Ionic equations and formulae

In this lesson, we will learn:

• To recall the definition of a metathesis (double replacement) reaction.
• Different methods to write metathesis reactions involving precipitate products.
• How to construct balanced net ionic equations from full formula equations.

• Recall that when ionic species react, salts are produced. If two salts react, a metathesis reaction occurs. This is also known as a double replacement reaction.
  • You can spot these reactions; look for two salts as the reactants. Salts can be spotted by using a table of common ions and charges. Most chemistry textbooks have back page for this; a salt has two oppositely charged ions or groups in a ratio that makes the compound neutral overall.
    For example the reaction of potassium phosphate, K₃PO₄ and calcium sulfide, CaS:
    • The potassium ion has a single positive charge (formula K⁺) and the phosphate ion has a 3- charge (formula PO₄^3-). To balance, the formula of potassium phosphate is K₃PO₄ to give an overall neutral compound.
    • The calcium ion has a double positive charge (formula Ca²⁺) and the sulfide ion has a double negative charge (formula S^2-). As such, the formula of calcium sulfide is CaS which gives an overall neutral compound.
    So four ions that make up the two salts have been identified: K⁺, Ca²⁺, S^2- and OH^-. In a metathesis reaction, ions are exchanged between the two salts reacting together. The products can be predicted by swapping the anions:
    
    K₃PO₄ + CaS $\enspace$→$\enspace$ K S + CaPO₄
    
    As with before, the formula of the molecule will allow charges to balance and give a neutral molecule. To find this formula you can ‘cross’ the charges of the ions that make the compound.
    
    With the formulae found, we just need to balance the equation now.
    When balancing you can treat polyatomic ions as one combined entity; counting the individual atoms in the ions isn’t necessary as the ratio of atoms within the group should not change when balancing equations anyway:
    
    K₃PO₄ + CaS $\enspace$→$\enspace$ K₂ S + Ca₃(PO₄)₂
    
    There are 3 Ca atoms in the products to 1 in the reactants, and 2 PO₄ groups in the products to 1 in the reactants. Balancing for these:
    
    2K₃PO₄ + 3CaS $\enspace$→$\enspace$ K₂ S + Ca₃(PO₄)₂
    
    There are now 6 K atoms in the products to 2 in the reactants and 3 S atoms in the reactants to 1 in the products.
    Balancing for these gives the full equation:
    
    2K₃PO₄ + 3CaS $\enspace$→$\enspace$ 3K₂ S + Ca₃(PO₄)₂

• When metathesis reactions occur, sometimes one of the products formed is an insoluble solid, and comes out of solution. This is called a precipitate. In a real chemistry experiment spotting a precipitate is easy because the solution turns cloudy; the precipitate is blocking light passing through the solution. Because metathesis reactions normally use solutions, we assume the reactants and products are aqueous so if a precipitate is formed, you MUST make it clear with an (s) state symbol. There are a few ways to write metathesis reaction equations:
  • A formula equation, or full equation, involves writing all reactants and products of the reaction in their neutral molecular form. This is the normal way of writing an equation, like we did above with K₃PO₄ and CaS reacting.
    
    
  • A full ionic equation is written by displaying all the dissolved species – the dissolved ions! Full ionic equations include spectator ions; these are the ions still in an aqueous state in the products (which did not react). For example in the reaction with silver nitrate, AgNO₃ and sodium chloride, NaCl.
    We assume aqueous state unless given otherwise:
    
    Ag⁺ + NO₃^- + Na⁺ + Cl^- $\enspace$→$\enspace$ AgCl_(s) + Na⁺ + Cl^-
    
  • A net ionic equation is an ionic equation where only the reacting ionic species are included; spectator ions are ignored and stoichiometry is in its simplified form. This is a simplified version of the equation that can be written after the balanced full equation has been determined.
    
    Ag⁺ + Cl^- $\enspace$→$\enspace$ AgCl_(s)