Horvath Practice Problems 57

Horvath Practice Problems 57 - C 6 H 4 NO 2 [ HC 6 H 4 NO 2...

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All of the OH reacts with the H + from niacin. Nowth isleaveson lyC 6 H 4 NO 2 , but since this is the conjugate base to a weak acid we must set-up a base dissociation reaction C 6 H 4 NO 2 ­ HC 6 H 4 NO 2 + OH . The K b for this reaction is obtain from the K a for niacin and K w , K b = K w K a =7 . 14 × 10 10 . (14.3) We will need to setup and ICE chart but f rstwemustdeter - mine the concentration of C 6 H 4 NO 2 . To do this we see, 0 . 025 LHC 6 H 4 NO 2 × 0 . 10 mol HC 6 H 4 NO 2 L × 0 . 10 mol C 6 H 4 NO 2 0 . 10 mol HC 6 H 4 NO 2 =0 . 0025 mol C 6 H 4 NO 2 But we also had 25 mL from the NaOH solution so the total volume is now 50 mL thus, £ C 6 H 4 NO 2 ¤ = 0 . 0025 mol C 6 H 4 NO 2 0 . 050 L =0 . 05 MC 6 H 4 NO
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Unformatted text preview: C 6 H 4 NO 2 [ HC 6 H 4 NO 2 ] OH I . 05 C x x x E . 05 x x x . So, we have K b = 7 . 14 10 10 = ( x ) ( x ) (0 . 05 x ) Rearranging and using the quadratic formula we get x = [ OH ] = 5 . 97 10 6 . From this pOH = log 5 . 97 10 6 = 5 . 22 . Hence, pH = 14 5 . 22 = 8 . 78 . The reaction between a weak base and a strong acid would be done in a similar manner. 56 REACTIONS WITH WEAK ACIDS AND WEAK BASES...
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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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