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Old 04-16-2006, 04:36 PM   #80
shanianwang
Junior Member
 
Join Date: Oct 2005
Posts: 220
Since there's supposed be a lot of equilibrium on the test, here's one:


Calculate the pH at the equivalence point in the titration of 50 mL of .10 M methylamine (CH3NH2) with a .20M HCl.

Kb of CH3NH2 (methylamine): 4.4*10^-4 or Ka of CH3NH3+ : 2.3*10^-11



HINT: This is a weak base, strong acid titration, so you need to eventually find the pH for the hydrolysis of CH3NH3+ with ICE charts and approximation.
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