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Lecture

# CHM110 Experiment Lab-4.docx

8 Pages
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Department
Chemistry
Course Code
CHM110H5
Professor
Judith C Poe

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Description
Lab 4 - Identification of an Unknown Acid by pH Titration 29 October 2012 5 November 2012 Purpose: To experimentally determine, through titration, the K of an unknown acid in order to a find out what the acid is. This will be determined by finding the K far the following reaction: - + HA + H O2 A +H O 3 where - + K a [A ][H O3]/[HA] From there, it will be compared with the K vaaues of other acids. Experimental Method: Refer to pages 68-70 in Course Manual for CHM110. In the beginning of the titration of the unknown acid, 0.4 mL of NaOH was added for the first reading instead of 0.25 mL, and proceeded as described in the manual from there. Additionally, in the titration of the unknown acid, the portion where NaOH was supposed to be added drop by drop was skipped. Results: Part 1: Figure 1 Titration Curve for HCl 14 12 10 8 pH 6 4 2 0 0 5 10 15 20 25 30 35 40 45 50 Volume NaOH Added (mL) Figure 2 Titration Curve for Unkown Acid 14 12 10 8 pH6 4 2 0 0 5 10 15 20 25 30 35 40 Volume NaOH Added (mL) Figure 3 pH/V 8 7 6 5 ∆pH 4 ∆ V 3 2 1 0 -1 0 5 10 15 20 25 30 35 40 Average Volume Interval Calculations: 1. pH at equivalence for HCl=7. Based on Figure 1, when pH =7, ≈25.9 mL of NaOH was added. V b 0.0259 L C b 0.1002 M V a 0.0250 L C V =C V b b a a C aC Vb/Vb a C a 0.1002(0.0259)/0.0250 C = 0.103 M a 2. Based Figure 3, the equivalence point was reached at 24.5mL. C a =a V b b C aC Vb/Vb a C a 0.1002M(0.0245L)/(0.025L) C a 0.0981 M 3. K at equivalence: pH at equivalence based on Figure 2 ≈ 8 + [H 3 ] = -log8 -8 = 10 [ ][ ] K a [ ] K w [ ][ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] ( [ ] )[ ] [ ] K a [ ] [ ] - + At Equivalence, [OH ]>>[H O ] becau3e it is a basic solution. [ ] [ ] [ ] [ ] [ ] ( )
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