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%
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