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Precipitation Reactions

Precipitation Reactions. Not all ionic compounds will separate into ions in solution. the ions are held together so tightly that the water molecules cannot allow them to separate. These ionic compounds are said to have LOW SOLUBILITY. yellow precipitate. Pb(NO 3 ) 2(aq). 2 KI (aq).

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Precipitation Reactions

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  1. Precipitation Reactions Not all ionic compounds will separate into ions in solution. • the ions are held together so tightly that the water molecules cannot allow them to separate. These ionic compounds are said to have LOW SOLUBILITY yellow precipitate Pb(NO3)2(aq) 2 KI(aq) PbI2(s) 2 KNO3(aq) • when Pb2+ and I- are exposed to each other they form an insoluble (or low soluble) compound

  2. Pb(NO3)2(aq) 2 KI(aq) PbI2(s) 2 KNO3(aq) Recall: • the subscript (aq) means dissolved in water • ionic compounds dissociate into ions in water The Complete Ionic Equation would be: Pb2+(aq) 2 NO3-(aq) 2 K+(aq) 2 I-(aq) PbI2(s) 2 K+(aq) 2 NO3-(aq) The ions not involved in the precipitation reaction are known as “spectator ions” K+and NO3- The Net Ionic Equation for this reaction is: Pb2+(aq) 2 I-(aq) PbI2(s)

  3. to determine which combinations of ions have low solubility use the table on p. 459 e.g: Balance the following double replacement reaction and determine whether or not a precipitate will form. NiCl2(aq) Na2CO3(aq) NiCO3(s) 2 NaCl(aq) according to the solubility table the precipitate must be NiCO3 the net ionic equation for this reaction is: Ni2+(aq) CO3-(aq) NiCO3(s)

  4. Precipitation Problem 40.0 mL of 1.0M NaCl is mixed with 10.0 mL of 1.0M Pb(NO3)2. a) What is the identity of the precipitate? b) What will the mass of the precipitate be? c) Find the concentration of the excess ion. d) Find the concentrations of the spectator ions. Solution: a) Balance the equation; use your solubility table to determine the precipitate. 2 NaCl(aq) Pb(NO3)2(aq) PbCl2(s) 2 NaNO3(aq) net ionic equation: Pb2+(aq) 2 Cl-(aq) PbCl2(s)

  5. Precipitation Problem Solution Con’t b) # moles of each ion: NaCl(aq) Na+(aq) Cl-(aq) M = 1.0M 0.040mol 0.040mol V = 0.040L mol = 0.040mol Pb(NO3)2(aq) Pb2+(aq) 2 NO3-(aq) M = 1.0M 0.010mol 0.020mol V = 0.010L mol = 0.010mol net ionic equation: Pb2+(aq) Pb2+(aq) 2 Cl-(aq) PbCl2(s) 0.010mol 0.040mol We have: Which one do we run out of first? Pb2+ If we use all of the Pb2+, how much Cl- do we need 0.010mol Pb2+ 0.020mol Cl- 0.010mol PbCl2 How much PbCl2 do we get out?

  6. Precipitation Problem Solution Con’t mass of PbCl2 produced: 278.1g 0.010mol = 2.78 g of PbCl2 1mol c) Concentration of the excess ion: Moles of Cl- remaining = Moles before reaction – Moles used = 0.040mol – 0.020mol = 0.020mol Cl- Concentration of Cl- remaining = Moles remainingTotal Volume = 0.020mol0.050L = 0.40M d) Concentration of the spectator ions: [Na+] = Moles remainingTotal Volume [NO3-] = Moles remainingTotal Volume = 0.040mol0.050L = 0.80M = 0.020mol0.050L = 0.40M

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