When an aqueous solution of sodium hydroxide, NaOH, is added to an aqueous solution of chromium(III) nitrate, Cr(NO3)3, a precipitate of chromium(III) hydroxide, Cr(OH)3, forms. Write a balanced net ionic equation for this reaction.

Answer 1

Cr(NO3)3(aq) + 3NaOH(aq) ==> Cr(OH)3(s) + 3NaNO3(aq)

Now you convert that balanced molecular equation into a balanced ionic equation.
Cr^3+(aq) + 3(NO3^-)(aq) + 3Na^+(aq) + 3OH^-(aq) ==> Cr(OH)3(s) + 3Na^+(aq) + 3NO3^-(aq) and here are the rules you follow.
1. Gases remain as molecules.
2. Solids remain as molecules.
3. Weak electrolytes remain as molecules.
4. All others are separated into ions.
[Note: There are no gases. The solid is Cr(OH)3 so that is NOT separated into ions. There are no weak electrolytes so all of the others are shown as the ions.

The last step is to cancel ions that are common(that is the same) on both sides. That is the Na^+ and the NO3^-. What is left?
Cr^3+(aq) + 3OH^-(aq) ==> Cr(OH)3(s)
It's really a simple three step process.
1. Write and balanced the molecular equation.
2. Using the rules above, separate into the COMPLETE balanced ionic equation.
3. Cancel ions common to both sides.