Representations of Solutions: Study Guide (Chemistry 3.8)

What a particle diagram of a solution must show

In Topic 3.7 you described solutions with numbers: 0.25 M CaCl₂, for example. A particle diagram describes the same solution as a picture of a tiny volume of it. Exam questions ask you both to draw these diagrams and to read information from them. A good diagram communicates two things.

1. Concentrations. The relative numbers of each kind of particle must be right. In CaCl₂(aq) there must be twice as many Cl⁻ ions as Ca²⁺ ions. If you compare two diagrams, the number of particles per unit volume shows which solution is more concentrated.
2. Interactions. The way particles are arranged must reflect the forces between them, which you met in Topic 3.1. Solvent molecules near an ion point their oppositely charged end towards it. A molecular solute that can hydrogen bond is drawn with hydrogen bonds to the solvent.

Use a key, labels or distinct shapes for each particle type. Colour alone is not enough, because diagrams are often printed in black and white.

Drawing ionic solutes

When an ionic compound dissolves in water, its ions separate and spread out among the water molecules. Follow four rules:

• Separate the ions. NaCl(aq) is drawn as free Na⁺ and Cl⁻ ions, not as Na–Cl pairs and not as a lump of solid.
• Use the formula ratio. For Na₂CO₃(aq) draw two Na⁺ for every CO₃²⁻. A diagram with 6 Na⁺ and 3 CO₃²⁻ is correct; 4 Na⁺ and 4 CO₃²⁻ is not.
• Keep polyatomic ions whole. SO₄²⁻, NO₃⁻, CO₃²⁻ and NH₄⁺ are held together by covalent bonds, which do not break on dissolving. Draw each one as a single unit with its charge.
• Check the charges. Total positive charge must equal total negative charge in the box. For 2 Ca²⁺ and 4 Cl⁻: +4 and −4.

How water surrounds ions

A water molecule is polar. The oxygen carries a partial negative charge (δ−) and each hydrogen a partial positive charge (δ+). Each dissolved ion is surrounded by several water molecules, held by ion–dipole attractions:

• Around a cation (such as Na⁺), the water molecules point their oxygen atoms inwards.
• Around an anion (such as Cl⁻), they point a hydrogen atom inwards.

The reason is simple electrostatics: opposite charges attract. The positive Na⁺ attracts the δ− end of each water molecule; the negative Cl⁻ attracts a δ+ end. If a diagram shows hydrogen atoms pointing at Na⁺, it shows a repulsion, not an attraction, and it is wrong.

Drawing molecular solutes

A molecular solute does not break into ions when it dissolves. Glucose, sucrose, ethanol and methanol stay as whole molecules among the water molecules. What you show instead is how they interact with water:

• Methanol in water (CH₃OH). Methanol has an O–H group. Draw hydrogen bonds (dashed lines) both ways: from the H of methanol’s O–H to the O of a water molecule, and from the H of a water molecule to methanol’s O. The CH₃ end does not hydrogen bond.
• Iodine in hexane (I₂ in C₆H₁₄). Both are nonpolar, so the only attractions are London dispersion forces. There is no particular orientation to show; the I₂ molecules are simply spread evenly among the hexane molecules.

The same counting rules apply. If a diagram represents 0.20 M glucose and 0.10 M methanol in the same volume, it should show twice as many glucose molecules as methanol molecules.

Reading concentration from a diagram

A box in a diagram stands for a definite volume of solution. The concentration it shows is:

concentration ∝ number of solute particles ÷ volume represented

Two consequences follow.

• If two boxes represent equal volumes, the box with more solute particles is more concentrated.
• If the volumes are different, you must divide by the volume before you compare. A bigger box with more particles can still be the more dilute solution.

Dilution is easy to draw: keep the same number of solute particles and make the box represent a larger volume. Doubling the volume halves the concentration.

A four-point check for any solution diagram

Before you hand in a drawing, or when you judge someone else’s, run through these questions in order:

1. Are the solute particles the right kind? Ions for ionic compounds (with polyatomic ions whole), whole molecules for molecular compounds.
2. Is the ratio right? Count each kind of particle and compare with the formula. Then add up the charges: they must total zero.
3. Are the interactions right? Water’s O towards cations, H towards anions; hydrogen bonds drawn between an H on O (or N) and a lone pair on O (or N) of the other molecule.
4. Does the amount match the concentration? If the question gives a concentration or a comparison, check that the number of particles per volume shows it, including after any dilution or mixing.

A diagram can be neat and still lose credit on any one of these. Steps 2 and 3 are the easiest to get wrong, so check them twice.

Worked example 1: from a diagram to molarity

Question. A particle diagram represents 0.250 L of potassium sulfate solution, K₂SO₄(aq). It shows 6 K⁺ ions and 3 SO₄²⁻ ions, each drawn as a single unit, with water molecules oriented around them. Each ion drawn stands for 0.0100 mol of that ion. (a) Is the ratio of ions correct? (b) Find [K⁺], [SO₄²⁻] and the molarity of K₂SO₄.

1. Ratio. K₂SO₄ gives 2 K⁺ per SO₄²⁻. 6 : 3 = 2 : 1, so the ratio is correct. Charge check: 6(+1) + 3(−2) = 0.
2. Moles. n(K⁺) = 6 × 0.0100 = 0.0600 mol; n(SO₄²⁻) = 3 × 0.0100 = 0.0300 mol.
3. Concentrations. [K⁺] = 0.0600 mol ÷ 0.250 L = 0.240 M; [SO₄²⁻] = 0.0300 mol ÷ 0.250 L = 0.120 M.
4. Salt molarity. One SO₄²⁻ per formula unit, so [K₂SO₄] = [SO₄²⁻] = 0.120 M.

Answer. (a) Yes, 2 : 1 with zero total charge. (b) [K⁺] = 0.240 M, [SO₄²⁻] = 0.120 M, K₂SO₄ is 0.120 M.

What else the diagram should show. The water molecules nearest each K⁺ should have O facing the ion; those nearest each SO₄²⁻ should have H facing one of the sulfate’s oxygen atoms.

Worked example 2: finding the errors in a student’s diagram

Question. A student is asked to draw CaCl₂(aq). The student’s box shows 3 units, each a Ca atom joined to two Cl atoms, with water molecules around them. Around every unit, the water molecules point their H atoms inwards. Identify the errors and describe a correct diagram.

1. Ions not separated. CaCl₂ is ionic. In solution there are no Ca–Cl units: the Ca²⁺ and Cl⁻ ions move apart, each surrounded by its own water molecules.
2. Charges missing. The particles should be labelled Ca²⁺ and Cl⁻, not drawn as neutral atoms.
3. Wrong orientation near the cation. Water near Ca²⁺ should point its O atom (δ−) towards the ion. H atoms facing Ca²⁺ would mean like charges facing each other.

A correct diagram shows, for example, 3 Ca²⁺ and 6 Cl⁻ ions spread out separately. Water molecules around each Ca²⁺ have O inwards; those around each Cl⁻ have one H inwards. Charge: 3(+2) + 6(−1) = 0.

Why the ratio matters. Keeping 3 calcium and 6 chlorine was the one thing the student got right. A diagram with equal numbers of Ca²⁺ and Cl⁻ would be wrong even with perfect orientation.

Worked example 3: comparing two diagrams

Question. Box A represents 0.50 L of a sugar solution and contains 6 sugar molecules. Box B represents 1.00 L of the same kind of sugar solution and contains 8 sugar molecules. Which solution is more concentrated, and to what volume would solution A have to be diluted to match B?

1. Particles per litre: A = 6 ÷ 0.50 L = 12 per L; B = 8 ÷ 1.00 L = 8 per L.
2. A is more concentrated, by a factor of 12 ÷ 8 = 1.5, even though B shows more molecules.
3. To match B, the 6 molecules of A must be spread through 6 ÷ 8 per L = 0.75 L.

Answer. A is more concentrated; diluting A to 0.75 L would make the two match.

Common misconceptions

• Drawing ionic compounds as pairs or lattices in solution. Dissolved ions are separate, each with its own shell of water molecules.
• Breaking up polyatomic ions. NO₃⁻ stays as one NO₃⁻ unit; the N–O bonds are covalent and do not break on dissolving.
• Splitting molecular solutes into ions. Glucose and ethanol dissolve as whole molecules.
• Water the wrong way round. O faces cations; H faces anions. Ask yourself, “do these ends attract?”
• Wrong ion ratio. Count from the formula and check that the charges sum to zero.
• “More particles drawn means more concentrated.” Only if the boxes represent the same volume.
• “Dilution removes solute particles.” A diluted diagram has the same number of solute particles in a larger volume.
• Beyond the scope. Colligative properties and calculations of molality, percent by mass or percent by volume are not assessed in this course. Molarity is the concentration you will calculate.

Where this leads

Particle diagrams of solutions return in Topic 4.2 (net ionic equations), Topic 4.3 (representing reactions) and in acid–base chemistry in Unit 8, where you will draw which species are present in a weak acid solution. Next, Topic 3.9 uses intermolecular interactions to explain how solution components can be separated: Separation of Solutions and Mixtures. Try the practice questions now, then use the revision notes and the checklist to consolidate. For molarity calculations, go back to Solutions and Mixtures.