Molarity as a Source of Reactant Moles

Introduction

A laboratory may supply an aqueous reactant as a measured volume and concentration rather than a solid mass. Molarity converts that solution measurement into moles of the named solute. The balanced reaction then relates those solute moles to another reactant or product.

Core explanation

Molarity c is defined as amount of solute divided by volume of solution : c = n/V. Rearranging gives n = cV, with c in mol L⁻¹ and V in liters. A 0.250 L portion of 0.400 mol L⁻¹ NaOH contains 0.100 mol NaOH formula units in the solution-composition model. The volume is not the volume of pure NaOH. It includes solvent and all dissolved components, but the concentration identifies how much NaOH corresponds to each liter of the homogeneous solution.

Most laboratory volumes are reported in milliliters. Convert 25.0 mL to 0.0250 L before multiplying by a concentration in mol L⁻¹. Then 0.200 mol L⁻¹ × 0.0250 L = 0.00500 mol solute. One can also use the equivalent unit mmol mL⁻¹: 0.200 mol L⁻¹ equals 0.200 mmol mL⁻¹, so 25.0 mL contains 5.00 mmol. Mixing mol L⁻¹ and milliliters without conversion and then labeling the result “mol” creates a thousand-fold error.

Take HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l). If 25.0 mL of 0.200 mol L⁻¹ HCl is used and enough NaOH is available, HCl amount is 0.00500 mol. The 1:1 equation predicts 0.00500 mol NaCl formed and 0.00500 mol NaOH consumed. If both solution quantities are supplied, calculate each separately and determine which limits the reaction. Do not assume that equal solution volumes contain equal moles unless their concentrations also match.

The formula of the named solute and its chemistry matter. For 25.0 mL of 0.100 mol L⁻¹ H₂SO₄, the amount is 0.00250 mol H₂SO₄ formula units, not automatically 0.00500 mol acid molecules. Complete neutralization with NaOH follows H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, so the required NaOH amount is 0.00500 mol. The factor of two comes from the balanced reaction, not from changing the acid's stated molarity. Some acid-base reactions are incomplete or stepwise under particular conditions; use the reaction specified by the problem.

Dilution changes concentration and volume but not the amount of solute, provided no solute is added, removed or reacted. If 10.0 mL of a stock solution is diluted to 100.0 mL, its solute amount remains the stock concentration times 0.0100 L; the final concentration becomes one-tenth of the stock concentration in the ideal volume account. In a reaction problem, use the concentration that corresponds to the volume actually measured. Combining a stock concentration with the post-dilution volume without a dilution calculation is a common error.

Molarity can depend on temperature because solution volume can expand or contract. For ordinary classroom calculations at stated conditions, use the provided concentration and volume. A real titration may require calibrated glassware and temperature control for higher precision. Those measurement issues affect numerical uncertainty, while the balanced coefficient ratio remains an exact relationship for the assumed reaction.

Step-by-step reasoning

1. Identify the solute named by the concentration and the volume of its solution portion. 2. Convert the volume to liters when concentration is in mol L⁻¹. 3. Multiply cV to obtain moles of that solute, preserving its chemical label. 4. Write and balance the reaction; apply the appropriate coefficient ratio. 5. Check other reactant supply and express the final answer in requested units.

Visual explanation

Draw a beaker marked “25.0 mL of 0.200 mol L⁻¹ HCl.” An arrow labeled “convert 25.0 mL → 0.0250 L; multiply cV” leads to “0.00500 mol HCl.” A second arrow labeled “balanced 1:1 HCl:NaCl” leads to “0.00500 mol NaCl.” This separates solution measurement from reaction chemistry.

Real-world analogy

A drink concentrate label might state grams of sugar per liter of prepared drink. Multiplying by a measured drink volume estimates sugar mass in that portion; the volume refers to the whole beverage, not to a puddle of pure sugar. Molarity works similarly but counts solute moles per liter of whole solution.

Real-world example

Suppose 50.0 mL of 0.300 mol L⁻¹ AgNO₃ reacts with excess NaCl according to AgNO₃ + NaCl → AgCl(s) + NaNO₃. The silver nitrate amount is 0.0500 L × 0.300 mol L⁻¹ = 0.0150 mol. The 1:1 ratio predicts 0.0150 mol AgCl precipitate, assuming full precipitation and sufficient chloride.

Why?

Why multiply concentration by solution volume before applying a reaction ratio? Balanced equations compare amounts of chemical species, not liters of mixture. The concentration tells how many moles of named solute occupy the measured solution volume, yielding the quantity the equation can use.

Common misconception

“Twenty-five milliliters of acid and twenty-five milliliters of base contain the same number of moles.” Equal volumes can have different concentrations. Even equal solute amounts may react at a non-1:1 ratio if the balanced equation requires multiple acid or base units.

Worked example

How many moles of NaOH are needed to fully neutralize 40.0 mL of 0.150 mol L⁻¹ H₂SO₄? First n(H₂SO₄) = 0.0400 L × 0.150 mol L⁻¹ = 0.00600 mol. Balance H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Then n(NaOH) = 0.00600 mol H₂SO₄ × (2 mol NaOH / 1 mol H₂SO₄) = 0.0120 mol NaOH. The calculation does not yet give NaOH solution volume. If the base concentration were provided, divide 0.0120 mol by that concentration to obtain the required solution volume in liters. The two hydrogen ions associated with complete neutralization are reflected in the balanced coefficient 2.

Quick check

1. How many moles of solute are in 25.0 mL of a 0.200 mol L⁻¹ solution? Answer: There are 0.00500 mol of the named solute after converting the solution volume to 0.0250 L.

Exam focus

Specify “liters of solution,” convert mL carefully and label the solute. Use n = cV once for each supplied solution. A polyprotic acid's mole amount is not multiplied by its proton count until the appropriate reaction or equivalent relation is applied.

Advanced insight

The relationship c = n/V is an intensive concentration measure under a given physical state. Dilution preserves solute amount while changing c and V, so c₁V₁ = c₂V₂ when no reaction or loss occurs. A reaction changes the solute amount and instead requires the balanced equation's coefficient relation.

Summary

Molarity links a solution's measured volume to moles of a named dissolved species: n = cV, with compatible units. Those moles can then enter a balanced reaction calculation. The solution volume is not pure solute volume, and equal volumes need not supply equal reacting amounts.

Practice questions

1. How many moles NaCl are in 0.100 L of 0.250 mol L⁻¹ NaCl solution? Answer: 0.0250 mol NaCl formula units. 2. Convert 15.0 mL to liters. Answer: 0.0150 L, preserving three significant figures. 3. How many moles HCl are in 15.0 mL of 0.400 mol L⁻¹ HCl? Answer: 0.00600 mol HCl from cV. 4. Does 0.0100 mol H₂SO₄ require 0.0100 mol NaOH for complete neutralization? Answer: No. It requires 0.0200 mol NaOH by the balanced 1:2 ratio. 5. What stays constant when a solution is diluted without solute loss? Answer: The amount in moles of dissolved solute stays constant while concentration decreases.