Unit 8: Precipitation Reactions
Learning to predict when solids form from solutions and how to prepare them.
8.7 Predicting the Solubility of Ionic Compounds (Rules)
When two ionic solutions are mixed, a solid may form. To predict this, we use a set of solubility rules. These are generalizations that help determine whether a given ionic compound will be soluble or insoluble in water.
| Rule | Soluble Compounds | Important Exceptions (Insoluble) |
|---|---|---|
| 1 | All compounds containing Group 1 ions ($Li^+, Na^+, K^+$) and the ammonium ion ($NH_4^+$). | None |
| 2 | All compounds containing the nitrate ion ($NO_3^-$). | None |
| 3 | Most compounds containing halide ions ($Cl^-, Br^-, I^-$). | Halides of Silver ($Ag^+$), Lead ($Pb^{2+}$), and Mercury(I) ($Hg_2^{2+}$). |
| 4 | Most compounds containing the sulfate ion ($SO_4^{2-}$). | Sulfates of Calcium ($Ca^{2+}$), Strontium ($Sr^{2+}$), Barium ($Ba^{2+}$), and Lead ($Pb^{2+}$). |
| 5 | Most compounds containing hydroxide ($OH^-$) and carbonate ($CO_3^{2-}$) ions are insoluble. | Compounds from Rule 1. $Ba(OH)_2$ and $Sr(OH)_2$ are also soluble. |
Solved Examples:
-
Is silver chloride (AgCl) soluble in water?
Solution: No. Rule 3 states that most chlorides are soluble, but silver chloride is a key exception. It is insoluble. -
Is potassium carbonate ($K_2CO_3$) soluble in water?
Solution: Yes. Rule 5 states that most carbonates are insoluble. However, Rule 1 takes precedence: all compounds containing Group 1 ions (like potassium, K⁺) are soluble. Therefore, it is soluble. -
Is barium sulfate ($BaSO_4$) soluble in water?
Solution: No. Rule 4 states that most sulfates are soluble, but barium sulfate is a key exception. It is insoluble.
8.8 Definition of Precipitation & Precipitates
A precipitation reaction is a type of chemical reaction in which two soluble ionic compounds in aqueous solution are mixed, and one of the products is an insoluble solid. This solid product is called the precipitate.
When the two solutions are mixed, the ions from both compounds are free to move and collide. If a combination of a cation from one solution and an anion from the other solution forms an insoluble compound (according to the solubility rules), those ions will bond together and fall out of the solution as a solid.
Example: Mixing Silver Nitrate and Sodium Chloride
When a clear solution of silver nitrate ($AgNO_3$) is mixed with a clear solution of sodium
chloride (NaCl), a cloudy white solid immediately forms.
Possible products are $NaNO_3$ and $AgCl$.
According to the rules, $NaNO_3$ is soluble, but $AgCl$ is insoluble.
Therefore, the white solid precipitate is silver chloride.
Solved Examples:
-
What is a precipitate?
Solution: A precipitate is an insoluble solid that emerges from a liquid solution during a chemical reaction. -
If you mix solutions of potassium iodide (KI) and lead(II) nitrate
($Pb(NO_3)_2$), a bright yellow precipitate forms. Identify the
precipitate.
Solution: The ions present are $K^+$, $I^-$, $Pb^{2+}$, and $NO_3^-$. The possible new compounds are $KNO_3$ and $PbI_2$. According to the solubility rules, all nitrates are soluble ($KNO_3$), but lead halides are insoluble. Therefore, the yellow precipitate is lead(II) iodide ($PbI_2$).
8.9 Writing Ionic Equations for Precipitation Reactions
We can represent precipitation reactions with ionic equations, which show the reacting ions more clearly.
- Molecular Equation: Shows all reactants and products as if they were
intact, undissociated molecules.
$AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)$ - Full Ionic Equation: Shows all strong electrolytes dissociated into
their ions. Insoluble solids, liquids, and gases are not dissociated.
$Ag^+(aq) + NO_3^-(aq) + Na^+(aq) + Cl^-(aq) \rightarrow AgCl(s) + Na^+(aq) + NO_3^-(aq)$ - Net Ionic Equation: Shows only the particles that are directly involved
in the chemical change. Ions that appear unchanged on both sides of the equation
(spectator ions) are cancelled out.
$$ Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s) $$
The net ionic equation is the most concise way to represent the chemical change that occurs in a precipitation reaction.
Solved Examples:
-
Identify the spectator ions in the reaction between silver nitrate and
sodium chloride.
Solution: The spectator ions are the sodium ion ($Na^+$) and the nitrate ion ($NO_3^-$), as they remain in solution and do not participate in forming the precipitate. -
Write the net ionic equation for the reaction between aqueous solutions of
barium chloride ($BaCl_2$) and sodium sulfate ($Na_2SO_4$).
Solution: The precipitate formed is barium sulfate ($BaSO_4$). The ions that form it are $Ba^{2+}$ and $SO_4^{2-}$.
Net Ionic Equation: $Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)$
8.10 Preparing Insoluble Salts by Precipitation
Precipitation provides a simple and direct method for preparing a pure sample of an insoluble salt.
The steps are as follows:
- Choose Reactants: Select two soluble ionic compounds that contain the required cation and anion. For example, to make insoluble lead(II) chloride ($PbCl_2$), you could choose soluble lead(II) nitrate ($Pb(NO_3)_2$) and soluble sodium chloride (NaCl).
- Mix Solutions: Mix the two aqueous solutions together. The precipitate will form instantly. It is good practice to use one reactant in excess to ensure all of the limiting reactant is precipitated.
- Filter: Separate the solid precipitate from the solution using filtration. The solid collected on the filter paper is the residue.
- Wash: Wash the precipitate on the filter paper with a small amount of distilled water to remove any soluble spectator ions.
- Dry: Carefully remove the filter paper and allow the solid precipitate to dry, for example, in a warm oven.
Solved Examples:
-
You want to prepare solid copper(II) carbonate ($CuCO_3$). Suggest two
soluble starting compounds.
Solution: You need a soluble salt containing $Cu^{2+}$ and a soluble salt containing $CO_3^{2-}$. Good choices would be copper(II) nitrate ($Cu(NO_3)_2$) or copper(II) sulfate ($CuSO_4$), mixed with sodium carbonate ($Na_2CO_3$) or potassium carbonate ($K_2CO_3$). -
Why is the precipitate washed with distilled water after
filtration?
Solution: The precipitate is washed to remove any soluble spectator ions that may be clinging to the surface of the solid. This ensures the final product is pure. -
Does it matter which reactant is in excess when preparing an insoluble
salt?
Solution: No, it does not significantly matter. Since the product is an insoluble solid that will be filtered off, any excess soluble reactant will simply be washed away with the filtrate. However, using an excess of one reactant ensures the complete precipitation of the other, maximizing the yield.