Result

Concentration and dilution

T-017Home CU-103Threads stoichiometry

Statement

Molarity \(c=n/V\) (moles of solute per litre of solution) is the standard measure of solution concentration. Diluting a solution by adding solvent changes its volume but not the moles of solute present, so \(c_1V_1=n=c_2V_2\) — the dilution equation — relates any solution's concentration and volume before and after dilution, and together with \(n=cV\), extends mole-based stoichiometry (mole-avogadro, limiting-reagent) to reactions carried out in solution.

Why it matters

Most real chemistry, from a school laboratory titration to industrial and biological chemistry, happens in solution rather than as pure solids, liquids, or gases mixed directly. Molarity plays exactly the same conceptual role for a dissolved solute that mass (mole-avogadro) plays for a weighed solid or gas volume (gas-stoichiometry) plays for a measured gas sample: a directly measurable macroscopic quantity (concentration times a measured volume) that converts straightforwardly to moles, the currency every stoichiometric calculation ultimately runs on.

Hypotheses

Diluting a solution changes only its total volume, not the moles of solute dissolved in it.This is simply conservation of matter applied to the solute specifically: adding pure solvent introduces no additional solute and removes none, so whatever moles of solute were present before dilution remain present after, now distributed through a larger volume. It assumes no solute is lost (e.g. by precipitation, evaporation, or reaction) during the dilution process itself. Solution volumes are treated as simply additive when mixing (e.g. combining a measured volume of stock solution with enough solvent to reach a target total volume).This is a good approximation for dilute aqueous solutions but not an exact physical law: mixing two liquids can produce a total volume slightly different from the sum of the individual volumes, due to intermolecular interactions between solute, solvent, and any other dissolved species (a phenomenon called volume of mixing or excess volume). For the concentrations typical of general chemistry laboratory work, this effect is small enough to be safely neglected; for very concentrated solutions or precision analytical work, it is not, and standard practice is to dilute "to volume" using a calibrated volumetric flask rather than by simply adding a calculated solvent volume, precisely to sidestep this assumption.

Proof

\[ c = \frac{n}{V} \quad \Longleftrightarrow \quad n = cV \]

Molarity is defined as moles of solute per litre of total solution volume (not per litre of solvent); rearranged, this gives the moles of solute present in any measured volume of a solution of known concentration. A

\[ n_{\text{before}} = n_{\text{after}} \quad (\text{Hypothesis: dilution conserves solute moles}) \]

Before dilution, the solute amount is \(n=c_1V_1\) (concentration and volume of the original, more concentrated solution); after adding solvent to reach a new total volume \(V_2\), the same moles of solute are now present in that larger volume, at the new, lower concentration \(c_2\), so \(n=c_2V_2\) as well. A

\[ c_1V_1 = c_2V_2 \]

Equating the two expressions for the conserved solute moles from Step 2 gives the dilution equation directly: the product of concentration and volume is unchanged by dilution, even though concentration and volume individually both change. A

\[ \text{Solution stoichiometry: } n_A = c_AV_A \ \Rightarrow\ \text{apply the same mole-ratio logic as mole-avogadro / limiting-reagent.} \]

Once moles of a dissolved reactant are obtained via Step 1, every stoichiometric technique already established for mass- or gas-volume-based reactants (mole ratios from a balanced equation, limiting-reagent identification) applies identically, since those techniques operate on moles regardless of how the moles were obtained. A

Result

\[ c=\frac{n}{V}, \qquad c_1V_1=c_2V_2\ (\text{dilution}), \qquad n=cV\ (\text{solution stoichiometry input}) \]

Reading. A single definition (molarity) yields both a direct dilution shortcut, useful whenever only concentrations and volumes (not moles) are given, and a route into ordinary mole-based stoichiometry for any reaction run in solution.

Scope. \(c_1V_1=c_2V_2\) applies only to dilution (adding solvent to an existing solution of the same solute); it does not apply when two different solutions are mixed together and allowed to react, where a full stoichiometric (limiting-reagent-style) calculation via Step 4 is required instead.

Corollaries & converses

Serial dilution (repeated dilution by a fixed factor) compounds multiplicatively: two successive \(1{:}100\) dilutions reduce concentration by a total factor of \(100\times100=10{,}000\), not \(200\) — a common source of error when the two steps' factors are added rather than multiplied.
Titration — adding a solution of known concentration (the titrant) to an unknown-concentration solution (the analyte) until a stoichiometrically exact reaction point is reached — is a direct application of Step 4: the titrant's known moles at the endpoint (\(n=cV\)) combined with the balanced equation's mole ratio gives the analyte's moles, and hence its unknown concentration.
Converse: a strong electrolyte's molarity (moles of the compound formula unit dissolved per litre) is generally not the same as the molarity of each dissociated ion in solution; a compound that dissociates into more than one ion per formula unit produces a correspondingly higher ionic concentration than the stated compound concentration (Worked examples/Problems).

Fails without

Drop conservation of solute moles (Step 2), e.g. by mistakenly treating dilution as changing the moles of solute: the dilution equation \(c_1V_1=c_2V_2\) would have no justification at all — it exists specifically because moles are conserved while volume changes, and without that conservation there would be no fixed relationship between the "before" and "after" concentration-volume products.
Confuse "moles of compound" with "moles of dissociated ions" for a strong electrolyte: using the compound's molarity directly as if it were the concentration of a specific ion (when that ion's stoichiometric coefficient in the dissociation is not 1) understates or overstates the true ionic concentration — directly the Converse's caution, and a genuine, frequently tested distinction (Problems).

Common errors

Adding serial dilution factors instead of multiplying them (Corollaries).
Forgetting that \(c_1V_1=c_2V_2\) applies to dilution specifically, not to mixing two different reacting solutions together, where the correct approach is Step 4's full stoichiometric calculation rather than a direct dilution-equation shortcut.
Computing the volume of solvent to add as if it equalled the target final volume \(V_2\) directly, rather than \(V_2-V_1\) (the target volume minus the volume of stock solution already present).
Treating the molarity of a dissolved ionic compound as automatically equal to the molarity of every ion it produces upon dissociation, without accounting for the dissociation stoichiometry (Fails without, second bullet).

Discussion

Molarity as a standard concentration unit, and the practice of preparing precise solutions by dilution from a concentrated stock, became routine laboratory technique alongside the broader 19th-century maturation of quantitative analytical chemistry (the same period, and often the same practitioners, responsible for establishing reliable atomic and molecular weights via the mole concept itself, mole-avogadro's Discussion). Volumetric analysis (titration) in particular became a standard, precise quantitative technique well before spectroscopic or other instrumental methods existed, and remains widely used today specifically because it requires no specialised instrumentation beyond calibrated glassware.

The additive-volume approximation (Hypotheses) is a genuine, if usually small, source of systematic error in solution preparation; the standard laboratory practice of diluting "to the mark" in a calibrated volumetric flask (filling with solvent until the total solution volume, not necessarily solvent volume, reaches the flask's calibrated line) is specifically designed to sidestep this approximation entirely — the true total volume is measured directly, rather than assumed from adding together the stock and solvent volumes separately.

Common misconception: that "molarity" and "moles" are interchangeable or that a concentration alone specifies an amount of substance. Concentration only specifies an amount once multiplied by a volume (\(n=cV\)); a small volume of a highly concentrated solution and a large volume of a dilute one can contain identical moles of solute despite very different concentrations.

Worked examples

\[ 25.0\,\text{g NaOH}\ (M=39.997\,\text{g/mol})\ \text{dissolved to make } 500.0\,\text{mL solution}: \quad n=\frac{25.0}{39.997}\approx0.6250\,\text{mol}, \quad c=\frac{0.6250}{0.5000}\approx1.250\,\text{M} \]

A direct mass-to-molarity calculation, chaining mole-avogadro's \(n=m/M\) with Step 1's \(c=n/V\); the solute mass is first converted to moles, then divided by the solution's total volume (in litres) to obtain molarity. A

\[ \text{Dilute } 6.00\,\text{M HCl stock to } 250.0\,\text{mL of } 0.500\,\text{M}: \quad V_1=\frac{c_2V_2}{c_1}=\frac{0.500\times250.0}{6.00}\approx20.83\,\text{mL} \]

A direct rearrangement of Step 3's dilution equation for the required stock volume; the volume of solvent to add is \(V_2-V_1\approx250.0-20.83\approx229.2\,\text{mL}\), not the full \(250.0\,\text{mL}\) (Common errors, third bullet). A

\[ 25.0\,\text{g NaOH in }500.0\,\text{mL}\approx1.250\,\text{M}; \qquad 20.83\,\text{mL of }6.00\,\text{M HCl}\ \xrightarrow{+229.2\,\text{mL water}}\ 250.0\,\text{mL of }0.500\,\text{M} \]

Reading. Molarity converts freely between a weighed solid dissolved to a known volume, and a concentrated stock diluted to a target concentration, using nothing beyond Steps 1 and 3.

Scope. Identical procedures apply to any solute and any concentration units expressed consistently as moles per litre.

Further example. A laboratory protocol requires preparing \(250.0\,\text{mL}\) of a \(0.200\,\text{M}\) \(\text{K}_2\text{Cr}_2\text{O}_7\) solution from a \(1.50\,\text{M}\) stock solution. Calculate the volume of stock solution required and the volume of distilled water to be added to reach the final volume.

\[ n_{\text{target}} = c_{\text{target}} \times V_{\text{target}} = 0.200\,\text{mol/L} \times 0.2500\,\text{L} = 0.0500\,\text{mol} \]

This calculates the total moles of solute required in the final solution using the definition of molarity \(c=n/V\). The volume is converted from mL to L to maintain unit consistency. A

\[ V_{\text{stock}} = \frac{n_{\text{target}}}{c_{\text{stock}}} = \frac{0.0500\,\text{mol}}{1.50\,\text{mol/L}} \approx 0.03333\,\text{L} = 33.33\,\text{mL} \]

Since dilution does not change the number of moles of solute, the moles in the stock aliquot must equal the target moles. Dividing by the stock concentration yields the required volume of stock. A

\[ V_{\text{water}} = V_{\text{target}} - V_{\text{stock}} = 250.0\,\text{mL} - 33.33\,\text{mL} = 216.7\,\text{mL} \]

The volume of solvent added is the difference between the final desired volume and the volume of the concentrated stock used, assuming volumes are additive (standard approximation for dilute aqueous solutions). A

\[ V_{\text{stock}} = 33.33\,\text{mL}; \qquad V_{\text{water}} = 216.7\,\text{mL} \]

Reading. To prepare the solution, one measures \(33.33\,\text{mL}\) of the \(1.50\,\text{M}\) stock and adds it to a volumetric flask, then adds water until the total volume reaches the \(250.0\,\text{mL}\) mark.

Scope. This calculation holds for any dilute aqueous solution where volume additivity is a valid approximation and the stock concentration is known accurately.

Problems

In a titration, \(25.00\,\text{mL}\) of an unknown-concentration HCl solution is exactly neutralised by \(32.50\,\text{mL}\) of \(0.100\,\text{M}\) NaOH, via \(\text{HCl}+\text{NaOH}\to\text{NaCl}+\text{H}_2\text{O}\) (a \(1{:}1\) mole ratio). Find the concentration of the HCl solution.
Solution
Moles of NaOH at the endpoint: \(n=cV=0.100\times0.03250=0.003250\,\text{mol}\). By the \(1{:}1\) stoichiometry, moles of HCl \(=0.003250\,\text{mol}\) as well. Concentration: \(c_{\text{HCl}}=\dfrac{0.003250}{0.02500}=0.1300\,\text{M}\).
A \(2.00\,\text{M}\) stock solution is diluted by taking \(1.00\,\text{mL}\) and adding solvent to make \(100.0\,\text{mL}\), then taking \(1.00\,\text{mL}\) of that solution and again adding solvent to make \(100.0\,\text{mL}\). Find the final concentration.
Solution
Each dilution step reduces concentration by a factor of \(100.0/1.00=100\). Two successive steps multiply (not add) these factors: total dilution factor \(=100\times100=10{,}000\). Final concentration: \(\dfrac{2.00}{10{,}000}=2.00\times10^{-4}\,\text{M}\).
How many millilitres of \(12.0\,\text{M}\) stock \(\text{H}_2\text{SO}_4\) are needed to prepare \(1.00\,\text{L}\) of \(0.150\,\text{M}\,\text{H}_2\text{SO}_4\), and how much water must be added?
Solution
\(V_1=\dfrac{c_2V_2}{c_1}=\dfrac{0.150\times1000}{12.0}=12.5\,\text{mL}\) of stock. Water added: \(1000-12.5=987.5\,\text{mL}\) (bringing the total solution, not just the added water, to \(1.00\,\text{L}\), per the additive-volume approximation in Hypotheses).
A solution is prepared to be \(0.10\,\text{M}\) in \(\text{CaCl}_2\), a strong electrolyte that fully dissociates as \(\text{CaCl}_2\to\text{Ca}^{2+}+2\text{Cl}^-\). State the concentration of \(\text{Ca}^{2+}\) and of \(\text{Cl}^-\) ions in this solution, and explain why they are not both equal to \(0.10\,\text{M}\).
Solution
Each formula unit of \(\text{CaCl}_2\) that dissociates produces exactly one \(\text{Ca}^{2+}\) ion and two \(\text{Cl}^-\) ions, so \([\text{Ca}^{2+}]=0.10\,\text{M}\) (same as the compound's stated molarity, matching its \(1{:}1\) dissociation coefficient) while \([\text{Cl}^-]=2\times0.10=0.20\,\text{M}\) (twice the compound's molarity, matching chloride's coefficient of \(2\)). This is exactly the distinction flagged in the Converse and Fails without: the "\(0.10\,\text{M}\,\text{CaCl}_2\)" label describes the dissolved compound's original formula-unit concentration, not the concentration of either individual ion once dissociation is accounted for.
A student needs to prepare \(500.0\,\text{mL}\) of a \(0.150\,\text{M}\) \(\text{Na}_2\text{CO}_3\) solution. They have a \(2.50\,\text{M}\) stock solution available. Calculate the volume of stock solution required and the volume of water to be added.
Solution
First, calculate the moles of \(\text{Na}_2\text{CO}_3\) required for the final solution: \[ n = c_{\text{final}} \times V_{\text{final}} = 0.150\,\text{mol/L} \times 0.5000\,\text{L} = 0.0750\,\text{mol} \] Next, determine the volume of the \(2.50\,\text{M}\) stock solution that contains this amount of solute: \[ V_{\text{stock}} = \frac{n}{c_{\text{stock}}} = \frac{0.0750\,\text{mol}}{2.50\,\text{mol/L}} = 0.0300\,\text{L} = 30.0\,\text{mL} \] Finally, calculate the volume of water to be added to reach the final volume: \[ V_{\text{water}} = V_{\text{final}} - V_{\text{stock}} = 500.0\,\text{mL} - 30.0\,\text{mL} = 470.0\,\text{mL} \] The student must mix \(30.0\,\text{mL}\) of stock solution with \(470.0\,\text{mL}\) of water.