AP Chemistry · Free-response skill

A precipitation calculation, step by step

The move. Work a reaction calculation in labeled steps: write the net ionic equation, find the limiting reactant, and compute a product mass and a leftover concentration with units and significant figures.

How to do it

  1. Start every calculation question with the balanced equation, written as a net ionic equation when ions in water react. Check atoms and charge before you use it.
  2. Turn every given into moles first: molarity times volume in liters, or grams divided by molar mass. Label each number with its unit and substance, such as 0.0150 mol Ba²⁺.
  3. Find the limiting reactant by one clear comparison through the mole ratio, and then compute every product from that reactant only. The amount of the other reactant left over is what you started with minus what was used.
  4. Round only the final answer, to the significant figures your data support, and give its unit.

Practice prompt Try it

A student mixes 25.0 mL of 0.400 M AgNO₃(aq) with 35.0 mL of 0.200 M K₂CrO₄(aq). A red-brown solid, silver chromate, Ag₂CrO₄ (331.74 g/mol), forms. Assume the volumes add to 60.0 mL and the solid forms completely.

Answer each part below. Show your work and include units.

  1. Write the balanced net ionic equation for the formation of the red-brown solid.
  2. Show by calculation which reactant runs out first.
  3. Find the mass of Ag₂CrO₄ that forms.
  4. Find the concentration of chromate ions, CrO₄²⁻, still dissolved in the 60.0 mL mixture after the reaction.

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Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 5 points

Model answer, annotated

(a) 2Ag⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄(s). The K⁺ and NO₃⁻ ions are spectators, so they are left out.

  • Point 1 Two silver ions per chromate ion balance the atoms and the charge (0 on each side).

(b) Moles Ag⁺ = 0.400 M × 0.0250 L = 0.0100 mol. Moles CrO₄²⁻ = 0.200 M × 0.0350 L = 0.00700 mol. By the 2 : 1 ratio, 0.0100 mol Ag⁺ needs only 0.00500 mol CrO₄²⁻, and 0.00700 mol is present, so Ag⁺ (the AgNO₃) runs out first.

  • Point 2 Names the limiting reactant plainly.
  • Point 3 Shows both mole amounts and the comparison through the 2 : 1 ratio.

(c) From the limiting Ag⁺: 0.0100 mol Ag⁺ × (1 mol Ag₂CrO₄ / 2 mol Ag⁺) = 0.00500 mol Ag₂CrO₄. Mass = 0.00500 mol × 331.74 g/mol = 1.66 g Ag₂CrO₄.

  • Point 4 Uses the limiting reactant, the mole ratio and the molar mass, and gives 3 significant figures with the unit.

(d) CrO₄²⁻ used = 0.00500 mol, so 0.00700 − 0.00500 = 0.00200 mol CrO₄²⁻ is left. Concentration = 0.00200 mol ÷ 0.0600 L = 0.0333 M CrO₄²⁻.

  • Point 5 Subtracts the amount used from the starting amount and divides by the total volume of the mixture.