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SATII化学试题分析—Net Ionic Equations

Net ionic equations are equations that show only the soluble, strong electrolytes reacting (these are represented as ions) and omit the spectator ions, which go through the reaction unchanged. When you encounter net ionic equations on the SAT II Chemistry test, you’ll need to remember the following solubility rules, so memorize them! Also keep in mind that net ionic equations, which are the bare bones of the chemical reaction, usually take place in aqueous environments. Here are those solubility rules:

Most alkali metal compounds and compounds are soluble.
Cl^-, Br^-, I^- compounds are soluble, except when they contain Ag⁺, , or Pb²⁺.
F^- compounds are soluble, except when they contain group 2A metals.
, , , and CH₃COO^- compounds are soluble.
compounds are soluble, except when they include Ca²⁺, Sr²⁺, Ba²⁺, Ag⁺, Pb²⁺, or .
, , , , S^2-, OH^-, and O^2- compounds are insoluble.
Group 2A metal oxides are classified as strong bases even though they are not very soluble.

The two solubility rules that you will use the most are numbers 1 and 4. You must memorize that all group 1A metal and ammonium compounds are soluble. As soon as you see a compound

, Li, Na, K, Rb, Cs, or Fr, you should know that it’s soluble. Also, all nitrates are soluble—look at the end of the compound. If it ends in

, you know that it’s soluble.

What’s the big deal with solubility? Well, if the ion is soluble, it won’t form a precipitate, and this means it doesn’t react and should be left out of the net ionic equation. The key is first to write the compound’s chemical formula and then determine if it’s soluble. If it is soluble, then ionize it—if it isn’t, don’t ionize it; leave it as a molecule.

Here are some additional rules about common reaction types that you should be familiar with for the exam:

If an insoluble precipitate or gas can be formed in a reaction, it probably will be.
Oxides (except group 1A) are insoluble, and when reacted with water, they form either acids (nonmetal oxides) or bases (metal oxides).
There are six strong acids that completely ionize: HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄. All other acids are weak and are written together, as molecules.
The strong bases that ionize are oxides and hydroxides of group 1A and 2A metals. All other oxides and hydroxides are considered weak and written together, as molecules.

Now try writing some net ionic equations, using the rules above.

Example

Write the net ionic equation for a mixture of solutions of silver nitrate and lithium bromide.

Explanation

Ag⁺ +

+ Li⁺ + Br^-

This is a double replacement reaction. Both compounds are soluble, so everything ionizes. If anything is formed, it will come from recombining the “inside” two ions with the “outside” two ions to make LiNO₃ and AgBr. If either of them is insoluble, a precipitate will be formed, and the ions that react to form it will be in our net ionic equation; the other ions are spectators and should be omitted! As we said, the two possible products are lithium nitrate and silver bromide. Since halides are soluble except those containing silver, mercury, or lead, we have a precipitate of silver bromide, and our net ionic equation looks like this:

Ag⁺ + Br^-

AgBr

Example

Hydrochloric acid and sodium hydroxide are mixed. Write the net ionic equation.

Explanation

This is a mixture of a strong acid and a strong base, so each ionizes completely.

H⁺ + Cl^- + Na⁺ + OH^-

The two possible compounds formed are sodium chloride, which is soluble, and water, which is molecular; thus water is the only product in our net ionic equation.

H⁺ + OH^-

H₂O

Example

Chlorine gas is bubbled into a solution of potassium iodide; write the net ionic equation.

Explanation

This one is a single replacement, so you need to consider the activity series. Since halogens are involved, you can determine their activity by using the periodic table: Cl is more active than I.

Cl₂ + K⁺ + I^-

Remember that halogen is diatomic and that all potassium compounds are soluble. The resulting compound is also soluble, so K⁺ is a spectator and is left out of the final equation.

Cl₂ + I^-