Horvath Practice Problems 75

# Horvath Practice Problems 75 - This leaves 0.00250.1X mol...

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This leaves 0 . 0025 0 . 1 X mol of niacin and 0 . 1 X mol of C 6 H 4 NO 2 The concentrations acetic acid and acetate ion are 0 . 0025 0 . 1 X 0 . 025 + X and 0 . 1 X 0 . 025 + X respectively. Now we setup an ICE chart [ HC 6 H 4 2 ] £ H + ¤£ C 6 H 4 2 ¤ I 0 . 0025 0 . 1 X 0 . 025+ X 0 0 . 1 X 0 . 025+ X C xxx E 0 . 0025 0 . 1 X 0 . 025+ X xx 0 . 1 X 0 . 025+ X + x Using in the K a we can solve for x which gives the H + concen- tration, K a =1 . 4 × 10 5 = ¡ 0 . 1 X 0 . 025+ X + x ¢ ( x ) ¡ 0 . 0025 0 . 1 X 0 . 025+ X x ¢ . At the equivalence point and equal amount of NaOH has fully reacted with the niacin. The total volume is now 50 mL and there are 0.0025 mol of acetate ion, which is 0 . 0025 0 . 050 =0 . 05 M We use this in an ICE chart for the conjugate base reaction
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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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