Mixed Solution Problem Set

Introduction

Solution calculations become difficult when several ideas appear in one story. The first decision is not which formula to memorize; it is what physically happened. Was solvent added, were two solute-containing liquids mixed, did a reaction consume material, or did a solubility limit leave solid behind?

Core explanation

If only solvent is added to a known aliquot and the named solute is conserved, use c₁V₁ = c₂V₂. If two solutions of the same stable solute are mixed, add their separate cV mole contributions and divide by final volume. If different solutes react, find each starting cV, apply the balanced equation and limiting behavior, and divide any remaining dissolved species by final volume. If excess solid is added, compare amount added with solubility at the stated temperature and solvent basis; only the dissolved part appears in liquid concentration.

Consider a two-stage example. A student dilutes 20.0 mL of 0.500 M NaCl to 100.0 mL and then mixes 50.0 mL of that diluted solution with 50.0 mL of 0.100 M NaCl. First dilution gives 0.100 M. The 50.0 mL aliquot contains 0.00500 mol, and the other 50.0 mL contains 0.00500 mol. Assuming 0.1000 L final volume, mixture concentration is 0.0100/0.1000 = 0.100 M. The numerical equality with each input is not a universal mixing rule; it occurs because both mixed liquids happen to have 0.100 M concentration.

Now contrast an acid-base case. If 25.0 mL of 0.100 M HCl mixes with 10.0 mL of 0.100 M NaOH, the initial amounts are 0.00250 mol and 0.00100 mol. A 1:1 reaction leaves 0.00150 mol acid-equivalent. Assuming final 0.0350 L, formal excess acid concentration is about 0.0429 M. Adding or averaging the original molarities would ignore consumption.

Solubility problems have a different denominator. A limit of 30 g per 100 g water means 15 g capacity for 50 g water. Adding 18 g leaves 3 g solid after equilibrium. If asked for percent by mass of liquid solution , use 15/(50 + 15) × 100 ≈ 23.1%, not 18/68 or 30%. Underlying method choice determines which material belongs in the numerator and denominator.

A good answer labels every species and condition. A gas solubility calculation needs partial pressure and temperature. A mass-percent-to-molarity conversion needs density and molar mass. Missing data should be identified rather than filled with an unstated assumption.

Step-by-step reasoning

1. Describe the physical or chemical change in one sentence. 2. Choose conservation, mixing, reaction or solubility-limit accounting accordingly. 3. Convert all input amounts to a consistent mass or mole basis. 4. Find the final dissolved amount of the named species. 5. Divide by the appropriate final solution or solvent basis.

Visual explanation

Draw a four-way decision tree: “only solvent added” → dilution; “same solute liquids mixed” → add moles; “reaction” → balanced stoichiometry; “excess solid” → solubility limit. Each path ends at “name final species and denominator.”

Real-world analogy

A bank balance requires different arithmetic for a transfer, two deposits, a purchase or an account limit. Applying “add everything” to every story fails. Chemical solution problems likewise require identifying the actual event.

Real-world example

Preparing a reagent for a precipitation demonstration may involve diluting stock, mixing it with another reactant and collecting solid. Each stage uses a different balance: conserved solute in dilution, reaction ratios in precipitation and dissolved amount afterward.

Why?

Why is method selection more reliable than formula hunting? Similar-looking numbers can represent different processes. The event tells which amount is conserved and which can change.

Common misconception

“Every solution problem uses c₁V₁ = c₂V₂.” That relation needs a conserved named solute during a dilution. Mixing or reaction can require a different balance.

Worked example

A 0.200 M CaCl₂ stock supplies 10.0 mL to a flask made up to 100.0 mL. Formal CaCl₂ concentration becomes 0.0200 M. Under ideal complete dissociation, chloride is 0.0400 M. The first step is dilution of CaCl₂ formula amount; the second is ion stoichiometry. Reversing the order or calling 0.0400 M the CaCl₂ concentration would confuse species.

Quick check

1. Which method applies when NaCl and AgNO₃ solutions form AgCl solid? Answer: Use initial cV amounts and the balanced precipitation reaction before calculating any remaining dissolved-ion concentrations.

Exam focus

Annotate the problem with its process and requested species before arithmetic. Include assumptions about final volume, complete reaction and equilibrium where needed.

Advanced insight

Real systems may combine dilution, reaction and equilibrium at once. A full treatment can require mass balance, charge balance and equilibrium constants; the simple decision tree identifies the first model, not every advanced correction.

Summary

Choose a method from the event: dilution conserves solute, mixing adds contributions, reaction changes species amounts, and solubility limits dissolved material. A named species and correct denominator complete each solution.

Practice questions

1. What method follows when 10 mL of stock is made to 100 mL with solvent? Answer: Dilution: conserve the stock portion's solute moles and use the final 100 mL solution volume. 2. What method follows when two NaCl solutions of different concentration mix? Answer: Calculate NaCl moles in each with cV, add them, then divide by measured or permitted final volume. 3. What method follows when added solid exceeds a g-per-100-g-water limit? Answer: Scale the solubility limit to solvent mass; excess remains a separate phase and is excluded from liquid concentration.