Formula, Total & Net Ionic Equations

To be able to properly write formula, total and net ionic equations, you must know: elements and charges, polyatomic ions and charges, weak and strong acids/bases, solubility rules and be able to balance equations.

FORMULA UNIT EQUATION
This is the type of equation we have been dealing with in previous chapters. Basically it is a displacement reaction.
Step 1: If no products are given, then determine the products.
Step 2: Balance the equation.

Given: lead (II) hydroxide + hydrochloric acid à
1) Pb(OH)2 + HCl à PbCl2 + H2O
2) Pb(OH)2 + 2 HCl à PbCl2 + 2 H2O

TOTAL IONIC EQUATION
As the name states, this equation shows ions (charges) for everything that is soluble or ionizes/dissociates 100%.
Step 3: Look at each compound in the formula unit and ask yourself: Is it soluble? Strong acid? Strong Base?
- If yes to any of the three questions, assign charges to each element in the compound
- If no, then leave the compound as it is

3) Pb(OH)2 + 2H+1 + 2 Cl-1 à Pb+2 + 2 Cl-1 + 2 H2O
(assume HOT water)

Question: Where did the coefficient 2 come from in front of H+1 and Cl-1?
- Coefficient is the number we use to balance equations in the Formula Unit Equation. In the Total Ionic Equation, the coefficients come from 1) the coefficients in the formula unit equation and 2) from the subscript assigned when creating a compound. If you have both a coefficient and a subscript to account for on the same compound, you would multiply the two together:
formula unit equation 2 CaBr2 à total ionic equation 2 Ca+2 + 4 Br-1

Spectator ions do not change their charge (remember ions are charged particles, not compounds). Whatever the charge was on the reactant side it will be the same charge on the product side of the equation. Spectators are present in the solution, but do not participate in the reaction. That is why they have no change in charge. Think of it like a sports fan: if you are a spectator, you are involved in the overall event because you are physically present and viewing the sport, but you do not participate in the action of the sport.

NET IONIC EQUATION
Step 4: Look at each ion (not compound) in the Total Ionic Equation. Cancel out the identical spectator ions that are on both the reactant and product side.
Step 5: What remains (doesn’t cancel) constitutes the Net Ionic Equation. You will always have something that doesn’t cancel. If everything cancels, go back and check your work, something is wrong.
**** Net ionic equations for ALL reactions of strong acids with strong soluble bases that form soluble salts and water is: H+ + OH- à H2O (l)
In other words, if you can’t find anything else, your net ionic equation will be: H+ + OH- à H2O (l)


4) Pb(OH)2 + 2H+1 + 2 Cl-1 à Pb+2 + 2 Cl-1 + 2 H2O
5) Pb(OH)2 + 2H+1 à Pb+2 + 2 H2OPRACTICE PROBLEMS

Write balanced formula unit, total ionic and net ionic equations for the following reactions. Assume all reactions occur in water or in contact with water.

1) acetic acid + sodium hydroxide à



2) calcium hydroxide + hydrosulfuric acid à



3) Ba(NO3)2 + K2CO3 à



4) lead (II) hydroxide + carbonic acid à



5) Our bones are mostly calcium phosphate. Calcium chloride reacts with potassium phosphate to form calcium phosphate and potassium chloride.





ANSWERS

1) Formula Unit: HC2H3O2 + NaOH à NaC2H3O2 + H2O
Total Ionic: HC2H3O2 + Na+1 + OH-1 à Na+1 + C2H3O2-1 + H2O
Weak acid
Net Ionic: HC2H3O2 + OH-1 à C2H3O2-1 + H2O

2) Formula Unit: Ca(OH)2 + H2S à CaS + 2 H2O
Total Ionic: Ca+2 + 2OH-1 + 2H +1 + S-2 à Ca+2 + S-2 + 2 H2O
Net Ionic: 2OH-1 + 2H +1 à 2 H2O
OH-1 + H +1 à H2O

3) Formula Unit: Ba(NO3)2 + K2CO3 à BaCO3 + 2KNO3
Total Ionic: Ba+2 + 2NO3-1 + 2K +1 + CO3-2 à BaCO3 + 2K+1 + 2NO3-1
Insoluble
Net Ionic: Ba+2 + CO3-2 à BaCO3

4) Formula Unit: Pb(OH)2 + H2CO3 à PbCO3 + 2H2O
Total Ionic: Pb(OH)2 + H2CO3 à PbCO3 + 2H2O
Insoluble Weak Acid Insoluble
& Insoluble
Net Ionic: Pb(OH)2 + H2CO3 à PbCO3 + 2H2O

5) Formula Unit: 3CaCl2 + 2K3PO4 à Ca3(PO4)2 + 6KCl
Total Ionic: 3Ca+2 + 6Cl-1 + 6K+1 + 2 PO4-3 à Ca3(PO4)2 + 6K+1 + 6Cl-1
Insoluble
Net Ionic: 3Ca+2 + 2 PO4-3 à Ca3(PO4)2