determine the percent ionization of the following solutions of formic acid at 25

Question

determine the percent ionization of the following solutions of formic acid at 25 degrees centigrade (C) : a) 0.016 M, b) 5.7 X 10^-4, c)1.75 M

Explanation / Answer

Ka = 1.7 * 10^-4 HCOOH -------> HCOO- + H+ 0.016-X-----------X--------X Ka = 1.7 * 10^-4 = X^2 / 0.016-X => X = 1.566 * 10^-3 M % dissociation = 1.566 * 10^-3 * 100 / 0.016 = 9.79 % b) Ka = 1.7 * 10^-4 = X^2 / 5.7*10^-4 - X => X = 2.38 * 10^-4 M % dissociation = 2.38 * 10^-4 * 100 / 5.7 * 10^-4 = 41.75 % c) Ka = 1.7 * 10^-4 = X^2 / 1.75 - X => X = 0.0172 % dissociation = 0.0172 * 100 / 1.75 = 0.98 %

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