E-04.

I'd check out out your quadratic formula, I think it should be:
                    x2    +  1.5x10-5 x  - 1.125E-7 = 0
That will lead the concentration of H+ above.

I think your answer about the buffers isn't bad.  You might improve it by stating that since acetic acid is a weak acid, its conjugate base is relatively strong and thus able to neutralize added acid.  HCl is a strong acid and its conjugate base is very weak and thus has little ability to neutralize any additional acid.

For this equation if you get messy quadratic formula numbers that means your x value is very very tiny.
In this case, you can just drop the the x in .0075-x.
Then you're left with x2 / (0.0075)= 1.5 x 10-5 square root both sides to findyour x.
x/.0075 = squareroot 1.5 X10^-5
thats how youll find your x value! hope this helps!




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