Limiting Reactants in Mixed Solutions

Introduction

Mixing two measured solutions creates a familiar limiting-reagent problem with one extra step for each input. Convert each concentration and added volume into solute moles, then compare those mole supplies against their balanced coefficients. The larger solution volume need not contain more reacting material.

Core explanation

For H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l), suppose 40.0 mL of 0.200 mol L⁻¹ H₂SO₄ is mixed with 50.0 mL of 0.250 mol L⁻¹ NaOH. Acid amount is 0.0400 L × 0.200 mol L⁻¹ = 0.00800 mol H₂SO₄. Base amount is 0.0500 L × 0.250 mol L⁻¹ = 0.0125 mol NaOH. Divide by coefficients: acid capacity is 0.00800/1 = 0.00800 mol reaction, and base capacity is 0.0125/2 = 0.00625 mol reaction. NaOH limits despite its larger solution volume and larger raw mole amount. The theoretical Na₂SO₄ amount is 0.00625 mol.

Acid consumed is 0.00625 mol, leaving 0.00800 − 0.00625 = 0.00175 mol H₂SO₄ in the ideal complete-neutralization account. Water formed is 2 × 0.00625 = 0.0125 mol according to the balanced equation. This reaction bookkeeping is separate from calculating the final concentration of leftover acid. For concentration, one would need the final solution volume and an appropriate solution model; it would generally be wrong to divide leftover acid moles by only the original 40.0 mL acid portion.

The factor of two between H₂SO₄ and NaOH comes from the specified complete-neutralization equation. One mole of acid formula amount does not automatically mean one mole of every ionic species at all conditions. In an introductory acid-base calculation, use the balanced chemical reaction rather than inventing a one-to-one rule from equal concentrations or visual solution volumes. For a different acid, base or endpoint, the relevant reaction may have a different ratio.

Mixed precipitation solutions follow the same workflow. For Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq), convert Pb(NO₃)₂ and KI solution data separately to moles. Compare lead-salt amount divided by one against KI amount divided by two. The smaller capacity sets theoretical PbI₂. A high KI concentration does not guarantee KI excess if its added volume is tiny; concentration and volume both matter.

Be careful with what the concentration describes. A bottle labeled 0.100 mol L⁻¹ CaCl₂ gives moles of CaCl₂ formula amount per liter of solution; in the simple dissociation model that equals moles of Ca²⁺ but twice that amount of Cl⁻. If the reaction uses chloride, apply the internal subscript after cV. If it uses calcium, the relationship is one-to-one. The solution's molarity is not automatically the concentration of each ion at the same numerical value.

Measured mixing volumes may not be perfectly additive, and chemical reactions may change solution composition. Introductory product-mole calculations need only the original portion volumes for cV, not the final mixed volume. If a final concentration is requested, a problem should provide or authorize an approximation for final volume. Actual acid-base equilibria, incomplete precipitation and side reactions can also complicate ideal results; the balanced stoichiometric calculation gives a maximum for the stated pathway.

Step-by-step reasoning

1. Write and balance the reaction with correct aqueous or solid state labels. 2. Convert each added solution volume to liters and calculate its named solute moles by cV. 3. Convert formula amounts to relevant ion amounts if the net ionic equation is used. 4. Divide each reacting amount by its coefficient and select the smallest capacity. 5. Use that extent for product and leftover moles; use final-volume data only if concentration is asked.

Visual explanation

Draw two input beakers. The first reads “40.0 mL × 0.200 M = 0.00800 mol H₂SO₄”; the second reads “50.0 mL × 0.250 M = 0.0125 mol NaOH.” Beneath, put capacity cards “÷1 = 0.00800” and “÷2 = 0.00625.” Arrow from the smaller card to “NaOH limits; 0.00625 mol Na₂SO₄ maximum.”

Real-world analogy

Two delivery trucks may carry different numbers of ingredients per liter of container space. A larger truck does not necessarily carry enough of its ingredient for the assembly ratio. First determine the actual number delivered from amount per liter and liters, then compare with the recipe's per-product requirements.

Real-world example

Mix 20.0 mL of 0.150 mol L⁻¹ AgNO₃ with 30.0 mL of 0.0800 mol L⁻¹ NaCl. Silver amount is 0.00300 mol, chloride amount 0.00240 mol. In Ag⁺ + Cl⁻ → AgCl(s), chloride limits, so theoretical AgCl is 0.00240 mol and 0.00060 mol Ag⁺ remains in the ideal account.

Why?

Why not compare only concentrations or only volumes? The reacting supply is their product cV. A concentrated solution can supply little solute if only a small portion is added, while a dilute solution can supply more if its volume is large. Coefficients then set how each supply converts into reaction progress.

Common misconception

“Both solutions are 0.100 M, so equal milliliters always neutralize.” Equal molarity and volume give equal formula mole amounts, but a reaction such as H₂SO₄ + 2NaOH requires twice as many base moles as acid moles for complete neutralization.

Worked example

Mix 25.0 mL of 0.300 mol L⁻¹ Pb(NO₃)₂ with 40.0 mL of 0.250 mol L⁻¹ KI. Balanced equation: Pb(NO₃)₂ + 2KI → PbI₂(s) + 2KNO₃. Lead-salt amount is 0.0250 × 0.300 = 0.00750 mol. KI amount is 0.0400 × 0.250 = 0.0100 mol. Capacities are 0.00750/1 = 0.00750 mol reaction and 0.0100/2 = 0.00500 mol reaction. KI limits; PbI₂ theoretical amount is 0.00500 mol. Pb(NO₃)₂ consumed is 0.00500 mol, leaving 0.00250 mol formula amount. If M(PbI₂) ≈ 461.0 g mol⁻¹, theoretical dry mass is 2.31 g to three significant figures. These amounts are based on the starting solution portions; no final mixed volume was needed.

Quick check

1. Which limits 0.00800 mol H₂SO₄ and 0.0125 mol NaOH in H₂SO₄ + 2NaOH → products? Answer: NaOH limits because dividing its amount by two gives 0.00625, below the acid's 0.00800 capacity.

Exam focus

Convert both mL values to liters before multiplying by molarity. Label each result with its solute, then normalize by coefficients. A leftover mole amount and its concentration are different answers; final concentration requires the final solution volume.

Advanced insight

The capacity method treats each initial dissolved solute amount as an inventory constraint. In a real aqueous equilibrium, free-ion concentrations may change continuously and some product may redissolve. The ideal limiting calculation still gives a stoichiometric ceiling, while equilibrium data determine the actual final distribution.

Summary

For mixed solutions, calculate each input solute amount by concentration times its own added volume. Divide those mole amounts by their reaction coefficients to find the limit, then predict product and leftovers. Neither solution volume nor concentration alone identifies the limiting reactant, and final concentration needs separate volume information.

Practice questions

1. What amount of H₂SO₄ is in 40.0 mL of 0.200 mol L⁻¹ solution? Answer: 0.00800 mol H₂SO₄ formula amount. 2. What amount of NaOH is in 50.0 mL of 0.250 mol L⁻¹ solution? Answer: 0.0125 mol NaOH formula amount. 3. Which limits those solutions under H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O? Answer: NaOH limits because its normalized reaction capacity is 0.00625 mol. 4. What is the theoretical Na₂SO₄ amount for that mixture? Answer: 0.00625 mol Na₂SO₄ by its coefficient of one. 5. Can the original 40.0 mL acid volume alone determine the final leftover acid concentration? Answer: No. The final mixed solution volume or an authorized approximation is also needed.