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CHM110H5 (162)
Judith C Poe (117)
Lecture

CHM-110-Lab4-Report.docx

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Department
Chemistry
Course
CHM110H5
Professor
Judith C Poe
Semester
Fall

Description
P a g e | 1 I. Purpose: The purpose of this experiment was to identify the identity of a given unknown acid which was one of the derivatives of chloro-acetic acid or acetic acid by experimentally determining the valueAof K for the reaction. Then, the value of A of the reaction was calculated by the following equation: [ ][ ] II. Experimental Method: Refer to laboratory manual .  No deviations were taken from the lab manual. III. Collection of data Refer to appendix IV. Results and Calculations The data for HCl titration was obtained from Lily-Ann Diep and Parmvir Banwait. 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 base added (mL) Figure 1 – Titration curve for Hydrochloric acid. From the graph, it appears that the pH at equivalence point is around 7. P a g e | 2 Volume of base added Vs. Change in pH per unit volume 25 20 15 V ∆ 10 pH/ ∆ 5 0 0 5 10 15 20 25 30 35 40 45 50 -5 Volume of base added (mL) Figure 2 – Derivative of pH titration curve of HCl plotted against the volume of base added. It appears that the equivalence point was reached when about 25.45 mL of base was added to the mixture. Titration Curve for Unknown acid # 193 14 12 10 8 pH 6 4 2 0 0 10 20 30 40 50 60 Volume of NaOH added to the mixture (mL) Figure 3 – Titration curve for Unknown acid #193. The equivalence point of this curve is slightly higher than 7 at around 7.5 P a g e | 3 Volume of Base added Vs. change in pH/unit of volume [Unknown acid #193] 40 35 30 25 ∆V0 ∆pH 10 5 0 -5 0 10 20 30 40 50 60 Volume of Base (mL) Figure 4 – Derivative of the titration curve of the unknown acid. The equivalence point is reached when about 25 mL of NaOH was added to the solution in the beaker. 1) Calculating concentration of HCl Since a strong acid (HCl) and a strong base (NaOH) were titrated, the pH at equivalence for HCl is equal to 7 -2  Volume of Base when pH is 7 (V ) B 25.45 mL (2.545 x 10 L)  Initial concentration of Base (C ) = 0.1002 M B -2  Volume of Acid (V ) A 25.00 mL (2.500 x 10 L)  Volume of Solution (V Soln. = 0.02545 L + 0.02500 L = 0.05045 L By using Since the reaction is at equilibrium, [ ] [ ] Therefore, the Initial Concentration of HCl can be calculated by the following method. Thus, the initial concentration of HCl was found to be 0.1020 Mol/L P a g e | 4 2) Calculating concentration of unknown acid From Figure 3, the equivalence point of the unknown acid titration was reached when 24.93 mL of NaOH was added.  Volume of Base at Equivalence point (V ) ≅ 2B.93 mL (2.493 x 10 L) -2  Initial concentration of Base (C ) B 0.1002 M  Volume of Acid (V ) = 25.00 mL (2.500 x 10 L) -2 A  Volume of Solution
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