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Horvath Practice Problems 78

# Horvath Practice Problems 78 - NaHC 6 H 6 O 6 NaOH ­ Na 2...

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18 TITRATIONS PART II With lecture and the group work problems we have tackled titrations with monoprotic acids and bases. Now we will complete the picture with the titration of weak polyprotic acids (or bases) with strong bases (or acids) Example : Calculate the titration curve when 25 ml of a 0.10 M solution of vitamin C (H 2 C 6 H 6 O 6 ) is titrated with 0.10 M NaOH. The K a 1 = 7 . 9 × 10 5 and K a 2 = 1 . 6 × 10 12 . Now the titration process will experience two equivalence points. Because K a 1 À K a 2 we take separate the process into two parts: Reaction with the fi rst equivalent of acid H 2 C 6 H 6 O 6 + NaOH ­ NaHC 6 H 6 O 6 + H 2 O and the reaction with the second equivalent of acid
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Unformatted text preview: NaHC 6 H 6 O 6 + NaOH ­ Na 2 C 6 H 6 O 6 + H 2 O The problem is now like what we have done for monoprotic acids so we will not go through every detail. Instead, we will simply calculate the pH at each equivalence point. The f rst equivalence point is controlled by the K a 1 = 7 . 9 × 10 − 5 reaction. So we take the NaOH to fully react with the f rst proton of vitamin C. This occurs at 25 mL of NaOH added. Or, . 025 L × . 10 mol HC 6 H 6 O − 6 1 L = 0 . 0025 mol HC 6 H 6 O − 6 . TITRATIONS PART II 77...
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