Concentrations After Reaction

Introduction

When two solutions react, adding their initial moles is not enough to describe the final mixture. Some reactants are consumed and products may appear or precipitate. The amount balance must follow the balanced equation first; only then should the remaining dissolved amount be divided by final volume.

Core explanation

For an acid-base example, HCl + NaOH → NaCl + H₂O has a 1:1 mole ratio. Mix 50.0 mL of 0.200 M HCl with 30.0 mL of 0.100 M NaOH. Initial HCl is 0.0500 × 0.200 = 0.0100 mol; NaOH is 0.0300 × 0.100 = 0.00300 mol. The base is limiting and consumes 0.00300 mol HCl. HCl excess is 0.0100 − 0.00300 = 0.00700 mol. If final volume is 0.0800 L and complete reaction is assumed, formal excess HCl concentration is 0.00700/0.0800 = 0.0875 M.

The c₁V₁ = c₂V₂ dilution relation cannot be applied directly to HCl here because HCl moles are not conserved. Water and the other solution dilute the mixture, but reaction also consumes acid. A useful sequence is “initial amounts → reaction extent → final amounts → final volume.” This sequence works for precipitation and other aqueous reactions as well.

Not every concentration in the final liquid represents a molecule with the original formula. Strong acid and base solutions are often modeled using ions. If a question asks for chloride concentration after the HCl/NaOH reaction, chloride is a spectator ion and may remain at its initial total amount even when H⁺ is consumed. If it asks for NaCl formula-equivalent concentration, use product stoichiometry. Name the species carefully instead of saying only “the concentration.”

For a precipitation reaction, an insoluble product leaves the dissolved phase. Mixing AgNO₃ and NaCl can make AgCl(s); the solid is not included in the dissolved chloride or silver-ion concentration. Any excess dissolved ion is found using the initial cV amounts and 1:1 precipitation ratio. Real equilibria leave tiny residual concentrations, but a complete-precipitation school model often neglects them unless solubility data are supplied.

Final volume is not automatically the sum of starting volumes in every real mixture. Use a supplied measured value, or use volume additivity only when the exercise permits it. If an acid-base reaction releases heat, temperature may affect volume and equilibrium; introductory calculations usually stipulate conditions that make a simplified final volume reasonable.

Step-by-step reasoning

1. Write and balance the reaction for the named species. 2. Convert each starting solution's cV to moles. 3. Identify the limiting reactant and subtract reacted amounts. 4. Identify which final species is requested and whether it remains dissolved. 5. Divide its final moles by the final solution volume.

Visual explanation

Draw two incoming arrows labeled 0.0100 mol HCl and 0.00300 mol NaOH. A reaction box removes 0.00300 mol from each, leaving 0.00700 mol acid-equivalent. A final arrow divides by 0.0800 L.

Real-world analogy

Combining two inventories and then using matched pairs changes what remains. Counting the original boxes together without subtracting assembled pairs overstates leftover parts. A reaction balance plays the role of the assembly rule.

Real-world example

A titration beyond equivalence leaves excess titrant. Calculating its concentration requires the delivered titrant moles minus the amount consumed by analyte, followed by division by the mixture's final volume.

Why?

Why is a plain concentration average wrong after reaction? The numerator for a reactant changes chemically. Averaging can account for volume proportions but cannot remove moles consumed by a balanced reaction.

Common misconception

“Mixing two solutions only dilutes both.” If their solutes react, the initial species can be consumed and new dissolved or solid products form.

Worked example

Mix 20.0 mL of 0.100 M AgNO₃ with 30.0 mL of 0.100 M NaCl, assuming AgCl precipitates completely and final liquid volume is 50.0 mL. Initial silver is 0.00200 mol and chloride 0.00300 mol. Silver limits; 0.00200 mol chloride is consumed, leaving 0.00100 mol dissolved chloride. Its formal final concentration is 0.00100/0.0500 = 0.0200 M. The AgCl solid is excluded from liquid concentration.

Quick check

1. Why must reaction stoichiometry be applied before final concentration? Answer: Reactant moles can be consumed, so initial cV amounts do not necessarily equal the final dissolved moles being requested.

Exam focus

Use a mole table or clearly labeled initial, change and final amounts. State the final-volume assumption and the species counted.

Advanced insight

If the reaction reaches an equilibrium rather than going fully to completion, final concentrations require an equilibrium calculation. The complete-reaction subtraction is an explicit model, not a universal property of all aqueous chemistry.

Summary

Reaction mixtures require chemical accounting before volume division. Find initial cV amounts, use balanced coefficients and limiting behavior, then divide the final dissolved species amount by final solution volume.

Practice questions

1. Mix 0.010 mol HCl and 0.006 mol NaOH in 0.100 L final solution. Find formal excess acid concentration. Answer: A 1:1 reaction consumes 0.006 mol HCl; 0.004 mol remains, giving 0.040 M. 2. Why is a precipitated solid excluded from dissolved-ion concentration? Answer: It is a separate solid phase, so its amount is not part of the ions dispersed in the liquid. 3. What do you need to know after calculating remaining moles? Answer: The final liquid solution volume, measured or validly assumed, is needed for molarity.