Planning a Dilution

Introduction

Planning a dilution begins with the target: a required concentration and final volume. The stock aliquot must contain exactly the solute amount that the final solution needs. Solving V₁ = c₂V₂/c₁ gives a volume to measure, but the result also needs a feasibility and handling check.

Core explanation

Suppose a working solution should be 0.0500 M in 250.0 mL, and stock is 1.00 M. Target moles are c₂V₂ = 0.0500 × 0.2500 = 0.01250 mol. Stock volume delivering those moles is n/c₁ = 0.01250/1.00 = 0.01250 L, or 12.50 mL. Transfer that stock portion into a suitable vessel and add solvent to a final 250.0 mL solution, then mix. The equation c₁V₁ = c₂V₂ condenses these two amount steps.

A dilution target must be below the stock concentration if only solvent is being added. If calculation gives V₁ greater than final V₂, something is wrong for a simple dilution: the target may be more concentrated than the stock, or units may have been mixed. A stock that is 0.020 M cannot be diluted with water to make a 0.100 M solution. Concentration requires removing solvent or adding solute, which is a different operation.

Precision matters in choosing apparatus. An aliquot of 0.04 mL from a 100 M hypothetical stock would be difficult to measure accurately with ordinary teaching-lab pipettes. A two-step serial dilution may be more reliable. Record final volume, not an approximate amount of added water. When stock and solvent mix with volume contraction, direct addition of nominal volumes may miss the target; bring the mixture to a calibrated mark.

When preparing several solutions from one stock, calculate each aliquot independently unless the later solutions are intentionally made from earlier ones. A volume taken from a homogeneous stock has the stock concentration even though the stock bottle contains less total liquid afterward. Be sure the concentration units refer to the same solute: one stock may be labeled by formula units while a target is specified by ion amount, requiring a stoichiometric conversion.

Chemical handling may affect the practical plan. Strongly exothermic dilution requires the correct safety procedure and cooling before final volume adjustment; some solutions are unstable or incompatible with the proposed solvent. These considerations do not alter the basic mole balance, but they determine whether the simple plan can be executed as assumed.

Step-by-step reasoning

1. Calculate target moles from c₂ and final V₂ in litres. 2. Divide by stock c₁ to get V₁. 3. Confirm V₁ is smaller than V₂ for an ordinary dilution. 4. Select suitable volume apparatus and transfer the aliquot. 5. Make up to final volume, mix, and label the solution.

Visual explanation

Draw a calculation chain: target 0.0500 M × 0.2500 L → 0.01250 mol; divide by 1.00 M stock → 0.01250 L. An arrow shows that aliquot entering a 250.0 mL volumetric flask.

Real-world analogy

To make a fixed-size drink of chosen sweetness from syrup, first calculate how much syrup contains the needed sugar, then add water until the final drink reaches its target size. Simply adding a guessed water amount is less reliable.

Real-world example

Instrument calibration often needs several known dye concentrations. A concentrated stock can supply carefully measured aliquots to separate volumetric flasks. The concentration record should include stock identity, aliquot volume and final volume.

Why?

Why calculate target moles before stock volume? It makes the conservation explicit: the stock portion must deliver exactly the solute amount required in the final liquid.

Common misconception

“Use the calculated V₁ of stock and add V₂ of water.” V₂ is the final total solution volume. Adding that much solvent would over-dilute the stock.

Worked example

Prepare 500.0 mL of 0.0200 M NaCl from 0.250 M stock. Target amount is 0.0200 × 0.5000 = 0.0100 mol. Required stock is 0.0100/0.250 = 0.0400 L = 40.0 mL. Transfer 40.0 mL and make up to 500.0 mL. The dilution factor is 0.250/0.0200 = 12.5, matching 500.0/40.0.

Quick check

1. What stock volume gives 100 mL of 0.10 M from 1.0 M stock? Answer: V₁ = 0.10 × 100/1.0 = 10 mL stock, then solvent is added to 100 mL final solution.

Exam focus

Check the inequality c₂ < c₁ and V₁ < V₂. Label all four c and V terms before substitution.

Advanced insight

When desired aliquots are too small for accurate direct measurement, an intermediate solution can reduce relative volumetric uncertainty. Its concentration must be calculated from the first dilution before planning the next.

Summary

Dilution planning determines the stock portion that carries the target moles. Divide target c₂V₂ by stock c₁, verify feasibility, then make the final solution to the calibrated volume. The calculated final volume is not an added-solvent volume.

Practice questions

1. Find stock volume for 250 mL of 0.040 M from 0.500 M stock. Answer: V₁ = 0.040 × 250/0.500 = 20 mL of stock, made up to 250 mL. 2. Can 0.20 M stock be diluted with pure water to make 0.50 M? Answer: No. Adding solvent can only reduce concentration when solute amount is conserved. 3. Why might a serial dilution be preferable to measuring a tiny direct aliquot? Answer: A very small measured volume can have large relative error; two accurately measurable transfers may be more reliable.