272_Exp4BC

# 20 ml x x 1 y2 y1 y1 y x2

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Unformatted text preview: it was for the first equivalent point, € but we can still determine where it should appear. Furthermore, we can also determine the pH at this equivalence point. The amount of titrant required to reach the second equivalence point will be equal to the volume required to reach the first equivalence point (15.7 mL). At this point, C2HO4- has been completely deprotonated. Only the oxalate ion remains (C2O4- 2). The concentration of this ion is determined from the amount of oxalic acid (in moles) at the start of the titration divided by the total volume (25.00 mL initially, plus the 15.7 mL needed to reach the first equivalence point, plus the 15.7 mL needed to reach the second equivalence point). We can use an ICE table to find [OH- , and then use this concentration to find pOH and pH: − − C2O4 2 + H 2O ⇔ C2 HO4 + OH − I 0.027 M 0 0 C -x +x + x E 0.027 - x x x K b = 1.6 × 10−10 x2 0.027 − x = [OH ] = 2.07 × 10−6 1.6 × 10−10 = € pOH = 5.67 pH = 14 − pOH = 8.32 This technique can be used to determine the pH at the equivalence point. At points other than the equivalence point, the solution exists as a mixture of the protonated (acid) and € deprotonated (conjugate base) forms; in other words, as a buffer. The pH at any point can then be determined from the Henderson- Hasselbalch equation. At the ½ equivalence point (at the volume halfway to the equivalence point), there is an equal concentration of protonated and deprotonated forms, so the pH = pKa (prove this to yourself with the...
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## This lab report was uploaded on 04/06/2014 for the course CHEM 272 taught by Professor Dr.brooks during the Summer '08 term at Maryland.

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