Mixing Two Solutions of One Solute

Introduction

If two solutions contain the same dissolved chemical, mixing them combines their solute moles. The final molarity is not usually the simple average of the two printed molarities. It depends on how much of each solution is used and on the final mixture volume. Mole conservation is the reliable starting point; a volume sum is a separate assumption that should be checked.

Core explanation

Let solution 1 have molarity c₁ and volume V₁, and solution 2 have c₂ and V₂. Their initial solute amounts are n₁ = c₁V₁ and n₂ = c₂V₂, with volumes in litres. If the solute neither reacts nor escapes, final solute moles are n f = c₁V₁ + c₂V₂. Final molarity is c f = n f/V f, where V f is the actual final solution volume. If the question explicitly allows additive volumes, V f = V₁ + V₂ and c f = (c₁V₁ + c₂V₂)/(V₁ + V₂).

The additive-volume form is a volume-weighted mean of c₁ and c₂. It must fall between those two concentrations when both volumes are positive and no solute is lost. For equal volumes, it becomes their arithmetic mean. For unequal volumes, the larger portion pulls the final result closer to its own concentration. A final c outside the initial range can be valid only if volume change, reaction, solvent loss, or another condition changes the simple averaging model.

For example, 200 mL of 0.250 M NaCl contains (0.200 L)(0.250 mol L⁻¹) = 0.0500 mol NaCl. Another 300 mL of 0.100 M NaCl contains 0.0300 mol. Total NaCl is 0.0800 mol. If final volume is taken as 500 mL, c f = 0.0800/0.500 = 0.160 M. The simple unweighted average (0.250 + 0.100)/2 = 0.175 M is wrong because the two portions have unequal volumes.

Actual liquid volumes can contract or expand when mixed, so a precise problem may report a measured final volume. If the same two solutions occupy 0.490 L after mixing, c f = 0.0800/0.490 ≈ 0.163 M. The solute mole count is unchanged; only its volume denominator changes. Concentration can also shift with temperature because solution volume changes. Never replace a provided measured V f with V₁ + V₂ just because the arithmetic seems convenient.

The method applies to a conserved species. If one solution contains an acid and the other a base, they are not two solutions of one conserved solute; reaction stoichiometry must be applied first. Even if both solutions are labelled with the same element in different chemical forms, the target species may change by reaction. Clarify which chemical amount is conserved.

Step-by-step reasoning

1. Confirm that both portions contain the same solute species and that no reaction, precipitation or loss is specified. 2. Convert V₁ and V₂ to litres and calculate n₁ = c₁V₁ and n₂ = c₂V₂. 3. Add solute amounts, retaining enough digits for the final rounding. 4. Use the stated measured final volume, or V₁ + V₂ only when volume additivity is an allowed approximation. 5. Divide total moles by final litres and check the answer against the two starting concentrations.

Visual explanation

Draw two beakers labelled 0.200 L × 0.250 M = 0.0500 mol and 0.300 L × 0.100 M = 0.0300 mol. Arrows lead to a third beaker with 0.0800 mol above it. Under that beaker, write its measured final volume; dividing 0.0800 mol by that number of litres gives the final M.

Real-world analogy

Combining two groups with different average exam scores requires counting the students in each group. The combined average is weighted by group sizes, not the plain average of the two averages unless group sizes match. Mixing conserved solute similarly weights each molarity by the amount of solution used, with final volume determining the exact denominator.

Real-world example

A technician combines two batches of a buffer component made at slightly different concentrations. They calculate each batch's solute moles before mixing, then measure the final volume and report the new concentration. If the buffer chemistry changes pH or speciation, further equilibrium calculations are needed; the simple balance only tracks the conserved analytical solute amount.

Why?

Why add moles rather than molarities? Molarity is a ratio with litres in the denominator; adding two ratios without considering their sample sizes has no direct conservation meaning. Solute moles are the actual counted chemical amount that transfers into the mixed vessel.

Common misconception

“Mixing 1 M and 2 M always makes 1.5 M.” That is true under additive volumes only when equal volumes are used. A small drop of 1 M mixed into a large volume of 2 M leaves a concentration near 2 M, not 1.5 M.

Worked example

Mix 150.0 mL of 0.400 M KNO₃ with 350.0 mL of 0.100 M KNO₃. The first contributes 0.1500 × 0.400 = 0.0600 mol; the second contributes 0.3500 × 0.100 = 0.0350 mol. Total solute is 0.0950 mol. If the measured final volume is 495.0 mL, c f = 0.0950/0.4950 = 0.192 M. If the problem instead instructs us to assume additive volumes, V f = 500.0 mL and c f = 0.190 M. Both calculations use the same mole total but answer slightly different volume assumptions.

Quick check

1. Mix equal volumes of 0.20 M and 0.60 M solutions of one conserved solute, assuming additive volumes. What is the final molarity? Answer: Equal volumes give the arithmetic mean, (0.20 + 0.60)/2 = 0.40 M.

Exam focus

Show both cV mole calculations and label the source of V f. If no reaction occurs and volumes add, check that final concentration lies between the initial values. State when a measured final volume replaces the simple volume sum.

Advanced insight

For several portions of the same conserved solute, the relation generalises to c f = Σ(cᵢVᵢ)/V f. If V f = ΣVᵢ, the weights Vᵢ/ΣVᵢ sum to one. This reveals mathematically why the final concentration is bounded by the smallest and largest input concentrations under additive-volume conditions.

Summary

Conserve solute moles across mixing: n f = c₁V₁ + c₂V₂. Divide by the actual final solution volume. A volume-weighted average is valid only when volume additivity is stated or justified and no chemistry changes the solute amount.

Practice questions

1. Mix 100 mL of 0.50 M solute with 300 mL of 0.10 M solute. How many moles are present in total? Answer: 0.100 × 0.50 + 0.300 × 0.10 = 0.080 mol. 2. If final volume is 0.400 L, what is final molarity? Answer: c f = 0.080/0.400 = 0.20 M. 3. If measured final volume is 0.395 L instead, what changes? Answer: Solute moles remain 0.080, but c f = 0.080/0.395 ≈ 0.203 M.