# 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: x^{2} / (0.0075 - x)

Finally my quadratic equation: -x^{2} - 1.4_{E}^{-14}x + 1.05_{E}^{-6} = 0

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

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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?

Sunil

Mon, 2008-03-31 19:59

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.

spock

Mon, 2008-03-31 21:28

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!

ejswaim

Wed, 2010-03-24 14:42