Dilution of Solutions

Diluting a solution means adding solvent to lower its concentration. This single idea lets chemists take one concentrated "stock" solution off the shelf. They can then prepare an entire family of weaker solutions for different experiments. The reasoning rests on a conservation law. The number of moles of solute does not change when you add solvent. Only the volume it occupies changes. That conservation law is expressed in terms of Molarity and Solution Concentration. It gives the compact relation $M_1V_1 = M_2V_2$ . This is one of the most-used equations in the introductory laboratory. This article builds that relation from first principles. It works through fully numeric examples. It also flags the errors students make most often. These include unit mismatches and confusing "adding solvent" with "adding solute." The same logic underlies Acid-Base Titrations. There, precisely diluted solutions of known concentration are used to determine unknown concentrations.

Key takeaways

Dilution adds solvent only; the moles of solute present stay fixed.
The master equation is $M_1V_1 = M_2V_2$ , valid whenever no solute is created, destroyed, or added.
$M_1, V_1$ describe the concentrated stock before dilution; $M_2, V_2$ describe the dilute solution after.
Units of $V_1$ and $V_2$ must match (both mL or both L) — the equation does not care which, as long as they agree.
Dilution factor $= V_2/V_1 = M_1/M_2$ tells you how many times more dilute the new solution is.
Key idea: moles of solute are conserved during dilution, so $M_1V_1 = M_2V_2$ lets you solve for any one unknown volume or concentration.

The core insight

Dilution never changes how much solute you have — only how much space it occupies. Track moles, not concentration, and the arithmetic becomes automatic.

The physical picture

Picture a flask holding a small volume $V_1$ of a concentrated solution, say a bright blue copper sulfate stock solution. It contains a fixed number of moles of solute, $n = M_1V_1$ . Now you pour that stock into a larger flask and add water until the total volume becomes $V_2$ . The solute particles — the same fixed number of $\text{Cu}^{2+}$ and $\text{SO}_4^{2-}$ ions — are now spread through a larger volume of liquid, so the color visibly lightens. Nothing was added or removed except solvent. The moles of solute before dilution must equal the moles of solute after:

n_{\text{before}} = n_{\text{after}}

Since moles equal molarity times volume, this becomes the dilution equation. The physical picture below shows the same fixed number of solute particles (represented as dots) occupying a small volume at high concentration, then a larger volume at low concentration.

Figure

Same moles of solute, before and after adding solvent

Notice the dot count is identical in both flasks: six solute particles, representing the same fixed number of moles. Only the container volume changed, from 25 mL to 150 mL, so the concentration dropped by the same factor the volume grew — a factor of six.

The core relations

The moles of solute in the concentrated (stock) solution:

n = M_1 V_1

where: $M_1$ is the molarity of the stock solution, $V_1$ is the volume of stock measured out. The moles of solute in the dilute (final) solution:

n = M_2 V_2

where: $M_2$ is the molarity after dilution, $V_2$ is the total volume after adding solvent. Since the same $n$ appears in both, the dilution equation follows directly:

M_1 V_1 = M_2 V_2

where: $M_1, V_1$ = concentration and volume before dilution; $M_2, V_2$ = concentration and volume after dilution; $V_1$ and $V_2$ must be expressed in the same volume unit. A useful companion quantity is the dilution factor $D$ , the ratio by which concentration drops:

D = \frac{V_2}{V_1} = \frac{M_1}{M_2}

where: $D$ has no units — it is a pure ratio (e.g., "diluted tenfold" means $D = 10$ ).

Derivation AFrom conservation of moles to

M_1V_1 = M_2V_2

Goal: show that mixing a stock solution with pure solvent forces $M_1V_1 = M_2V_2$ , using nothing but the definition of molarity and the fact that solvent addition cannot create or destroy solute.

Step 1. Molarity is defined as moles of solute per liter of solution:

M = \frac{n}{V}

Rearranged, the moles of solute present in any solution of known molarity and volume are

n = MV

Step 2. Before dilution, a volume $V_1$ of stock solution at concentration $M_1$ is measured out. By Step 1, the moles of solute in this portion are

n_{\text{stock}} = M_1 V_1

Step 3. Solvent (pure water, with no solute) is added until the total volume becomes $V_2$ . Because pure solvent contributes zero moles of solute, the total moles of solute in the final solution are unchanged from Step 2:

n_{\text{final}} = n_{\text{stock}}

Step 4. By Step 1 applied to the final solution, the moles of solute can also be written as $M_2 V_2$ , where $M_2$ is the new, lower concentration and $V_2$ is the new, larger volume:

n_{\text{final}} = M_2 V_2

Step 5. Combining Step 2, Step 3, and Step 4, the two expressions for the same quantity of moles must be equal:

M_1 V_1 = M_2 V_2

Limiting-case check. If no solvent at all is added, $V_2 = V_1$ , and the equation forces $M_2 = M_1$ — concentration is unchanged, exactly as expected when nothing is diluted. If the solution is diluted to twice its original volume ( $V_2 = 2V_1$ ), the equation gives $M_2 = M_1/2$ : the concentration exactly halves, matching physical intuition that spreading the same moles over twice the volume halves the concentration.

Worked examples

Example 1 — Preparing a dilute HCl solution from concentrated stock. A student has a stock bottle of hydrochloric acid labeled 11.6 M. The task is to prepare 250.0 mL of a 0.500 M HCl solution. How many milliliters of the concentrated stock must be measured out? Identify the known and unknown quantities:

M_1 = 11.6\ \text{M}, \quad M_2 = 0.500\ \text{M}, \quad V_2 = 250.0\ \text{mL}, \quad V_1 = \,?

Apply the dilution equation and solve for $V_1$ :

M_1 V_1 = M_2 V_2

V_1 = \frac{M_2 V_2}{M_1}

V_1 = \frac{(0.500\ \text{M})(250.0\ \text{mL})}{11.6\ \text{M}}

V_1 = \frac{125.0\ \text{mL}\,\cdot\,\text{M}}{11.6\ \text{M}} = 10.78\ \text{mL}

So the procedure is: measure 10.78 mL of the 11.6 M stock acid, transfer it into a 250.0 mL volumetric flask, and add water up to the calibration mark to reach a total volume of 250.0 mL. The moles of HCl are conserved throughout:

n = (11.6\ \text{M})(0.01078\ \text{L}) = 0.1250\ \text{mol}

n = (0.500\ \text{M})(0.2500\ \text{L}) = 0.1250\ \text{mol} \quad \checkmark

Both routes give 0.1250 mol, confirming the dilution preserves the amount of solute.

Example 2 — Finding the resulting concentration after a specific dilution. A chemist takes 18.0 mL of a 2.40 M sodium hydroxide solution and dilutes it by adding water until the total volume is 75.0 mL. What is the new concentration? Identify knowns and unknown:

M_1 = 2.40\ \text{M}, \quad V_1 = 18.0\ \text{mL}, \quad V_2 = 75.0\ \text{mL}, \quad M_2 = \,?

Solve the dilution equation for $M_2$ :

M_1 V_1 = M_2 V_2

M_2 = \frac{M_1 V_1}{V_2}

M_2 = \frac{(2.40\ \text{M})(18.0\ \text{mL})}{75.0\ \text{mL}}

M_2 = \frac{43.2\ \text{M}\,\cdot\,\text{mL}}{75.0\ \text{mL}} = 0.576\ \text{M}

The dilution factor is $D = V_2/V_1 = 75.0/18.0 = 4.17$ , and indeed $2.40\ \text{M}/4.17 = 0.576\ \text{M}$ , consistent with the ratio form of the equation. As a limiting-case sanity check: if instead only a trace of water were added so that $V_2 \to V_1$ , the formula would correctly return $M_2 \to M_1 = 2.40$ M, showing the equation behaves properly at the boundary of "no dilution."

Limiting cases & common traps

Situation	What happens	Why
$V_2 = V_1$ (no water added)	$M_2 = M_1$	No dilution occurred; moles and volume both unchanged
$V_2 \to \infty$	$M_2 \to 0$	Same finite moles spread through unbounded volume
Mixing two different solutes	$M_1V_1 = M_2V_2$ does not apply directly	The equation assumes a single conserved solute, not a reaction or mixture of different species
Volumes in mismatched units (mL vs L)	Wrong numeric answer, often off by 1000×	The equation requires $V_1, V_2$ in the same unit; $M_1,M_2$ do too
Assuming volumes are simply additive when mixing two solutions of the same solute	Sometimes true only approximately for dilute aqueous solutions	Real mixing can show small volume contraction/expansion; volumetric flasks avoid this by fixing final $V_2$ directly

The chemist behind it

The dilution relation itself is not attributed to a single discoverer — it is a direct consequence of the definition of molar concentration, formalized as chemists standardized quantitative laboratory technique through the eighteenth and nineteenth centuries. But the person most responsible for making molar concentration a precise, universal laboratory language was Wilhelm Ostwald, who systematized physical chemistry methods, including standardized solution preparation, at a time when analytical chemistry badly needed reproducible quantitative conventions.

Origins · 1 minWilhelm Ostwald and how this was discovered

Wilhelm Ostwald

1853–1932 · Founder of physical chemistry; pioneered standardized volumetric and concentration methods used throughout analytical and physical chemistry

Ostwald co-founded physical chemistry as a discipline, studying reaction rates, equilibrium, and electrolytic dissociation with a rigor that demanded precisely known, reproducibly diluted solutions. His laboratory manuals helped fix the now-universal practice of preparing standard solutions by careful dilution from concentrated stock, the exact technique this equation describes. He received the Nobel Prize in Chemistry in 1909 for his work on catalysis, chemical equilibria, and reaction velocities — all fields that depend on knowing solution concentrations exactly.

Ask the tutor: How would the dilution equation change if I combine two different solutions of the same solute instead of adding pure water?

Does

M_1V_1 = M_2V_2

work if I add another solute-containing liquid instead of pure water?

No — the derivation assumes the only thing added is solvent with zero solute. If you mix two different solutions of the same solute together, you can still use mole conservation, but you must add the moles from each solution separately: $M_1V_1 + M_2V_2 = M_3V_3$ , where $V_3$ is the combined final volume.

Do the volumes have to be in liters?

No. Any consistent volume unit works, since the units simply cancel in the ratio. Milliliters are common in the lab because burets and volumetric flasks are calibrated that way, but you must use the same unit for $V_1$ and $V_2$ within one calculation.

Why doesn't the mass of solvent appear anywhere in the equation?

The dilution equation tracks moles of solute and total solution volume, not solvent mass. Molarity is defined per liter of total solution, so only the total final volume matters, regardless of how much solvent contributed to reaching it.

Can this equation be used for percent concentration or molality instead of molarity?

The same mole-conservation logic can be adapted to other concentration units, but the exact algebraic form $M_1V_1 = M_2V_2$ is specific to molarity because it relies directly on the definition $M = n/V$ . Molality, which is defined per kilogram of solvent, requires a different treatment since adding solvent changes the denominator in a different way.

Why is it safer to add stock solution to water rather than water to concentrated stock?

This is a laboratory safety practice, most critical for concentrated acids, because the dilution can release significant heat. Adding concentrated acid slowly to a larger volume of water lets that heat dissipate into the larger mass of liquid; adding water to concentrated acid can cause localized boiling and dangerous splattering. The $M_1V_1 = M_2V_2$ calculation itself is unaffected by the order of addition — it only tells you the final concentration, not the safe procedure.