# Percent Ionization of a Weak Acid

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```					Percent Ionization of a Weak Acid To determine percent ionizations, i.e. what percent of acid molecules ionize, the following method applies. Example: What is the percent ionization of the acetic acid (CH3COOH) in a 2.0M aqueous solution? Ka (acetic acid) = 1.8 x 10-5 CH3COOH + H2O ⇄ H3O+ + CH3COOStart: 2.0M -1 x 10-7 M 0 Rxn: -x -+x +x Equil: (2.0-x) M xM xM [H3O+][CH3COO-] (x)(x) Ka = -------------------------- = ---------- = 1.8 x 10-5 [CH3COOH] (2.0-x) x2 = 1.8 x 10-5(2.0-x) x2 = 3.6 x 10-5 - (1.8 x 10-5)x x2 + (1.8 x 10-5)x - 3.6 x 10-5 = 0 (a Quadratic Equation) -b ± √ b2 - 4ac Quadratic solution is: x = -------------------------2a = -1.8 x 10-5 ± √ (1.8x10-5)2 -4(1)(-3.6x10-5) ---------------------------------------------------2(1) = -1.8 x 10-5 ± √ 3.24x10-10 + 1.44x10-4 ------------------------------------------------2

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x = -1.8 x 10-5 ± √ 1.44x10-4 --------------------------------2 = (-1.8 x 10-5 + 0.0120)/2 = 0.005991 So, x = 0.005991, thus [H3O+] at equilibrium = 0.0060 M 0.0060 M Percent ionization = -------------- x 100 = 0.30% 2.0 M Compare this to 0.42% for a 1.0 M acetic acid soultion (Table 10.3, page 260)

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Lingjuan Ma