Dilution Conserves Solute Amount

Introduction

Adding solvent makes a solution less concentrated without creating or destroying the solute. This conservation gives the familiar equation c₁V₁ = c₂V₂. The equation is useful because each side represents the same solute moles, but it applies only when the named solute is conserved through the preparation.

Core explanation

Molarity is c = n/V, so a volume V of stock solution contains n₁ = c₁V₁ moles. After pure solvent is added and the mixture made to a final volume V₂, n₂ = c₂V₂. If no solute is lost, added, precipitated or consumed, n₁ = n₂. Therefore c₁V₁ = c₂V₂. Volumes may both be in mL for this particular equation if their units match, because the volume units cancel when solving the concentration ratio; use litres when calculating actual moles with c in mol L⁻¹.

If 20.0 mL of 0.500 M solution is diluted to 100.0 mL final volume, c₂ = 0.500 × 20.0/100.0 = 0.100 M. The concentration falls by a factor of five while the total solute amount remains 0.0100 mol. The 100.0 mL is final solution volume, not 100.0 mL of water added. A final volume of 100.0 mL usually requires less than 100.0 mL added solvent because the stock already occupies volume.

The dilution factor can be written V₂/V₁ = c₁/c₂ when solute amount is conserved. It applies to a defined solute; a stock with several solutes dilutes each conserved one by the same final-to-aliquot volume ratio. But if a reaction occurs, the concentration of a reactive species cannot be found from dilution alone. Mixing acid with base consumes some original acid and base; a reaction balance must precede final concentration.

The equation is also unsuitable if material precipitates during dilution or if volatile solute escapes. A prepared solution can be contaminated or transferred incompletely, changing the actual solute amount. In ideal school problems these effects are excluded, but stating the conservation assumption helps prevent mechanical misuse of the formula.

Step-by-step reasoning

1. Identify the same named solute before and after dilution. 2. Check that only solvent is added and no solute is lost or transformed. 3. Mark V₁ as the stock portion and V₂ as final solution volume. 4. Use c₁V₁ = c₂V₂ and solve for the unknown. 5. Check that a true dilution gives c₂ lower than c₁.

Visual explanation

Draw a small dark-colored stock portion labeled c₁,V₁ poured into a larger calibrated flask. Show the same number of red solute dots distributed across the larger final V₂ and label the lighter solution c₂.

Real-world analogy

One spoonful of syrup contains a fixed amount of sugar. Adding water spreads the same sugar through more drink, reducing sugar per cup. The analogy captures conservation but not exact laboratory volume measurement.

Real-world example

A laboratory may dilute a concentrated dye stock for a spectrophotometer. If the dye is stable and transferred fully, a measured stock aliquot and final flask volume set the working concentration without weighing dye again.

Why?

Why does c₂ decrease when V₂ increases? The solute mole numerator remains fixed and the solution-volume denominator grows. This is the meaning behind the equation, not a separate chemical reaction.

Common misconception

“V₂ is the amount of water added.” It is the final total solution volume. Using added-water volume can give the wrong dilution factor.

Worked example

What final volume is needed to dilute 15.0 mL of 0.800 M stock to 0.120 M? V₂ = c₁V₁/c₂ = 0.800 × 15.0/0.120 = 100 mL. The stock contains 0.800 × 0.0150 = 0.0120 mol; 0.0120 mol in 0.100 L is 0.120 M. Make the solution up to 100 mL, not by adding exactly 100 mL water.

Quick check

1. Does dilution change the number of solute moles in the transferred portion? Answer: No, provided only solvent is added and the solute is neither lost nor chemically changed.

Exam focus

State conservation of named solute and label initial aliquot versus final volume. A lower target concentration should require a larger final volume.

Advanced insight

Mixing can change total volume nonadditively. The cV relation remains valid if V₂ is measured as the final volume, which is why calibrated glassware is preferable to adding calculated separate volumes.

Summary

The dilution equation follows from cV = n and unchanged solute moles. It uses stock portion volume and final solution volume. Reaction, precipitation or loss breaks its simple conservation assumption.

Practice questions

1. Dilute 25.0 mL of 0.400 M to 200.0 mL. Find c₂. Answer: c₂ = 0.400 × 25.0/200.0 = 0.0500 M. 2. Is adding 200 mL water to 25 mL stock the same as making up to 200 mL final? Answer: No. The first gives roughly 225 mL final or another measured value; the second ends at 200 mL total solution. 3. Why not use c₁V₁ = c₂V₂ for acid remaining after adding reactive base? Answer: Some acid is consumed by reaction, so its moles are not conserved; first calculate the reacted and remaining amounts.