Horvath Practice Problems 53

# Horvath Practice Problems 53 - Usually Ka1 Ka2 Ka3 . For...

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Usually K a 1 À K a 2 À K a 3 . For polyprotic acids (or bases) we must consider separate equi- librium reactions H 2 A ­ H + + HA (13.6) HA ­ H + + A 2 where, for this case, K a 1 = £ H + ¤£ HA ¤ [ H 2 A ] , (13.7) K a 2 = £ H + ¤£ A 2 ¤ £ HA ¤ Example 1 : Calculate the pH of a 0.012 M solution of niacin (HC 6 H 4 NO 2 ). The K a =1 . 4 × 10 5 . We f rst write the appropriate equilibrium reaction HC 6 H 4 NO 2 ­ H + + C 6 H 4 NO 2 . (13.8) Next we setup an ICE chart, [ HC 6 H 4 NO 2 ] £ H + ¤£ C 6 H 4 NO 2 ¤ I 0 . 012 0 0 C xx x E 0 . 012 xx x . From K a we get the equation, K a =1 . 4 × 10 5 = ( x )( x ) (0 . 012 x ) (13.9) Rearranging and solving for x using the quadratic formula we get x = £ H + ¤ =4 . 1 × 10 5 .
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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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