Dilution

Dilution lowers a solute's concentration by adding solvent. For conserved amount of substance, molar concentration satisfies

n = c_{1} V_{1} = c_{2} V_{2}, c_{2} = c_{1} \frac{V _{1}}{V _{2}} .

(1)

Both

V_{1}

and

V_{2}

are chemical solution volumes;

V_{2}

is the final total, not the added solvent.^[ref]

The proof is simply the definition

n = c V

applied before and after. Its useful assumptions are full transfer of the selected portion, no loss or gain of the counted solute, and complete mixing. An ideal chemical solution approximation is not required for this inventory equation. If chemical equilibrium redistributes dissolved chemical species, use a conserved analytical concentration or chemical element inventory rather than assuming each individual species is conserved. This conservation statement does not by itself preserve the pH: a chemical equilibrium may change as the chemical solution is diluted.

Table of contents
- Serial dilution Dilution

Serial dilution ( $c_{k} = c_{0} \prod_{j = 1}^{k} (v_{j} / V_{j})$ )

 0  0

Serial dilution repeatedly transfers a measured portion of one well-mixed chemical solution and dilutes that portion to a new final volume. With transferred volume

v_{j}

and final volume

V_{j}

at step

j

, conserved solute gives

n_{j} = c_{j - 1} v_{j}, c_{j} = \frac{n _{j}}{V _{j}} = c_{j - 1} \frac{v _{j}}{V _{j}}, c_{k} = c_{0} \prod_{j = 1}^{k} \frac{v _{j}}{V _{j}} .

(1)

This product follows by repeated substitution. Each step conserves the solute in the portion transferred into its receiving flask, not the entire inventory in the preceding flask. Uniform mixing makes the selected fraction representative.

Continue the original solution mass and volume bookkeeping example. A

10.00 mL

portion of the

0.2000 mol L^{- 1}

stock contains

0.002000 mol

of analytical sodium chloride, corresponding to

0.11688 g

before rounding. After dilution to

100.0 mL

, its analytical concentration is

0.02000 mol L^{- 1}

. A new

5.000 mL

portion of that mixed chemical solution contains

0.0001000 mol

; dilution to

50.00 mL

gives

0.002000 mol L^{- 1}

. The two factors

10.00/100.0

and

5.000/50.00

multiply to

1/100

Figure 1.
Final chemical solution volume and transferred solute in a serial dilution
. Original illustrative account, not experimental data. Calibration marks denote final contained volumes; arrows denote measured portions transferred. The first arrow carries $0.002000 mol$ ; the second carries $0.0001000 mol$ . Each receiving flask conserves the inventory in its incoming portion. Equal dilution factors give successively tenfold lower analytical concentrations. **Details**.

The remaining stock holds

0.05000 - 0.002000 = 0.04800 mol

; the remaining first dilution holds

0.002000 - 0.0001000 = 0.001900 mol

. Including the final flask restores the starting inventory:

0.04800 + 0.001900 + 0.0001000 = 0.05000 mol .

(2)

This is a direct check that making a dilute final sample does not destroy the solute. Most of it remains in earlier vessels.

Serial dilution makes a small final concentration accessible using measurable portions instead of a single extremely small transfer. Its limitation is that systematic errors propagate: if each transferred volume is

1%

below its intended value while final volumes are correct, two steps give

0.9 9^{2} = 0.9801

times the intended concentration, a

1.99%

deficit. Repeated readings alone do not remove that bias. The gain in manageable volume therefore depends on calibrated transfer and representative mixing.

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Dilution

Serial dilution ( $c_{k} = c_{0} \prod_{j = 1}^{k} (v_{j} / V_{j})$ )

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Dilution

Serial dilution (ck​=c0​∏j=1k​(vj​/Vj​))

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