Positively explain explain how to experimentally

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Unformatted text preview: q ) + H 2O(l ) Al ( H 2O ) 4 (OH ) 2 + (aq ) + H 2O (l ) Al ( H 2O)3 (OH )3 ( s ) + H 2O (l ) Al ( H 2O ) 2 (OH ) 4 − (aq ) + H 2O (l ) As the charge on the ion becomes more negative , it becomes more increasingly difficult to remove a positively charged proton! positively Explain Explain how to experimentally separate the two solids Al(OH)3 and Fe(OH)3 from each other based on from solubility. Al (OH )3 strong base Al (OH )3 ( s ) + OH − (aq ) Al (OH ) 4 − (aq ) → Fe(OH )3 Fe(OH )3 ( s ) + OH − (aq ) ≠ Filtration Aqueous ( Al (OH ) 4 − ) Al (OH ) 4 (aq ) → Fe(OH )3 ( s ) Solid ( Fe(OH )3 ) − Precipitation Precipitation and Separation of Ions Ions can be separated from one another based on their based salt solubility. A mixture of 0.10M Zn2+(aq) and 0.10M Cu2+(aq) K sp = 6.0 × 10-37 • CuS As H2S is added to the solution, • ZnS K sp = 2.0 × 10-25 which ion will be precipitated firstly? Cu 2+ (aq ) + S 2− (aq) CuS ( s) K sp =[Cu 2+ ][S 2− ]= 6.0 × 10-37 [S 2− ]min = 6.0 × 10-37 0.10 = 6.0 × 10-36 M Zn 2+ (aq ) + S 2− (aq ) [S2- ]min = 2.0 × 10-25 / 0.10 = 2.0 × 10-24 M ZnS ( s ) K sp =[Zn 2+ ][S2- ]= 2.0 × 10-25 Could the two metal ions be separate separate completely? What is the [Cu2+] when Zn2+ begin to precipitate? [Cu 2+ ]=6.0 × 10-37 / 2.0 × 10−24 = 3.0 × 10−13 M << 10 −6 M Question Question (1) on Cation Analysis • A solution contains 1.0×10-2 M each of Ag+ , Pb2+ and Hg22+ ions. 1.0 When NaCl is added to the solution, What concentration of Cl- is What necessary to begin precipitating each salt? Which will precipitate firstly firstly? AgCl ( K sp = 1.8 ×10-10 ) PbCl2 (K sp =1.7 ×10-8 ) Hg 2 Cl2 (K sp =1.2 × 10-18 ) AgCl ( s ) PbCl2 (s) Ag + (aq ) + Cl − (aq ) K sp = [ Ag + ][Cl − ] Pb 2+ (aq) + 2Cl- (aq) Hg 2 Cl2 (s) Hg 2 2+ (aq)+2Cl- (aq) K sp = [Pb 2+ ][Cl- ]2 K sp = [Hg 2 2+ ][Cl- ]2 = 1.0 ×10−2 × x = (1.0 × 10-2 )(x ) 2 =(1.0 × 10-2 )(x) 2 = 1.8 ×10-10 =1.7 × 10-8 =1.2 ×10-18 x = 1.3 × 10-3 M x =1.1× 10-8 M x = 1.8 ×10−8 M [Cl-] > 1.8 ×10-8 M, M, AgCl will ppt Cl-> 1.3×10-3M, PbCl2 will ppt Cl- > 1.1 ×10-8M, Hg2Cl2 will ppt...
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This document was uploaded on 02/26/2014.

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