We can then setup an ICE chart as[HC6H4NO2][NH3]£C6H4−2¤£+4¤I0.050.0500C−x−xxxE0.05−x0.05−xxxThus,K=2.5×10+4=(x)(x)(0.05−x)(0.05−x)Rearranging a solving forxgivesx=0.0496.So,[HC6H4NO2]=0.0004M[3.0004M£C6H4NO−2¤.0496M£+4¤.0496MWork Group1. What is thefnal pH if 25 ml of a 0.10 M ammonia solu-tion is reacted with 25 ml of 0.10 M HCl? For ammonia
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This document was uploaded on 11/01/2011 for the course CHM 2046 at University of Florida.