Exercise 1. An aqueous silver nitrate solution is mixed with an aqueous solution of potassium carbonate.

a) Write the chemical formulas for silver nitrate and potassium carbonate

AgNO₃, K₂CO₃

b) Write the balanced molecular (overall) equation for the reaction.

2 AgNO₃ (aq) + K₂CO₃ (aq) → Ag₂CO₃ (s) + 2 KNO₃ (aq)

c) Write the ionic equation

2 Ag⁺(aq) + 2 NO₃^–(aq) + 2 K⁺(aq) + CO₃^2-(aq) → Ag₂CO₃(s) + 2 K⁺(aq) + 2 NO₃^–(aq)

d) Write the net ionic equation

2 Ag⁺(aq) + CO₃^2- (aq) → Ag₂CO₃(s)

e) What are the spectator ions

K⁺ (aq) and NO₃^– (aq)

Back to Precipitation Reactions
 
Exercise 2. An aqueous solution of sodium oxalate is mixed with an aqueous solution of calcium chloride.

a) Write the balanced molecular (overall) equation for the reaction.

Na₂C₂O₄ (aq) + CaCl₂ (aq) → 2 NaCl (aq) + CaC₂O₄ (s)

b) Write the ionic equation

2 Na⁺ (aq) + C₂O₄^-2 (aq) + Ca²⁺ (aq) + 2 Cl^– (aq)→ CaC₂O₄ (s) + 2 Na⁺ (aq) + 2 Cl^– (aq)

c) Write the net ionic equation

Ca²⁺ (aq) + C₂O₄^-2 (aq) → CaC₂O₄ (s)

d) What are the spectator ions?

Na⁺ and Cl^–

Back to Precipitation Reactions

 
Exercise 3. An aqueous solution of iron(III) chloride is mixed with an aqueous solution of lead(II) nitrate.

a) Write the balanced molecular (overall) equation for the reaction.

2 FeCl₃ (aq) + 3 Pb(NO₃)₂ (aq) → 3 PbCl₂ (s) + 2 Fe(NO₃)₃ (aq)

b) Write the ionic equation

2 Fe³⁺ (aq) + 6 Cl^– (aq) + 3 Pb²⁺ (aq) + 6 NO₃^– (aq) → 3 PbCl₂ (s) + 2 Fe³⁺ (aq) + 6 NO₃^– (aq)

c) Write the net ionic equation

3 Pb²⁺ (aq) + 6 Cl^– (aq)→ 3 PbCl₂ (s)
Divide coefficients by 3 and the net ionic equation is:
Pb²⁺ (aq) + 2 Cl^– (aq)→ PbCl₂ (s)

d) What are the spectator ions?

Fe³⁺ and NO₃^–

Back to Precipitation Reactions

 
Exercise 4. Write the net ionic equation for the following. Identify any spectator ions.

Fe(ClO₄)₂ (aq) + (NH₄)₃(PO₄)₂ (aq) → ??

3 Fe⁺² (aq) + 2 PO₄^3- (aq) → Fe₃(PO₄)₂ (s)

The spectator ions are NH₄⁺ and ClO₄^–

Back to Precipitation Reactions

 
Exercise 5. If aqueous solutions of KBr and ZnCrO₄ are mixed, will a precipitate form? If so, write the net ionic equation, and identify any spectator ions.

The molecular equation is:

KBr(aq) + ZnÇrO₄(aq) → K₂CrO₄(aq) + ZnBr₂ (aq)

Notice that both products, K₂CrO₄ and ZnBr₂ are both soluble. There is no reaction–no precipitate is formed.

You can determine this from the ionic equation. The ions all cancel.

Zn²⁺ (aq) + CrO₄^2- (aq) + K⁺ (aq) + Br^– (aq) → Zn²⁺ (aq) + CrO₄^2- (aq) + K⁺ (aq) + Br^– (aq)

Back to Precipitation Reactions

 
Exercise 6. A chemist mixes 50.00 mL of aqueous 0.100 M K₂SO₄ with 45.00 mL of aqueous 0.150 M BaCl₂. What is the mass, in g, of the precipitate formed?

Step 1. Write the balanced molecular equation.

K₂SO₄ (aq) + BaCl₂ (aq) → BaSO₄ (s) + 2 KCl (aq)

Step 2. Next, determine the moles of K₂SO₄ and BaCl₂ by multiplying the volume, in L, by the molarity. The molar mass of K₂SO₄ is 174.259 g/mol and the molar mass of BaCl₂ is 208.23 g/mol.

moles_K2SO4= 0.100 M x 0.05000 L= 0.005000 moles K₂SO₄
moles_BaCl2= 0.150 M x 0.04500 L= 0.006750 moles BaCl₂

Step 3. Our limiting reactant is K₂SO₄. We can now use our mole ratios from the balanced chemical equation to determine the moles and then the grams of BaSO₄ formed. The molar mass of BaSO₄ is 233.38 g/mol. The molar ratio of K₂SO₄ to BaSO₄ is 1:1.

\(\displaystyle 0.005000\;mol K_2SO_4\times\frac{1\;mol\;BaSO_4}{1\;mol\;K_2SO_4}\;=\;{0.005000\;mol\;BaSO_4}\)

Step 4. We can now calculate the number of grams BaSO₄ formed.

\(\displaystyle 0.005000\;mol BaSO_4\times\frac{233.38\;g\;BaSO_4}{1\;mol\;BaSO_4}\;=\;\mathbf{1.28\;g\;BaSO_4}\)

Back to Precipitation Reactions

 
Exercise 7. Determine if a precipitation reaction will occur if aqueous solutions of the substances in each pair are mixed. If a precipitation reaction does occur, write a net ionic equation and identify the spectator ions.

a) KOH and HClO₄

The molecular equation is:

KOH (aq) + HClO₄(aq) → K⁺ (aq) + ClO₄^–(aq) + H²O (l)
We see this is the reaction between a strong acid and a strong base that produces a soluble salt and water
No precipitate forms. We have the net ionic equation:
H⁺ (aq) + OH^– (aq) → H₂O (l)

b) Hg(NO₃)₂ and Na₃PO₄

The molecular equation is:

3 Hg(NO₃)₂(aq) + 2 Na₃PO₄(aq) → Hg₃(PO₄)₂ (s) + 6 NaNO₃ (aq)

The net ionic equation is:
3 Hg²+ (aq) + 2 PO₄^3- (aq) → Hg₃(PO₄)₂ (s)
The spectator ions are Na⁺ and NO₃^–

c) CoCl₂ and Pb(NO₃)₂

The molecular equation is:
CoCl₂ + Pb(NO₃)₂ → PbCl₂ (s) + Co(NO₃)₂ (aq)

The net ionic equation is:

Pb²⁺ (aq) + Cl^– (aq) → PbCl₂ (s)
The spectator ions are Co²⁺ and NO₃^–

d) Sr(OH)₂ and BaCl₂

The molecular equation is:

Sr(OH)₂ + BaCl₂ → SrCl₂ (aq) + Ba(OH)₂ (aq)

Both SrCl₂ and Ba(OH)₂ are soluble, therefore NO REACTION

Back to Precipitation Reactions

 
Exercise 8. A chemist mixes 150.0 mL of 0.100 M Na₂SO₄, 75.0 mL 0.350 M CoCl₂, and 150.0 mL of 0.250 M SrS.

a) Which salts (ionic compounds) will precipitate out of solution?

Ions in each solution are Na⁺ and SO₄^2-; Co²⁺ and Cl^–; Sr²⁺ and S^2-; When the solutions are mixed, CoS and SrSO₄ will precipitate.

b) Assuming complete precipitation of insoluble salts, what is the concentration, in M, of each ion that remains in solution?

1. Find the number of moles of each ion

Moles of Na₂SO₄ = 0.1500 L x 0.100 M = 0.0150 mol Na₂SO₄
Moles of Na⁺ = 2 x 0.0150 mols = 0.0300 mol Na⁺
Moles of SO₄^2- = 0.0150 mol SO₄^2-

Moles of CoCl₂ = 0.0750 L x 0.350 M = 0.0263 mol CoCl
Moles of Co²⁺ = 0.0263 mol Co²⁺
Moles of Cl^– = 2 x 0.0263 mols = 0.0526 mol Cl^–

Moles of SrS = 0.1500 L x 0.250 M = 0.0375 mol SrS
Moles of Sr²⁺ = 0.0375 mol Sr²⁺
Moles of S^2- = 0.0375 mol S^2-

2. The precipitation reactions are:

1) Co²⁺ (aq) + S^2- (aq) → CoS (s)

2) Sr²⁺ + SO₄^-2 (aq) → SrSO₄ (s)

For reaction 1) we have 0.0263 mol Co²⁺ and 0.0375 mol of S^2-. The limiting reactant is Co²⁺. The amount of S^2- required for reaction is:

\(\displaystyle 0.0263\;mol Co^{2+}\times\frac{1\;mol\;S^{2-}}{1\;mol\; Co^{2+}}\;=\;{0.0263\;mol\;S^{2-}}\)

Moles of S^2- unreacted is 0.0375 mol – 0.0263 mol = 0.0112 mol S^2-

For reaction 2 we have 0.0375 mol Sr²⁺ and 0.0150 mol SO₄^2-. The limiting reactant is SO₄^2-. The amount of Sr²⁺ required for reaction is

\(\displaystyle 0.0150\;mol SO_4^{2-}\times\frac{1\;mol\;Sr^{2+}}{1\;mol\; SO_4^{2-}}\;=\;{0.0150\;mol\;Sr^{2+}}\)

The amount of unreacted Sr²⁺ is 0.0375 mol – 0.0150 mol = 0.0225 mol Sr²⁺

3. Now we can determine the concentration, M, of each ion in solution.

All of the SO₄^2- and Co²⁺ has been reacted, therefore,

[SO₄^2-] = [Co²⁺] = 0.

The total volume of solution is:

V_total = 150.0 mL + 75.0 mL + 150 mL = 375.0 mL = 0.3750 L

\(\displaystyle [{S^{2-}]\;=\;\frac{0.0112\;mol}{0.3750\;L}\;=\;0.0299\;M\;}\)

 
\(\displaystyle [{Sr^{2+}]\;=\;\frac{0.0225\;mol}{0.3750\;L}\;=\;0.0600\;M\;}\)

 
\(\displaystyle [{Na^{+}]\;=\;\frac{0.0300\;mol}{0.3750\;L}\;=\;0.0800\;M\;}\)

 
\(\displaystyle [{Cl^{-}]\;=\;\frac{0.0526\;mol}{0.3750\;L}\;=\;0.140\;M\;}\)

Back to Precipitation Reactions

 

 
Exercise 9. A chemist has 4 aqueous solutions, each solution containing the following pairs of ions. The solutions were prepared with sodium and potassium salts.

Solution 1. Cl^– and NO₃^–
Solution 2. Ca²⁺ and NH₄⁺
Solution 3. Fe²⁺ and Ba²⁺
Solution 4. SO₄^2- and CO₃^2-

The chemist has the following salts available: Pb(NO₃)₂, NaOH, CaS, Na₂SO₄, K₂C₂O₄, and CsNO₃. For each solution, indicate the salt that would separate the ions using a precipitation reaction. Write the net ionic equation for each.

For solution 1. Adding Pb(NO₃)₂ would precipitate PbCl₂. The chloride would be removed leaving the NO₃^– ions. The net ionic equation is:

Pb²⁺ (aq) + 2 Cl^– (aq) → PbCl₂ (s)

For solution 2. K₂C₂O₄ would precipitate CaC₂O₄. The calcium ion would be removed leaving the NH₄⁺ ions. The net ionic equation is:

Ca²⁺ (aq) + C₂O₄^2- (aq) → CaC₂O₄ (s)

For solution 3. Na₂SO₄ would precipitate BaSO₄. The barium ion would be removed leaving the Fe²⁺ ions. The net ionic equation is:

Ba²⁺ (aq) + SO₄^2- (aq) → BaSO₄ (s)

For solution 4. CaS would precipitate CaCO₃. The carbonate ions would be removed leaving the sulfate ions. The net ionic equation is:

Ca²⁺ (aq) + CO₃^2- (aq) → CaCO₃ (s)

Back to Precipitation Reactions

 
Exercise 10. A chemist mixed 250.00 mL of aqueoud 0.450 M Na₃PO₄ and with 300.00 mL of aqueous 0.355 M CaCl₂? She recovered 10.39 g of a precipitate. What is the percent yield?

The balanced molecular equation is:

2 Na₃PO₄ (aq) + 3 CaCl₂ (aq) → Ca₃(PO₄)₂ (s) + 6 NaCl (aq)

Next we find the number of moles of CaCl₂ and Na₃PO₄

moles CaCl₂ = 0.30000 L x 0.355 M = 0.1065 mol CaCl₂

moles Na₃PO₄ = 0.25000 L x 0.450 M = 0.1125 mol Na₃PO₄

Our limiting reactant is CaCl₂

\(\displaystyle 0.1065\;mol CaCl_2\times\frac{1\;mol\;Ca_3(PO_4)_2}{3\;mol\;CaCl_2}\;=\;{0.0355\;mol\;Ca_3(PO_4)_2}\)
 

The molar mass of Ca₃(PO₄)₂ is 310.18 g/mol
 

\(\displaystyle 0.0355\;mol\;Ca_3(PO_4)_2\times\frac{310.18\;g\;Ca_3(PO_4)_2}{1\;mol\;Ca_3(PO_4)_2}\;=\;{11.0\;g\;Ca_3(PO_4)_2}\)

The percent yield

\(\displaystyle percent\;yield\;=\;\frac{10.39\;g}{11.0\;g}\;\times\;100\;=\;\mathbf {94.5\;\%}\)

 
 
Back to Precipitation Reactions
Back to General Chemistry 1 Study Guides