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Solubility Lesson 6 Separating Ions. Positive ions react with negative ions to give a precipitate if they have low solubility. A precipitate can be separated.

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Presentation on theme: "Solubility Lesson 6 Separating Ions. Positive ions react with negative ions to give a precipitate if they have low solubility. A precipitate can be separated."— Presentation transcript:

1 Solubility Lesson 6 Separating Ions

2 Positive ions react with negative ions to give a precipitate if they have low solubility. A precipitate can be separated from a solution by filtration. The precipitate will be on the filter paper while the soluble ions will go through. To separate ions one ion must be low solubility and all others high.

3 Separating Positive Ions 1.SeparateBa 2+ &Pb 2+ Start at the top of page 4 and try Cl - first

4 Ba 2+ 1 st add Cl -, it reacts with Pb 2+ (low) and not Ba 2+ (high) 2 nd add SO 4 2-, it reacts with Ba 2+ (low)

5 Pb 2+ NO 3 - Ba 2+ NaCl Na + Cl - PbCl 2 Filter

6 Separating Positive Ions Cl - does not exist on its own Add NaCl because it is soluble- always use Na + for negative ions 1.SeparateBa 2+ &Pb 2+ i.AddNaClFilter out PbCl 2(s) Pb 2+ +2Cl -  PbCl 2(s) ii.AddNa 2 SO 4 Filter out BaSO 4(s) Ba 2+ +SO 4 2-  BaSO 4(s)

7 Separating Positive Ions 2.SeparateCu 2+,Mg 2+ &Sr 2+

8 Cu 2+ Mg 2+ Sr 2+ Cu 2+ Mg 2+ Cu 2+ Mg 2+

9 Separating Positive Ions 2.SeparateCu 2+,Mg 2+ &Sr 2+ i.AddNa 2 SO 4 Filter out SrSO 4(s) Sr 2+ +SO 4 2-  SrSO 4(s) ii.AddNa 2 SFilter out CuS (s) Cu 2+ +S 2-  CuS (s) iii.AddNaOHFilter out Mg(OH) 2(s) Mg 2+ +2OH -  Mg(OH) 2(s) SO 4 2- does not exist on its own Add Na 2 SO 4 because it is soluble- always use Na + for negative ions

10 Separating Negative Ions Remove the lowest negative ion on the chart first by adding a positive cation. 3.SeparateCl - &OH - Look for a cation that is low with OH - (lowest) and high with Cl -.

11 Ba 2+ Remove the lowest negative ion first by adding a positive cation. Look for a cation that is low with OH - (lowest) and high with Cl -. Look for a cation that is low with with Cl -

12 Separating Negative Ions 3.SeparateCl - &OH - ii.Add Ag + as AgNO 3 and filter out AgCl (s) Ag + +Cl -  AgCl (s) Ba 2+ does not exist on its own Add Ba(NO 3 ) 2 because it is soluble- always use NO 3 - to pair with positive ions i.Ba 2+ works so add Ba(NO 3 ) 2 and filter out Ba(OH) 2(s) Ba 2+ +2OH -  Ba(OH) 2(s)

13 Separating Negative Ions 4.SeparateCl - S 2- CO 3 2-

14 Ba 2+ Zn 2+ Look for a cation that is low with CO 2- and high with Cl - and S 2- Look for a cation that is low with S 2- and high with Cl - Look for a cation that is low with with Cl -

15 Separating Negative Ions 4.SeparateCl - S 2- CO 3 2- i.Ba 2+ works so add Ba(NO 3 ) 2 and filter out BaCO 3(s) Ba 2+ +CO 3 2-  BaCO 3(s) ii.Add Zn 2+ as Zn(NO 3 ) 2 and filter out ZnS (s) Zn 2+ +S 2-  ZnS (s) iii.Add Ag + as AgNO 3 and filter out AgCl (s) Ag + +Cl -  AgCl (s) Ba 2+ does not exist on its own Add Ba(NO 3 ) 2 because it is soluble- always use NO 3 - to pair with positive ions


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