Lab11_Identification of an Unknown Acid.pdf

The second derivative shows how the rate of change of

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The second derivative shows how the rate of change of the pH also changes as the volume of titrant increases. The second derivative helps us to more accurately determine the exact neutralization point. The neutralization point (value of volume of NaOH) is equivalent to the x-intercept of the second derivative. It is at this point that the rate of change of the pH (the first derivative) acquires a slope of 0 as a result of reaching its peak value. This indicated on the titration curve as the point with the greatest slope. From this value, we can then extract the half equivalence point since it is equal to the point with the x value that is half the x value of the neutralization point; in other words, the volume needed to get the titration to be at the half neutralization point is equal to half the volume of titrant needed to reach the neutralization point. The first neutralization region (the first proton that was being titrated) did not result in an actual x intercept but it did result in 2 local 22.28, 35 22.38, -23 -30.00 -20.00 -10.00 0.00 10.00 20.00 30.00 40.00 21 21.5 22 22.5 23 Δ(Δ pH/ Δ V )/ Δ V (pH/mL 2 ) Volume of NaOH (mL) 1st Neutralization Region of 2nd Derivative 44.5, 6.7 44.71, -0.83 44.93, -5.5 -8.00 -6.00 -4.00 -2.00 0.00 2.00 4.00 6.00 8.00 42.5 43 43.5 44 44.5 45 45.5 46 46.5 Δ(Δ pH/ Δ V )/ Δ V (pH/mL 2 ) Volume of NaOH (mL) 2nd Neutralization Region of 2nd Derivative
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extrema; we can apply the Median Value Theorem there must be a point along the graph with a value of x between the e xtrema’s x values that would result in a near zero value for y. 1?? ?????? ?𝑖??𝑎?𝑖??: 22.28?? ?𝑎?𝐻 + 22.38?? 2 = 22.3?? ?𝑎?𝐻 22.3?? − 0.15?? = 22.15?? ?? 0.20? ?𝑎?𝐻 ?? ?????𝑎?𝑖𝑧? ?ℎ? 1?? ?????? The 2 nd proton titration had a point very close to being an x intercept and it was in between the local extrema of this region. It was 44.71 mL. However, calculating the average results in the following: 44.5?? + 44.93?? 2 = 44.715?? ≈ 44.72?? ≈ 44.71?? 44.71?? − 0.15?? = 44.56 ?? As can be seen above, the actual resulting point that was near a y value ( Δ(Δ pH/ Δ V )/ Δ V (pH/mL 2 )) of 0 had the x value of 44.71mL which only deviates from the average between the extrema by 0.005mL; this is insignificant, so we can say that 44.71mL is the actual value, and thus, 44.56mL is the actual volume of titrant needed to fully neutralize the second proton. 44.56?? − 22.15?? = 22.41?? ?????? ?? ?𝑖??𝑎?? ?ℎ? 2?? ?????? 𝐻𝑎?? ?????𝑎?𝑖𝑧𝑎?𝑖?? ??𝑖?? 1: 22.15?? 2 = 11.075?? ≅ 11.08?? 𝐻𝑎?? ?????𝑎?𝑖𝑧𝑎?𝑖?? ??𝑖?? 2: 22.41?? 2 = 11.205?? ≅ 11.21?? 22.3?? + 11.21?? = 33.51?? The value of 11.21mL represents the volume of NaOH that needs to be added in order to reach the half equivalence point from the first neutralization point (which represents the end of the titration of the first proton and the start of the titration of the second proton). Thus, this volume can be added to the total volume at the first neutralization point in order to find the volume at the second half- equivalence point. Then, in the zoomed-in graphs below, the corresponding pH values of the half- equivalence points are marked and the buffer region is given.
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