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GroupProb11-16-06

# GroupProb11-16-06 - CH116 Fall 2006 Prof Sevian Solution to...

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CH116 Fall 2006 Prof. Sevian Solution to Group Problem 11/16/06 a) At this point, [HA] concentration is given and no OH¯ has been added yet. So this is the starting point of the titration. Use the K a reaction because this is a problem of HA dissociating slightly in water to make some H + , and we want to know pH. HA ( aq ) + H 2 O ( l ) ' A¯ ( aq ) + H 3 O + ( aq ) Initial 0.200 M 0 0 Change x + x + x Equilibrium 0.200 – x 0.200 x x Using the equilibrium constant expression: [ ] [ ] [ ] ( )( ) ( ) 4 2 3 10 9 . 1 200 . 0 200 . 0 HA A O H + × = = = x x x K a (which was given as the value for K a ) Solving for x , we get ( ) ( ) 3 4 10 16 . 6 10 9 . 1 200 . 0 × = × = x = [H 3 O + ] Therefore, pH = – log [H 3 O + ] = 2.21 Note that the small x approximation is justified because x turned out to be on the order of 10 –3 , so subtracting x from 0.200 would not make a significant change in the 0.200. b) You can recognize this as the titration midpoint because 0.200 moles of HA were initially present and now 0.100 moles of OH¯ have been added, so it’s halfway to adding 0.200 moles of OH¯ which would be the titration equivalence point.

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GroupProb11-16-06 - CH116 Fall 2006 Prof Sevian Solution to...

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