000E00 500E00 100E01 150E01 200E01 250E01 5 6 7 8 9 10 11 12 866 Trial 2

000e00 500e00 100e01 150e01 200e01 250e01 5 6 7 8 9

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0.00E+00 5.00E+00 1.00E+01 1.50E+01 2.00E+01 2.50E+01 5 6 7 8 9 10 11 12 8.66 Trial 2: Titration Curve Volume of NaOH, mL pH Value Figure 2: Titration curve of trial 2 In this graph, the approximate equivalence point is 22mL. The points before the equivalence point are acidic while the points after are basic.
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Another graph that can determine the equivalence point is the first derivative plot, wherein the change of the pH divided by the change of the volume is plotted with the volume of NaOH added in the solution. 0.00E+00 5.00E+00 1.00E+01 1.50E+01 2.00E+01 2.50E+01 0.00E+00 5.00E-01 1.00E+00 1.50E+00 2.00E+00 2.50E+00 3.00E+00 3.50E+00 4.00E+00 4.50E+00 4.16E+00 Trial 2: First Derivative Plot Volume of NaOH, mL ∆pH/∆V Figure 3: First derivative plot This graph gives a more accurate value of the equivalence point. In the equivalence point, the DpH/DV has the largest value, while the pre and post-equivalence DpH/DV values have almost equal changes. Lastly, the second derivative plot was then graphed to achieve a more accurate value of the equivalence point. 0.00E+00 5.00E+00 1.00E+01 1.50E+01 2.00E+01 2.50E+01 -8 -6 -4 -2 0 2 4 6 8 6 -6.32 Trial 2: Second Derivative Plot Volume of NaOH, mL ∆(∆pH)/∆V^2 Figure 4: Second Derivative plot
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As shown in the graph, a line passes through the x-axis, and the value in which y=0 is determined as the equivalence point. Since the graph doesn’t show the value of x when y=0, another graph was used to obtain the most accurate value of the equivalence point. In this case, equivalence point takes place when there are 21.7 mL of NaOH added in the analyte solution as shown in the graph below.
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  • Fall '17
  • Sir Jaden Smith
  • pH, KHP, Potentiometry

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