Acid-Base Buffers

Acid-Base Buffers - Date Performed: 3/20/06 Date Submitted:...

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Date Performed: 3/20/06 Date Submitted: 3/27/06 Instructor: Daniel Schoonover Acid-Base Buffers Objective The procedure of this lab is designed to ensure one understands how to determine the ionization constant of acetic acid. The experiment is intended as a study of the properties of acid-base buffers. Experimental Data pH probe calibrated with buffer solutioin to be a pH of 7.00 at a pH of 8.31 Part A: HC 2 H 3 O 2(aq) H + (aq) + C 2 H 3 O 2 - (aq) Initial concentration of acetic acid 1.00 M Solutions are made with different amounts of acetic acid and water to obtain the given molarities: Solution Concentration of acetic acid (M) pH value A 1.00 2.33 B 0.50 2.37 C 0.20 2.50 D 0.10 2.63 E 0.05 2.74 Ka of acetic acid given in text book 1.8 x 10 -5 Part B: Sodium acetate is a weak base Initial concentration of sodium acetate 0.40 M Acetic acid is a weak acid Initial concentration of acetic acid 0.40 M
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HCl is a strong acid Initial concentration of HCl 6.00 M NaOH is a strong base Initial concentration of NaOH 6.00 M Solution A contains an equal volume of the weak acid and weak base. Solution B contains 5.00 mL of both the weak acid and weak base and 10.0 mL of deionized water. Solution C contains 5.00 mL of deionized water. Solution addition 1 pH A0 nothing 4.50 A1 HCl 3.21 A2 NaOH 4.99 Solution addition pH B0 nothing 4.49 B1 HCl 1.42 B2 NaOH 11.9 Solution addition pH C0 nothing 4.62 C1 HCl 1.36 C2 NaOH 12.3 1 Addition refers to adding 4 drops of the stated compound for all the charts.
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Sample calculation Part A: Acetic acid concentration of solution B = (1.00 M x 25.0 mL) acetic acid 50.0 mL solution = 0.500 M acetic acid Equilibrium concentration H + in solution B = 10 -pH = 10 -2.37 = 4.27 x 10 -3 M Equilibrium concentration of acetic acid in solution B = 0.500 M – 4.27 x 10 -3 M = 0.496 M Dissociation constant (Ka) of solution B = (4.27 x 10 -3 ) 2 M 0.50 M = 3.65 x 10 -5 Average Ka value = (2.19 x 10 -5 + 3.65 x 10 -5 + 5.00 x 10 -5 + 5.50 x 10 -5 + 6.62 x 10 -5 ) / 5 = 4.59 x 10 -5 Standard deviation of average value = √((∑ i=1 i=4 (Ka i -Average Ka) 2 ) / (5-1)) = ± 1.71 x 10
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Acid-Base Buffers - Date Performed: 3/20/06 Date Submitted:...

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