Question

Calculate the fraction of acetate ions that undergohydrolysis in an aqueous solution with formalconcentration of 0.1000 M. What value does this fraction takes, if acetic acid is added to the solution so that theacid contributes a formal concentration of 0.1000 M?

Answer 1

Ans: Follow ICE method and Ka of acetic acid is fixed 1.75 * 10^-5

CH₃COOH -------> CH₃COO^- + H⁺

0.1-x M 0.1 + x M x M

Ka = [CH₃COO^-] [H⁺] / [CH₃COOH] = (0.1 + x) (x) / (0.1-x) = 1.75 * 10^-5

So, solving equation for x, x = 1.749 * 10^-5 M.

Now, α = x / M = 1.749 * 10^-5 / 0.1 = 1.749 * 10^-4

Thus, 1.749 * 10^-4 of acetate ion undergo hydrolysis in an aqueous solution.

Next,

CH₃COOH -------> CH₃COO^- + H⁺

0.1 M 0 M 0 M

0.1-x x M x M

Ka = [CH₃COO^-] [H⁺] / [CH₃COOH] = (x) (x) / (0.1-x) = 1.75 * 10^-5

So, solving equation for x, x = 1.134 * 10^-3 M

Now, α = x / M = 1.134 * 10^-3 / 0.1 = 1.134 * 10^-2

Thus, 1.134 * 10^-2 of acetic aicd is added to the solution.