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Reactions in solutions

Reactions in solutions. Precipitation reactions.

jorden-zimmerman
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Reactions in solutions

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  1. Reactions in solutions

  2. Precipitation reactions • When solutions of certain ionic compounds are mixed and the ions come in contact with one another, the positive ions from one can be so strongly attracted to the negative ions of the other that they cluster together in big groups to form crystals (solid precipitate). The solid formed is called a precipitate • We say that the ions have associated. • If the product is a gel we call it a gelatinous precipitate. • In order to determine whether a precipitate will form we must know which substances are soluble in water and which substances are insoluble.

  3. Steps involved in working out whether a precipitate will form Step one: Identify the products of the reaction by swapping the positive and negative ions of the reactants. Step two: Identify whether the products will be insoluble using solubility data. Step three: Write the correct formulae of the reactants and products Step four: Write the equation and balance it. Step five: Identify the precipitate and include the appropriate states of reactants and products.

  4. Example: Lead nitrate and Potassium iodide Step one: Identify the products of the reaction by swapping the positive and negative ions of the reactants. Lead nitrate + Potassium Iodide  Lead Iodide + Potassium nitrate Step two: Identify whether the products will be insoluble using solubility data • When Lead Pb2+ and Iodide I- ions combine they form an insoluble solid (precipitate) • When Potassium K+ and Nitrate NO3- ions combine they form a soluble solid.

  5. Example (cont) Step three: Write the correct formula of the reactants and products Pb(NO3)2 + KI  PbI2 + KNO3 Step four: Write the equation and balance it so that equal numbers of all atoms occur on both sides. Pb(NO3)2 + 2KI  PbI2 + 2KNO3 Step five: Identify the precipitate and include the appropriate states of reactants and products. Pb(NO3)2(aq) + 2KI (aq)  PbI2(s) + 2KNO3 (aq)

  6. Solubility guide for Common Ionic compounds

  7. Ionic equations • Ionic compounds will dissociate into their ions in solutions. • In a reaction between two ionic compounds you can write a chemical equation of ions. This will be a better representation on what is happening in solution. • Steps to writing an ionic equation for precipitation reactions Step one: Write any dissolved ionic compound as their separate ions. Don’t worry about the states Step two: Cross out any ions that appear on both sides. These ions have not participated in the reaction and are termed spectator ions Step three: Rewrite the equation deleting the spectator ions and add the states The ionic equation for the precipitation reaction check • The equation is balanced • The total charges on the left equals the total charge on the right (zero).

  8. Example Step one: Write any dissolved ionic compound as their separate ions. Don’t worry about the states Pb2+ + 2NO3- + 2K+ + 2I-  PbI2 + 2K+ + 2NO3- Step two: Cross out any ions that appear on both sides. These ions have not participated in the reaction and are termed spectator ions Pb2+ + 2NO3- + 2K+ + 2I-  PbI2 + 2K+ + 2NO3- Step three: Rewrite the equation deleting the spectator ions and add the states Pb2+(aq) + 2I-(aq)  PbI2(s) The ionic equation for the precipitation reaction check • The equation is balanced • The total charges on the left equals the total charge on the right (zero).

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