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Experiment 26 Titration Curves of Poly protic Acids To become familiar with cons

ID: 529305 • Letter: E

Question

Experiment 26 Titration Curves of Poly protic Acids To become familiar with consecutive equilibria, acid-dissociation constants, OBJECTIVE APPARATUS Apparatus AND CHEMICALS 150 mL beaker pH meter with electrodes I pt plastic bottle with a plastic lid. 100 mL beakers (3) buret, buret clamp, and ring stand 600 mL beaker 25 ml pipe wash bottle and wire gauze ring ring stand and ring Bunsen burner Chemicals sodium hydroxide solution potassium hydrogen phthalate (o.1 Mor 19 Mstock solution) phenolphthalein indicator unknown solution of a polyprotic solution buffer solution pH 4.0 acid approximately 0.1 M standard solution) I DISCUSSION In this experiment, you will an acid-base titration to determine the perform dissociation constants of an unknown polyprotic acid, HAA. Consider the triprotic acid H3POe. It undergoes the following dissociations in aqueous solution: H1PO4 H20 [HPO4 T1 The acid HsPO4 possesses three ionizeable protons, and for this reason, it is termed a triprotic acid Section 16.6). If you were to perform a titration of H,PO, with NaoH, the following reactions would occur in turn: [4] [5] NaH1PO4 +NaoH 16) Copyright c 2012 Pearson Education, Inc.

Explanation / Answer

From first part

first run

moles of KHP = 0.448/204.22

                       = 0.0022 mol

molarity of NaOH = 0.0022 mol/0.022 L

                             = 0.1 M

First titration data,

acid volume = 25 ml = 0.025 L

pKa1 = 2.115

Ka1 = 7.67 x 10^-3

pKa2 = 5.12

Ka2 = 7.59 x 10^-6

The experiment is condducted to determine the acid dissociation constants for the unknown acid. First the standardization of base NaOH solution is done by titrating against KHP a primary standard. The molarity of NaOH solution was found to be as 0.1 M

When this base was used for titration of unknown acid, the pH at half equivalence point = pKa is recorded.

Therefore the pKa1 and pKa2 as found by the titration are 2.115 and 5.12, respectively.

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