D_C_ Posted December 16, 2006 Share Posted December 16, 2006 What is the final pH of a solution obtained by mixing 200 ml of 0.400 M NH3 with 300 ml of 0.100 M HCl? (Kb = 1.8 x 10^-5) I used an ICE table and came up with 9.68. Is that correct? Should I have used two ICE tables instead of one? The first in the one direction and the second going back in the other direction? Link to comment Share on other sites More sharing options...
chemhelper Posted January 9, 2007 Share Posted January 9, 2007 To begin with, there are initially (0.2 L)(0.4M) = 0.08 mol NH3. (0.3 L)(0.1M) = 0.03 mol HCL is added, which completely ionizes. The 0.3 mol H+ neutralizes 0.03 mol NH3. Thus (0.08 - 0.03 mol) NH3 = 0.05 mol of NH3. The new concentration of NH3 is ( 0.05 mol ) / (0.2L + 0.3L) = 0.1M due to the volume of the HCL The reaction is now NH3(aq) -> NH4+ + OH- Kb = [NH4+][OH-]/[NH3] which yields 1.8 x 10^-5 = x^2 / (0.1-x) Solving for x=0.00133 mol/L and x=[OH-] -log [OH-] = pOH --> -log [0.00133] = 2.875 pOH + pH = 14 thus 14-2.875 = pH = 11.12 ---- Have homework questions in chemistry, physics or math? Who Likes Homework -- http://www.wholikeshomework.com Link to comment Share on other sites More sharing options...
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