March 2, 2009

March 2, 2009 - March 2, 2009 Solubility Product Common Ion...

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March 2, 2009 Solubility Product
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Common Ion [Ag][Cl] = 10 -10 If nothing else X = [Cl - ]= [Ag - ] X 2 =[Ag][Cl] = 10 -10 X = 10 -5 = [Cl - ] = [Na + ]
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start adding NaCl → Na + + Cl - in solution 10 -3 M Na + + 10 -3 M Cl - (10 -3 + 10 -5 )[Ag] ≈ (10 -3 )[Ag] = 10 -10 [Ag] = 10 -7 M originally 10 -5 M If 1L of original solution (1x10 -5 - .01x10 -5 )(1L) 1x10 -5 (1L) = 10 -5 moles of Ag.Cl out as AgCl
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When common ion, counterion with lowest K sp precipitates out. But – must do the calculation Dead sea almost saturated NaCl However “salt flats” are NaCO 3 • For illustration K sp (NaCl) = 1 K sp (Na 2 CO 3 ) = 0.6
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K sp (Na 2 CO 3 ) = 0.6 = [CO 3 ][Na] 2 = 0.6 [CO 3 ] = 0.6/(1) 2 = 0.6M Even though more chloride than CO 3 , Na 2 CO 3 precipitates first
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Fractional Precipitation K sp ( Ba 2+ ) = 10 -10 • K sp (Sr 2+ ) = 10 -5 for common ion A - and Ba = Sr = 0.1M BaA is less soluble add A - slowly BaA first when A - = 10 -10 /0.1 = 10 -9 M SrA when [A] = 10 -5 /0.1 = 10 -4 M
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Liesegang rings
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This note was uploaded on 02/26/2012 for the course CHEM 108 taught by Professor Dr.brennan during the Spring '09 term at Binghamton.

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March 2, 2009 - March 2, 2009 Solubility Product Common Ion...

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