Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of ea

Question

Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka= 1.3 x 10-5, Appendix D:)


Part A. 7.93

Explanation / Answer

A)since Ka = [C2H5COO-][H*]/[C2H5COOH] [C2H5COO-]=[H*]=x , [C2H5COOH] = 7.93×10^-2 - x => x^2 / 7.93×10^-2 - x = 1.3*10^-5 , solving gives x = 1.009*10^-3 percent ionization = x/7.93×10^-2 *100 = 1.272 % ionization B) using the same procedure as above gives [C2H5COO-]=[H*]=x , [C2H5COOH] = 2.04×10^-2 - x => x^2 / 2.04×10^-2 - x = 1.3*10^-5 , solving gives x = 5.085*10^-4 percent ionization = x/2.04×10^-2 *100 = 2.493 % ionization

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