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Unformatted text preview: i. m ionized = m effective-m stated 9. Percent Ionization: • % ionization = (m ionized /m stated ) x 100 a. (0.013 / 0.100) x 100 = 13% Concentrations: M HA that ionizes = % ionized x concentration of solution = X HA + H 2 O = H + + A-Initial: 0.0100M 0 M 0 M Changes: - (X) M + (X) M + (X) M Equilibrium: 0.0100 M – (X) M (X) M (X) M We already know that the % ionized = 0.130 so: Calculate K a : [H + ] [A-] / [HA] K a = [X] [X] / [0.0100 – X] M Ha that ionizes = % ionized x concentration of solution = X X = 0.13 x 0.0100 X = 0.0013 K a = [0.0013] 2 / [0.0100-0.0013] K a = 0.000019 or 1.9 x 10-5...
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- Spring '10
- 0m, dissociation, 13%, 0.0100 M, 0.0100M