HW 4 Solutions

# HW 4 Solutions - 53:050 Natural Environmental Systems...

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53:050 Natural Environmental Systems Solutions to Homework No. 4 – Spring 2011 1. Use the equilibrium expression given: K s = 10 -15.6 = [Pb 2+ ][OH - ] 2 [OH - ] = ] Pb [ 10 2 6 . 15 + - [Pb 2+ ] = Pb/mol mg 207,200 Pb/L mg 0.015 = 7.24 x 10 -8 M [OH - ] = 8 - 6 . 15 10 x 24 . 7 10 - = 5.89 x 10 -5 M [H + ] = ] [OH K w - = 5 - -14 10 x 5.89 10 = 1.70 x 10 -10 pH = -log[H + ] = -log(1.70 x 10 -10 ) = 9.77 It is unlikely that people would be comfortable with drinking water with a pH = 9.77, so this doesn’t look like a particularly good method for removing lead from drinking water. In addition, it turns out that the situation is much more complicated than this. Pb 2+ forms strong complexes such that it is very difficult to get the soluble lead below 1 mg/L. So, most definitely pH adjustment to precipitate Pb(OH) 2 (s) won’t work for drinking water. (FYI – I do not expect you to have known the part of this answer given in italics!) 2. Davis & Masten chapter 2, problem 2-28. A simple way to solve this problem is to calculate

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HW 4 Solutions - 53:050 Natural Environmental Systems...

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