Solubility and precipitation

A precipitate forms when a product of an aqueous reaction is insoluble. Precipitates typically form from reactions between ions in solutions of ionic compounds. The solubility rules are a guide that helps you determine which reaction products will form solid precipitates. Any reaction product that is insoluble will precipitate out of solution. Read the text aloud

solubility rules for common compounds

soluble compounds	insoluble compounds except with group 1 metal ions and NH₄⁺

group 1 metal ions Li⁺, Na⁺, K⁺, Rb⁺, Cs⁺	carbonates, CO₃²⁻
ammonium, NH₄⁺	hydroxides, OH⁻, except with Ba
acetate, C₂H₃O₂⁻ or CH₃COO⁻	chlorides of Pb, Ag and Hg
nitrates, NO₃⁻	bromides of Pb, Ag and Hg
sulfates, SO₄²⁻, except with Ba, Sr, and Pb	iodides of Pb, Ag and Hg
chlorides, except with Pb, Ag and Hg	sulfides, S²⁻
bromides, except with Pb, Ag and Hg	phosphates, PO₄³⁻
iodides, except with Pb, Ag and Hg

Lead(II) nitrate, Pb(NO₃)₂, and potassium iodide, KI solutions undergo the reaction: Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(?) + 2KNO₃(?). How can you tell what state the products are in? Use the solubility rules above to see why the reactants are aqueous and to predict the state of products formed. Nitrate compounds are soluble, so Pb(NO₃)₂ is dissolved in an aqueous solution (aq). For KI, potassium ions are soluble and so are iodides of potassium; potassium iodide is a soluble compound (aq). Iodides of lead are not soluble so PbI₂ will form a solid precipitate (s). Nitrates of potassium are soluble so KNO₃ will remain aqueous (aq). Any soluble compound dissociates into its ions in solution, and insoluble compounds precipitate out as a solid. Notice the solid PbI₂ at the bottom of the beaker shown below. Read the text aloud

Lead(II) iodide precipitates out the solution of lead(II) nitrate and potassium iodide.

Write the balanced equation for the reaction between magnesium chloride and sodium carbonate solutions. Include states for each compound.

Relationships	If two compounds react, a double replacement reaction will occur.
Solve	Write ions for all reactants: Mg²⁺ Cl⁻ Na⁺ CO₃²⁺ Criss-cross to get formulas; switch cations to get products: MgCl₂ + Na₂CO₃ → MgCO₃ + NaCl Balance the reaction: MgCl₂ + Na₂CO₃ → MgCO₃ + 2NaCl According to solubility rules, everything except MgCO₃ is soluble. MgCl₂(aq) + Na₂CO₃(aq) → MgCO₃(s) + 2NaCl(aq)
Answer	MgCl₂(aq) + Na₂CO₃(aq) → MgCO₃(s) + 2NaCl(aq)

Use the solubility rules to determine the states of the products in the following balanced reaction.
6CsOH(aq) + Sr₃(PO₄)₂(s) → 3Sr(OH)₂(?) + 2Cs₃PO₄(?)
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