Week 15
Mon 12/11: Notes 14.8-14.9 Due Wednesday Tues 12/12: In Class Problems Ch 14 #111,113,114,115,118,125 Wed 12/13: In Class Problems Ch 15 #15,17,31 Thurs12/14: Work on FRQs Fri 12/15: Try this FRQ problem do what you can and then work on the old FRQ handout 1) average score 4.5-5
a) three points CH3NH2 + H2O <===> CH3NH3+ + OH¯ Kb = ([CH3NH3+] [OH¯]) ÷ [CH3NH2] = 5.25 x 10¯4 CH3NH2 CH3NH3+ OH¯ I 0.225 0 0 C -x +x +x E 0.225 - x x x 5.25 x 10¯4 = [(x) (x)] / (0.225 - x) neglect the minus x to get 5.25 x 10¯4 = x2 / 0.225 x = [square root]((5.25 x 10¯4) (0.225)) [OH¯] = 1.09 x 10¯2 (Note: quadratic gives 1.06 x 10¯2) b) three points [CH3NH3+] = 0.0100 mol / 0.120 L = 0.0833 M 5.25 x 10¯4 = [(0.0833 + x) (x)] / (0.225 - x) = 0.0833x / 0.225 x = [OH¯] = 1.42 x 10¯3 mol/L pOH = 2.85 pH = 11.15 alternate solution using the Henderson-Hasselbalch Equation pH = pKa = log ([base] / [acid]) pKw = pKa + pKb pKa = 10.7 pH = 10.7 + log (0.225 / 0.0833) pH = 11.15 The solution using the pOH form is left to you, gentle reader. c) two points HCl must be added. 5.25 x 10¯4 = [(0.0833 + x) (0.0010)] / (0.225 - x) 1.18 x 10¯4 - 5.25 x 10¯4x = 8.33 x 10¯5 + 1.0 x 10¯3x x = 0.0228 mol/L 0.0228 mol/L x 0.120 L = 2.74 x 10¯3 mol HCl alternate solution based on Henderson-Hasselbalch Equation 11.00 = 10.72 + log ([base] / [acid]) log ([base] / [acid]) = 0.28 [base] / [acid] = 1.906 = (0.225 - x) / (0.0833 + x) x = 0.0227 mol / L 0.0227 mol / L x 0.120 L = 2.73 x 10¯3 mol HCl d) one point The [CH3NH3+] / [CH3NH2] ratio does not change in the buffer solution with dilution. Therefore, no effect on pH. |
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