Horvath Practice Problems 79

Horvath Practice Problems 79 - K b 2 which is K b 2 = K w K...

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Thus the concentration of HC 6 H 6 O 6 is 0 . 0025 mol HC 6 H 6 O 6 0 . 050 L =0 . 05 MHC 6 H 6 O 6 This HC 6 H 6 O 6 then goes back towards the acid via the conju- gate base reaction, HC 6 H 6 O 6 + H 2 O ­ H 2 C 6 H 6 O 6 + OH So we need the K b 1 which is K b 1 = K w K a 1 =1 . 27 × 10 10 The ICE chart is £ HC 6 H 6 O 6 ¤£ OH ¤ [ H 2 C 6 H 6 O 6 ] I 0 . 05 0 0 C xxx E 0 . 05 xx x Using the K b 1 equation and solving for x = £ OH ¤ =2 . 52 × 10 6 .Thu s , pH =1 4 ¡ log £ 2 . 52 × 10 6 ¤¢ =8 . 40 Now, for the second equivalence point we repeat the procedure. The second equivalence point occurs at the addition of a total of 50 mL of NaOH. So, £ C 6 H 6 O 6 ¤ = 0 . 0025 mol 0 . 075 L =0 . 033 M. Now we need the
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Unformatted text preview: K b 2 which is K b 2 = K w K a 2 = 6 . 25 × 10 − 3 The appropriate ICE chart for the reaction C 6 H 6 O 2 6 + H 2 O ­ HC 6 H 6 O − 6 + OH − is £ C 6 H 6 O − 6 ¤ £ OH − ¤ £ HC 6 H 6 O − 6 ¤ I . 033 C − x x x E . 033 − x x x Using the K b 2 equation and solving for x we get x = £ OH − ¤ = . 0116 . Thus pH = 14 − ( − log [0 . 0116]) 12 . 06 78 TITRATIONS PART II...
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This document was uploaded on 11/01/2011 for the course CHM 2046 at University of Florida.

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