Dilution with a Measured Final Volume

Introduction

Dilution changes concentration by adding solvent while leaving the amount of a conserved solute unchanged. A measured aliquot of stock solution is transferred and brought to a specified final volume. The key equation, c₁V₁ = c₂V₂, is a mole balance. Its V₂ is the final total solution volume , not the volume of solvent poured in.

Core explanation

If stock molarity is c₁ and the aliquot volume is V₁, initial solute moles are n₁ = c₁V₁. If solvent addition causes no reaction, precipitation, evaporation or transfer loss of solute, final moles n₂ equal n₁. With final molarity c₂ and final solution volume V₂, n₂ = c₂V₂. Equating the two yields c₁V₁ = c₂V₂. Volumes can both be in mL or both in L in this equation because their units cancel in the ratio, but molarity times litres is needed if displaying moles explicitly.

Suppose 25.0 mL of 2.00 M NaCl stock is transferred to a volumetric flask and diluted to 250.0 mL final volume. Initial amount is 2.00 mol L⁻¹ × 0.0250 L = 0.0500 mol. Final molarity is 0.0500/0.2500 = 0.200 M, or directly 2.00 × 25.0/250.0. The added water volume is approximately, but not always exactly, 225.0 mL. The concentration calculation requires only the calibrated final 250.0 mL, so volume contraction during mixing does not spoil the mole balance.

If a question says “add 250 mL water to 25 mL stock,” it has not necessarily specified V₂ = 250 mL. Under an explicit additive-volume approximation, V₂ ≈ 275 mL. Without that approximation or a measured final volume, exact final molarity may be unknown. This distinction matters especially when mixing concentrated solutions whose volumes are not exactly additive.

Dilution alone cannot increase molarity: if V₂ > V₁ and solute moles are fixed, c₂ = c₁V₁/V₂ < c₁. If a calculation gives c₂ above stock concentration, inspect volume roles, units and whether solvent was actually removed rather than added. For serial dilutions, each step has its own stock concentration and aliquot volume. Overall concentration ratio is the product of individual V aliquot/V final ratios, assuming no solute loss.

The equation fails when the measured “solute” changes chemically. Adding base to an acid is not dilution of unchanged acid moles; neutralisation must be solved by reaction stoichiometry. If a salt precipitates, the dissolved solute amount is no longer conserved even though total atoms may be. An aliquot may also lose material if transfer is incomplete, so laboratory technique underpins the ideal balance.

Step-by-step reasoning

1. Identify the species whose chemical amount is conserved during the operation. 2. Mark c₁ and V₁ for the stock aliquot, and identify the calibrated final solution volume V₂. 3. If needed, convert V₁ to litres and compute n = c₁V₁ as a check. 4. Rearrange c₁V₁ = c₂V₂ for the unknown and use matching volume units. 5. Check that a true dilution has V₂ greater than V₁ and c₂ lower than c₁.

Visual explanation

Sketch a small pipette delivering 25.0 mL stock into a 250.0 mL volumetric flask. A line on the flask neck marks the final solution volume. Draw 0.0500 mol solute dots in the pipette and the same count dispersed through the larger flask volume. Label added water separately from the 250.0 mL final volume.

Real-world analogy

Spreading a fixed number of seeds over a larger garden makes seeds per square metre lower without changing the number of seeds. A dilution spreads a fixed number of solute formula units through more solution volume. Counting the newly added soil rather than the final garden size would use the wrong denominator.

Real-world example

A student prepares a calibration standard by pipetting a measured amount of concentrated solution into a volumetric flask and filling to its mark. The flask mark defines V₂. If the student instead pours a guessed volume of water into a beaker, the final volume and concentration are less certain.

Why?

Why does concentration fall when solvent is added? Molarity divides unchanged solute moles by a larger final volume. The numerator stays constant under a proper dilution, while the denominator grows. This is the physical meaning behind c₁V₁ = c₂V₂.

Common misconception

“V₂ is the amount of water added.” V₂ is the final solution volume. Another error is to use the full stock bottle volume as V₁ when only a small aliquot was transferred. Follow the solute moles actually moved into the final flask.

Worked example

How much 1.50 M stock is needed to prepare 400.0 mL of 0.120 M solution? Solute moles required are (0.120 mol L⁻¹)(0.4000 L) = 0.0480 mol. Stock volume needed is V₁ = n/c₁ = 0.0480/1.50 = 0.0320 L = 32.0 mL. Measure 32.0 mL stock and dilute to a final total volume of 400.0 mL. Do not report that exactly 368.0 mL water must be added unless volume additivity is allowed; instead add solvent to the calibration mark.

Quick check

1. Dilute 10.0 mL of 1.00 M stock to 100.0 mL final volume. What is c₂? Answer: c₂ = (1.00 M)(10.0/100.0) = 0.100 M, assuming solute amount is conserved.

Exam focus

Label final solution volume and stock aliquot volume explicitly. Use cV conservation only for the same unchanged solute. A measured final volume is stronger information than an assumed sum of dispensed liquid volumes.

Advanced insight

For serial dilutions, c final/c initial = Π(V aliquot,i/V final,i). For example, two successive 1:10 dilutions give a 1:100 concentration ratio, although the total amount of stock used in the second step is only the aliquot from the first diluted solution. The product rule follows repeated mole conservation.

Summary

The dilution equation c₁V₁ = c₂V₂ expresses conservation of a specified solute's moles. V₂ is the calibrated final solution volume. Use it only when no reaction or loss changes solute amount, and check that a genuine solvent addition lowers molarity.

Practice questions

1. What final volume is needed to dilute 50.0 mL of 0.800 M stock to 0.200 M? Answer: V₂ = c₁V₁/c₂ = 0.800 × 50.0/0.200 = 200 mL final solution. 2. If 20.0 mL of 0.500 M stock is diluted to 250.0 mL, how many solute moles remain in the flask? Answer: n = (0.500)(0.0200) = 0.0100 mol; dilution leaves that amount unchanged. 3. Why is c₁V₁ = c₂V₂ unsuitable for acid mixed with a reactive base? Answer: Neutralisation consumes acid species, so the amount of unchanged acid solute is not conserved.