Understanding Solutions
Most chemistry happens in solution — reactions in water, blood, seawater. This guide covers molarity, dilution, solubility, and solution stoichiometry with step-by-step worked examples.
Concentration tells you how much solute is crammed into a given volume of solution. Chemists use molarity (mol/L) as the standard unit because moles connect directly to stoichiometry — once you have moles, you can use the balanced equation.
Chemical reactions happen between particles, not grams. A mole is a count of particles. Using mol/L means you can go directly from concentration and volume to mole ratios without an extra mass conversion step. It is the language of stoichiometry applied to solutions.
C = n / V n = C × V V = n / C
V must always be in litres. Convert mL ÷ 1000 first.
b) You have 2.0 M HCl and need 0.05 mol. What volume of solution do you need?
c) Why must the volume be in litres and not mL when using C = n/V?
a) V = 0.500 L. C = n/V = 0.25/0.500 = 0.50 mol/L
b) V = n/C = 0.05/2.0 = 0.025 L = 25 mL
c) The unit mol/L is defined with litres in the denominator. If you plug in mL you get mol/mL, which is 1000 times too large — your answer will be wrong by a factor of 1000. Always divide mL by 1000 first.
To make a solution of exactly the right concentration, you must dissolve the solute and then bring the total volume to a precise value. The critical tool is a volumetric flask — a flask with a single graduation mark at an exact volume.
Adding a fixed volume of water does not give a fixed total volume, because the solute itself takes up space. If you dissolve 20 g of NaOH in exactly 1.00 L of water, the total volume will be slightly more than 1.00 L. You must dissolve the solute first, then dilute to the mark — so the total volume of solution is exactly right.
1. Calculate the required mass of solute (m = n × M, where n = C × V)
2. Weigh out that mass precisely
3. Dissolve in a small amount of solvent in a beaker
4. Transfer to the volumetric flask and add solvent up to the graduation mark
Dissolve 20.0 g NaOH in about 500 mL water. Transfer to a 1.00 L volumetric flask. Add distilled water up to the 1.00 L mark. Stopper and mix.
Weigh 2.65 g Na₂CO₃. Dissolve in ~100 mL water. Transfer to a 250 mL volumetric flask. Dilute to the 250 mL mark.
b) Why do you add solvent up to the mark on a volumetric flask, rather than adding a fixed volume of water?
c) Why is a volumetric flask used rather than a measuring cylinder for this procedure?
a) n = 0.40 × 0.500 = 0.200 mol. m = 0.200 × 58.5 = 11.7 g
b) The solute itself occupies volume. If you dissolve the solute and then add exactly 500 mL of water, the total volume will be slightly more than 500 mL, making C slightly less than intended. Adding solvent to the exact mark ensures total solution volume is precisely 500 mL.
c) A measuring cylinder is accurate to only ±1–2 mL. A volumetric flask is calibrated to much tighter tolerances (±0.1–0.2 mL) for a single specific volume. For an accurate molar solution you need the precision a volumetric flask provides.
In practice you often have a concentrated stock solution and need a more dilute working solution. Since you are only adding solvent — not adding or removing solute — the number of moles of solute stays exactly constant.
Moles of solute = concentration × volume. Before dilution: n = C₁V₁. After dilution: n = C₂V₂. Because no solute is added or removed, these must be equal — so C₁V₁ = C₂V₂. It is nothing more than "moles before = moles after."
Moles of solute are constant throughout. V₂ is the total final volume — not the volume of water added.
Pipette 7.5 mL of the 6.0 M stock into a 300 mL volumetric flask. Add distilled water up to the 300 mL mark. Mix well.
b) You need 250 mL of 0.50 M from a 4.0 M stock. What volume of stock is needed?
c) A student adds 150 mL of water to 50 mL of stock and calls V₂ = 150 mL. What is the error?
a) C₂ = C₁V₁/V₂ = (3.0 × 0.100)/0.600 = 0.50 mol/L
b) V₁ = C₂V₂/C₁ = (0.50 × 0.250)/4.0 = 0.125/4.0 = 0.03125 L = 31.25 mL
c) V₂ is the total final volume, not the volume of water added. V₂ = 50 mL + 150 mL = 200 mL = 0.200 L, not 150 mL. Using 150 mL would give a concentration that is too high.
Not all ionic compounds dissolve equally. Solubility rules let you predict whether mixing two solutions will produce a precipitate (an insoluble solid). Electrolytes are substances that produce ions in solution — and ions are what carry electric current.
Dissolution is a competition between the lattice energy holding the ions together and the hydration energy gained when water surrounds the ions. If lattice energy wins, the compound stays solid. For compounds like AgCl, the Ag⁺–Cl⁻ attraction is so strong that water cannot pull them apart efficiently — so AgCl remains as a white precipitate.
Ba²⁺ with SO₄²⁻: solubility rule says SO₄²⁻ is insoluble with Ba²⁺ — this is an exception.
NaCl is a soluble ionic salt — it fully dissociates into Na⁺ and Cl⁻. Strong electrolyte.
Acetic acid is a weak acid — it only partially dissociates. Weak electrolyte.
Sugar is a molecular compound — it dissolves but does not produce ions. Nonelectrolyte.
b) Classify HCl(aq) as strong electrolyte, weak electrolyte, or nonelectrolyte. Explain.
c) Write the dissociation equation for Ca(OH)₂ dissolving in water.
a) Yes — Ag⁺ with Cl⁻ is insoluble (exception to "usually soluble halides"). Net ionic: Ag⁺(aq) + Cl⁻(aq) → AgCl(s) white precipitate.
b) HCl is one of the six strong acids — it fully dissociates into H⁺ and Cl⁻ in water. Strong electrolyte.
c) Ca(OH)₂(s) → Ca²⁺(aq) + 2 OH⁻(aq) — note the coefficient 2 in front of OH⁻.
In solution reactions, concentrations replace masses. The only new step is converting litres of solution to moles of solute using n = C × V. After that, the stoichiometry roadmap is exactly the same as for solid and gas reactions.
In a lab reaction, you measure volumes with a burette or pipette — not masses. Once you have n, you use the mole ratio from the balanced equation to get moles of the other substance, then convert to grams or to concentration if needed. The roadmap never changes; only the entry point does.
b) What is the number of moles in 20.0 mL of 0.500 M H₂SO₄?
c) What mass of AgCl (M = 143.3 g/mol) precipitates when 100 mL of 0.200 M AgNO₃ is mixed with excess NaCl? (AgNO₃ + NaCl → AgCl↓ + NaNO₃)
a) n(HCl) = 1.0 × 0.0500 = 0.050 mol. Mole ratio 1:1, so n(NaOH) = 0.050 mol. C(NaOH) = n/V = 0.050/0.0500 = 1.0 mol/L
b) n = C × V = 0.500 × 0.0200 = 0.0100 mol
c) n(AgNO₃) = 0.200 × 0.100 = 0.0200 mol. Mole ratio 1:1, so n(AgCl) = 0.0200 mol. m(AgCl) = 0.0200 × 143.3 = 2.87 g