Dilute a stock solution without losing the solute ledger

Track the transferred portion

A stock solution has a known concentration. An aliquot is a measured portion taken from it. In this model the stock is homogeneous and has concentration 2.00 mol/L. A 100 mL aliquot contains 2.00 mol/L × 0.100 L = 0.200 mol solute. This is the amount transferred, not the amount in the entire stock bottle.

Adding solute-free solvent and making the final solution volume 500 mL leaves those 0.200 mol unchanged under the stated assumptions. Final concentration is 0.200/0.500 = 0.400 mol/L. The larger volume spreads the same solute inventory over more solution; dilution does not make solute vanish. Background: dilution, OpenStax Chemistry 2e.

Derive the dilution equation

Use C1 and V1 for the stock concentration and transferred stock volume, and C2 and V2 for the final concentration and final solution volume. Before dilution, n = C1V1. After dilution, n = C2V2. If solute is conserved, C1V1 = C2V2, so C2 = C1V1/V2.

Convert volumes to litres when calculating moles. In the concentration ratio, V1 and V2 may both be in mL because their units cancel. Do not use one in mL and the other in L. The dilution factor is V2/V1; a factor of five lowers concentration by five when the stock concentration and transferred solute are otherwise unchanged.

This equation assumes complete transfer, uniform mixing and no reaction, precipitation or loss of the solute being counted. If these conditions fail, the conservation step needs evidence or revision. It is not a universal formula for every liquid-mixing problem.

Design a virtual preparation

The target is 500 mL of 0.300 mol/L solution from 2.00 mol/L stock. Solve for transferred volume: V1 = C2V2/C1 = 0.300 × 0.500/2.00 = 0.0750 L, or 75.0 mL. Use the controls to check that combination and capture the result.

A defensible written plan identifies the stock and its concentration, measures the 75.0 mL aliquot with suitable volumetric equipment, accounts for complete transfer into a vessel appropriate for setting the final volume, adds solvent to the final 500 mL volume and mixes thoroughly. Real glassware must be read with the appropriate meniscus convention and calibration conditions. The rectangular drawing is a shared volume scale; it does not simulate a pipette’s accuracy, a volumetric flask’s shape or your practical technique.

Do not interpret “dilute to 500 mL” as “add 500 mL solvent.” The latter does not establish a 500 mL final solution. Nor should you assume that separately measured liquid volumes always add exactly. Use the measured final solution volume in the concentration calculation.

Diagnose two different errors

Suppose the 75.0 mL aliquot was transferred correctly, but final volume was measured as 600 mL. Predict the concentration. Then compare a different failure: only 50.0 mL stock reached a correctly set 500 mL final volume. Identify which measurement would distinguish them.

Check the two failures

The first retains 0.150 mol and gives 0.150/0.600 = 0.250 mol/L. The second transfers 0.100 mol and gives 0.100/0.500 = 0.200 mol/L. Both are below target, but their solute amounts and final volumes differ. Records of transferred volume and final volume help distinguish them; a low concentration alone does not identify the cause.

Compare equal concentrations with different inventories

Use 50 mL diluted to 500 mL and 100 mL diluted to 1,000 mL. Both yield 0.200 mol/L, but the larger preparation contains twice the solute amount. If final volume equals transferred volume, the factor is one and concentration remains 2.00 mol/L; there has been no dilution.

Save the target preparation and a contrasting error case. Write the solute ledger and identify the assumption behind each equality. These are virtual preparations of a generic solute. They do not authorize an unsupervised protocol for a particular chemical or establish practical laboratory proficiency.