Precipitation Reaction

Precipitation Reactions §

Precipitate - insoluble solid that separates from solid Refer to Solubility Rules for these types of reactions

ppt = precipitation rxn = reaction

This is similar to double displacement reactions

Reactions that will create precipitate (when a solid forms)

When 2 aqueous solutions of iconic compounds are mixed, an insoluble substance forms

Pb(NO${}_3$)${}_2$ (aq) + 2 NaI (aq) -> PbI${}_2$ (s) + 2NaNO${}_3$ (aq) Molecular Formula

Pb${}^{2+}$ + 2NO${}_3$ ${}^{-}$ + 2 Na${}^{+}$ + 2I${}^{-}$ -> PbI${}_2$ (s) + 2Na${}^{+}$ + 2NO${}_3$ ${}^{-}$ Complete ionic equations (dissect every compound except the precipitate so everything that’s aqueous)

Pb${}^{2+}$ + 2I${}^{-}$ -> PbI${}_2$ (s) Net ionic equation

Na${}^{+}$ and NO${}_3$ ${}^{-}$ are spectator ions

Stoichiometry of ppt reactions §

100.0 mL of 0.100 M solution of barium chloride is mixed with 100.0 mL of 0.100 M solution of iron (III) sulfate

Step 1: what species are present?

Ba${}^{2+}$ Cl${}^{-}$ Fe${}^{3+}$ SO${}_4$ ${}^{2-}$

Step 2: write the balanced net ionic equation

Ba${}^{2+}$ + SO${}_4$ ${}^{2-}$ -> BaSO${}_4$ (s)

Step 3: Calculate the # of moles of each reactant

0.1000 L * 0.100 $\frac{mol}{L}$ BaCl${}_2$ = 0.0100 mol BaCl${}_2$ (rounded using Significant Digits)

0.100 mol BaCl${}_2$ * $\frac{1molB{a}^{2+}}{1molBaC{l}_2}$ = 0.0100 mol Ba${}^{2+}$

for every BaCl${}_2$ there’s 1 Ba${}^{2+}$ ion hence the 1:1 mole ratio

0.1000 L * 0.100 M Fe${}_2$(SO${}_4$)${}_3$ = 0.0100 mol Fe${}_2$(SO${}_4$)${}_3$

0.0100 mol Fe${}_2$(SO${}_4$)${}_3$ * $\frac{3molS{O}_4^{2-}}{1molF{e}_2(S{O}_4{)}_3}$ = 0.0300 mol SO_4$$^{2-}

Step 4: Determining the limiting reactant

0.0100 mol Ba${}^{2+}$ * $\frac{1molS{O}_4^{2-}}{1molB{a}^{2+}}$ = 0.0100 mol SO${}_4^{2-}$

0.0300 mol of SO${}_4$ ${}^{2-}$ available and need only 0.100 mol so Ba${}^{2+}$ is the L.R

Step 5: Calculate the # of moles of product

0.0100 mol of Ba ${}^{2+}$ * $\frac{1molofBaS{O}_4}{1molofB{a}^{2+}}$ = 0.0100 mol of BaSO${}_4$

Step 6: convert from moles to grams

0.0100 mol of BaSO${}_4$ * $\frac{233.38g}{1mol}$ = 2.33 grams of BaSO${}_4$