Lab Proposal - i. m ionized = m effective-m stated 9....

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1. Unknown number for acid solution: 2. Freezing point of water: a. Trial 1: 3.65 o C, 265 sec b. Trial 2: 3.63 o C, 259 sec 3. Freezing point of acid solution: a. Trial 1: 3.45 o C, 341 sec b. Trial 2: 3.41 o C, 300 sec 4. Freezing point depression, ∆T f , acid solution a. Trial 1: 3.65 o C- 3.45 o C = 0.20 o C b. Trial 2: 3.63 o C- 3.41 o C = 0.22 o C i. freezing point of water- freezing point of acid solution 5. van’t Hoff factor for the acid solution a. Trial 1: 0.200 o C / 0.186 o C = 1.075 b. Trial 2: 0.220 o C / 0.186 o C = 1.18 i. m/K f = i 6. Average van’t Hoff factor for the acid solution: a. 1.075 + 1.18 / 2 = 1.13 i. Trial 1 + Trial 2 / 2 7. 7. m effective : a. 1.13(0.100) = 0.113 i. m effective = i(m stated )
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8. m ionized : a. 0.113- 0.100 = 0.013
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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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This note was uploaded on 03/23/2011 for the course CHEM1212L CHEM1212L taught by Professor Stanton during the Spring '10 term at University of Georgia Athens.

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Lab Proposal - i. m ionized = m effective-m stated 9....

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