Mixing Solutions of One Solute

Introduction

Mixing two solutions containing the same stable solute is different from adding pure solvent. Both portions contribute solute. First add their moles; then divide by the final mixed volume. Using the dilution equation on only one portion would ignore solute from the other.

Core explanation

For two solutions of one conserved solute, total moles are n₁ + n₂ = c₁V₁ + c₂V₂. Final concentration is (c₁V₁ + c₂V₂)/V(final). If a school problem allows volumes to add, V(final) = V₁ + V₂. If mixing changes volume appreciably or the problem gives a measured final volume, use that actual value instead. The method assumes no chemical reaction, precipitation, evaporation or loss.

Mix 100.0 mL of 0.200 M NaCl with 200.0 mL of 0.0500 M NaCl. First portion supplies 0.200 × 0.1000 = 0.0200 mol; second supplies 0.0500 × 0.2000 = 0.0100 mol. Total is 0.0300 mol. If final volume is 0.3000 L, final c = 0.0300/0.3000 = 0.100 M. The result lies between the two starting concentrations, as expected for mixing same-solute solutions without loss and with positive volumes.

An unweighted average, (0.200 + 0.0500)/2 = 0.125 M, would be wrong here because the two portions have unequal volumes. A volume-weighted average is valid only under the volume-additivity condition and same-solute conservation. If equal volumes are mixed, the arithmetic mean does work under these assumptions.

If the solutions contain different solutes that do not react, calculate each named solute's amount separately, then divide each by the common final volume. If they react, use a balanced equation to determine remaining species before dividing. For instance, mixing NaCl(aq) and AgNO₃(aq) can form AgCl(s); merely averaging “salt concentrations” would miss precipitation.

Volume contraction can matter when mixing some liquids, notably certain solvent mixtures. Solute conservation still gives total moles, but a measured final volume is required for precise molarity. Mass-based concentration may sometimes avoid that volume issue if masses and compositions are known.

Step-by-step reasoning

1. Confirm both solutions contain the same named solute and no relevant reaction occurs. 2. Convert each volume to litres and find n = cV for each portion. 3. Add solute moles to obtain the conserved total. 4. Determine actual or permitted additive final volume. 5. Divide total moles by final litres and check that the result is plausible.

Visual explanation

Draw two differently shaded beakers with 20 red solute dots and 10 red dots. Pour both into a larger vessel showing 30 total red dots. Label each starting cV and the final 30-dot amount divided by measured total volume.

Real-world analogy

Mixing two drinks of different sweetness combines sugar from both. The final sweetness depends on each drink's sugar amount and the total drink volume, so averaging two sweetness labels without considering portion sizes can mislead.

Real-world example

A laboratory might combine remaining bottles of the same reagent at different concentrations. To label the mixture, record both concentrations and volumes, calculate each contribution and measure final volume if accuracy requires it.

Why?

Why does the result usually lie between the starting molarities? With additive positive volumes and conserved solute, the final concentration is a weighted average, so neither portion can drive it beyond both original values.

Common misconception

“Mixing is dilution of the stronger solution, so ignore solute in the weaker one.” The weaker solution still contributes solute. Count its cV amount.

Worked example

Mix 50.0 mL of 0.300 M glucose with 150.0 mL of 0.100 M glucose. Moles are 0.300 × 0.0500 = 0.0150 and 0.100 × 0.1500 = 0.0150. Total is 0.0300 mol. Assuming a 0.2000 L final volume, c = 0.0300/0.2000 = 0.150 M. Both portions contributed equal moles despite different concentrations and volumes.

Quick check

1. Is the final concentration the simple average when unequal volumes of two same-solute solutions mix? Answer: Generally no. Add cV moles and divide by final volume; the simple mean works only for equal volumes under the stated assumptions.

Exam focus

Write one mole contribution for each input before adding. Use the final measured volume when supplied, and check for reaction before applying a no-reaction mixture formula.

Advanced insight

The weighted-average result can be written c(final) = (c₁V₁ + c₂V₂)/(V₁ + V₂) only when final volume is additive. If it is not, the numerator remains conserved but the denominator must be measured.

Summary

Same-solute mixtures require an amount balance across all inputs. Calculate each cV, add moles and divide by the final volume. Equal-volume averaging is a special case, and reacting or nonadditive mixtures need more careful treatment.

Practice questions

1. Mix 100 mL of 0.10 M salt with 100 mL of 0.30 M salt; assume additive volume. Find final c. Answer: Moles are 0.010 and 0.030; total 0.040 mol in 0.200 L gives 0.20 M. 2. Why is the simple mean wrong for 50 mL of 0.10 M and 150 mL of 0.30 M? Answer: The 0.30 M portion has three times the volume and contributes more moles; a volume-weighted calculation is required. 3. What changes if the two solutions react to form a precipitate? Answer: Use the balanced reaction to find remaining dissolved amounts before calculating concentrations in the final liquid.