Solution Stoichiometry and Precipitation

Solution stoichiometry uses the molarity and volume of solutions together with a balanced equation to calculate the amounts of reactants and products, and a precipitation reaction is one in which two soluble solutions form an insoluble solid. In brief: mixing 25.0 mL of 0.200 M AgNO₃ with 30.0 mL of 0.150 M NaCl gives 0.00450 mol of AgCl, which is 0.645 g of white precipitate.

Chemists often work with reagents in solution because dissolved substances react quickly and their amounts are easy to measure by volume. Precipitation reactions are used in water treatment, in analysis and in the production of pigments. This article explains how to predict a precipitate with solubility rules, how to write net ionic equations and how to find the limiting reagent and the product in solution problems.

What is the method for solution stoichiometry?

1. Write a balanced equation for the reaction.
2. Calculate the moles of each reactant: n = M × V (with V in liters).
3. Use the mole ratio to find the limiting reagent.
4. Calculate the moles of product, then convert them to mass (m = n × M_molar) or concentration.
5. For the concentrations after the reaction, divide the moles by the total volume of the mixed solutions.

Which solubility rules predict precipitates?

These rules are guidelines, and the more precise criterion is the solubility product, K_sp.

How do you write a net ionic equation?

When silver nitrate and sodium chloride solutions are mixed, the molecular equation is AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq). Writing every soluble strong electrolyte as ions gives the complete ionic equation: Ag⁺ + NO₃⁻ + Na⁺ + Cl⁻ → AgCl(s) + Na⁺ + NO₃⁻. Removing the spectator ions (Na⁺ and NO₃⁻), which are present on both sides, leaves the net ionic equation:

Ag+(aq) + Cl−(aq) → AgCl(s)

Worked example: silver chloride

25.0 mL of 0.200 M AgNO₃ is mixed with 30.0 mL of 0.150 M NaCl. Find the mass of AgCl formed and the ion concentrations that remain.

1. n(Ag⁺) = 0.200 × 0.0250 = 0.00500 mol.
2. n(Cl⁻) = 0.150 × 0.0300 = 0.00450 mol.
3. The ratio is 1 : 1, so Cl⁻ is limiting and 0.00450 mol of AgCl forms.
4. m(AgCl) = 0.00450 × 143.32 = 0.645 g.
5. Excess Ag⁺ = 0.00500 − 0.00450 = 0.00050 mol. The total volume is 55.0 mL, so [Ag⁺] = 0.00050 / 0.0550 = 0.0091 M.
6. [Na⁺] = 0.00450 / 0.0550 = 0.082 M, and [NO₃⁻] = 0.00500 / 0.0550 = 0.091 M.

Worked example: lead(II) iodide

50.0 mL of 0.100 M Pb(NO₃)₂ is mixed with 50.0 mL of 0.300 M KI. The balanced equation is Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq).

1. n(Pb²⁺) = 0.100 × 0.0500 = 0.00500 mol.
2. n(I⁻) = 0.300 × 0.0500 = 0.0150 mol.
3. The reaction needs 2 mol of I⁻ per mol of Pb²⁺, or 0.0100 mol in this case, and 0.0150 mol is available, so Pb²⁺ is limiting.
4. n(PbI₂) = 0.00500 mol, and m = 0.00500 × 461.0 = 2.31 g of a bright yellow precipitate.
5. Excess I⁻ = 0.0150 − 0.0100 = 0.0050 mol, and [I⁻] = 0.0050 / 0.1000 L = 0.050 M.

Worked example: a titration

In a titration, a solution of known concentration is added to another until the reaction is complete. If 20.00 mL of NaOH solution neutralizes 25.00 mL of 0.1000 M HCl, what is the concentration of the NaOH?

1. HCl + NaOH → NaCl + H₂O (1 : 1).
2. n(HCl) = 0.1000 × 0.02500 = 2.500 × 10⁻³ mol = n(NaOH).
3. [NaOH] = 2.500 × 10⁻³ / 0.02000 = 0.1250 M.

Where are precipitation reactions used?

• Water treatment: phosphate and heavy metals are removed from wastewater by precipitation with lime or iron salts.
• Analysis: gravimetric analysis determines the amount of an ion by weighing a precipitate, such as BaSO₄ for sulfate or AgCl for chloride.
• Pigments: many colored compounds are made by precipitation.
• Qualitative tests: the formation of a white precipitate with silver nitrate indicates chloride ions.

Where does the word stoichiometry come from?

The German chemist Jeremias Benjamin Richter introduced the term stoichiometry in the 1790s, from the Greek words for element and measure. He studied the weights of acids and bases that neutralize one another. Later work by Joseph Proust and John Dalton on the laws of definite and multiple proportions put the calculation of reacting amounts on a firm footing.

Which mistakes are common?

• Forgetting to find the limiting reagent: the product is determined by the reagent that runs out first.
• Using the volume of only one solution: the concentrations after mixing use the total volume.
• Not converting mL to L: n = M × V requires liters.
• Ignoring the stoichiometric coefficients: PbI₂ needs two I⁻ for every Pb²⁺.
• Leaving spectator ions in the net equation: only the species that react appear.