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Predicting solubility. Using the table of solubilities we can now predict which of the products of a double replacement reaction will be insoluble (form.

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Presentation on theme: "Predicting solubility. Using the table of solubilities we can now predict which of the products of a double replacement reaction will be insoluble (form."— Presentation transcript:

1 Predicting solubility

2 Using the table of solubilities we can now predict which of the products of a double replacement reaction will be insoluble (form solids or precipitates : have low solubilities) and which will remain soluble.

3 Solubility of Ions Negative ions (anion)+ Positive ions (cation) Soluble Ie: > 0.1 mol/L Any anion+Alkali metal ions (Li +, Na +, K +,Rb +,or Cs + ) Soluble Any anion+Ammonium, NH 4 + Soluble Nitrate NO 3 - +Any cationSoluble Acetate CH 3 COO - +Any cation except Ag + soluble Chloride Cl - Bromide Br - Iodide I - ++++++ Ag +, Pb 2+, Hg 2 2+, or Cu + Any other cation Not soluble Soluble Sulfate,SO 4 2- ++++ Ca 2+, Sr 2+, Ba 2+, Ra 2+, Ag +, or Pb 2+ Any other cation Not soluble soluble Sulfide, S 2- ++++++ Alkali metals or NH 4 + Be 2+, Mg 2+, Ca 2+, Sr 2+, Ba 2+, or Ra 2+ Any other cation Soluble Not soluble Hydroxide, OH - ++++++ Alkali metals or NH 4 + Sr 2+, Ba 2+, or Ra 2+ Any other cation Soluble Slightly soluble Not soluble phosphate,PO 4 3- carbonate,CO 3 2- Sulfite,SO 3 2- ++++ Alkali metals or NH 4 + Any other cation Soluble Not soluble

4 Pb(NO 3 ) 2 (aq) + 2 KI(aq)  PbI 2 (s) + 2KNO 3 (aq) overall ionic equation, which separates all the ions out: Pb 2+ (aq) + 2 NO 3 - (aq) + 2 K - (aq) + 2 I - (aq)  PbI 2 (s) + 2 K + (aq) + 2 NO 3 - (aq) Pb 2+ (aq) + 2 I - (aq)  PbI 2 (s) Stays as a solid

5 The NO 3 - ion and the K + ion are called spectator ions.

6 Silver nitrate combines with potassium iodide. a) Nonionic or double replacement reaction AgNO 3 (aq)+KI (aq) → AgI (?)+KNO 3 (?)

7 Using the solubility table we see that nitrates are soluble with all positive ions and Iodide ion is insoluble with silver. AgNO 3 (aq)+KI (aq)→AgI (s)+KNO 3 (aq)

8 Now that we have predicted the solid that will form we can proceed to write the net ionic equation. Ag + (aq) + I - (aq) → AgI (s)

9 i)silver nitrate + potassium iodide ii)ammonium sulfide + lead(II) nitrate iii)zinc nitrate + sodium carbonate

10 Separating ions: Pb 2+ (aq) ; Ba 2+ (aq) ; Cu 2+ (aq) Add Cl - Ba 2+ and Cu 2+ soluble PbCl 2 ppt outs Add S 2- CuS ppt out Ba 2+ soluble Add PO 4 3- Ba 3 (PO 4 ) 2 ppt out

11 Using the solubility table create a flow chart that describes how to separate the following ions, if they are dissolved together in a solution. The last ion may remain in solution. a) Ag + (aq) ; Ba 2+ (aq) ; Mg 2+ (aq) b) Ba 2+ (aq) ; K + (aq) ; Zn 2+ (aq) c) Pb 2+ (aq) ; Al 3+ (aq) ; Sr 2+ (aq) d) Sr 2+ (aq) ; Mg 2+ (aq) ; Fe 3+ (aq)

12 1. Ionic Substances ( containing a metal or NH 4 - ion ) Ionic substances are composed of positive and negative ions. Arrhenius proposed that ionic substances come apart when they enter a water solution, and the ions are free to move about in the solution. The called this process dissociation. CaCl 2 (s) -------> Ca 2+ (aq) + 2 Cl - (aq) * Notice that the reactions are balanced for charge and # of atoms.

13 2. Acids (containing nonmetals with hydrogen written first in the formula) Certain molecular substances also form mobile ions when placed in solution. Arrhenius proposed that acids when placed in solution form mobile ions. He called this process ionization. H 2 SO 4 (l) -------> 2 H+ (aq) + SO 4 2- (aq) In both cases (dissociation and ionization) the end result is mobile ions in solution. Thus since ions are present and can move, both acids and ionic substances conduct electricity in solution and are termed electrolytes.

14 3. Molecular (contain only nonmetals) Molecular substance that are not acids on the other hand do not ionize nor dissociation and do not form ions in solution. Therefore these substances are non-electrolytes. CO 2 (g) ------> CO 2 (aq)


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