Reactions in Aqueous Solution — Science Sage Gr10

Dissolving Ionic Compounds in Water

Water is a polar molecule — it has a slightly negative oxygen end (δ⁻) and slightly positive hydrogen ends (δ⁺). When an ionic solid such as sodium chloride is placed in water, the polar water molecules surround and pull apart the positive and negative ions in the crystal lattice. This process is called dissociation: the ionic compound separates into free-moving, hydrated ions dispersed throughout the solution.

NaCl(s) → Na⁺(aq) + Cl⁻(aq)

The state symbol (aq) means "aqueous" — dissolved in water. Once dissociated, the ions are no longer locked in a fixed lattice; they are free to move independently through the solution, each surrounded by a "shell" of oriented water molecules (a process called hydration).

Rule of thumb: when a soluble ionic compound dissolves, write it as separate (aq) ions — this is how it actually exists in solution.

Why Do Ionic Solutions Conduct Electricity?

Electrical conductivity requires charged particles that are free to move. In solid NaCl, the ions are locked in place in a rigid lattice and cannot move — solid ionic compounds do NOT conduct electricity. But once dissolved in water, the Na⁺ and Cl⁻ ions become mobile charge carriers. When a voltage is applied, positive ions drift toward the negative electrode and negative ions drift toward the positive electrode, completing the circuit.

Electrolytes are substances whose aqueous solutions (or molten forms) conduct electricity because they contain free-moving ions. Non-electrolytes do not produce ions in solution, so their solutions do not conduct.

Most covalent (molecular) compounds — such as sugar (C₁₂H₂₂O₁₁) or ethanol (C₂H₅OH) — dissolve in water as whole, neutral molecules. No ions are formed, so there are no free charge carriers, and these solutions generally do not conduct electricity (they are non-electrolytes). A small number of covalent molecules, such as HCl, ionise fully or partially when dissolved and DO produce ions — these are the acids, which you will meet as electrolytes too.

Solution	Type of substance	Conducts?	Reason
NaCl(aq)	Ionic	Yes	Dissociates into free Na⁺ and Cl⁻ ions
CuSO₄(aq)	Ionic	Yes	Dissociates into free Cu²⁺ and SO₄²⁻ ions
Sugar solution	Covalent (molecular)	No	Dissolves as neutral molecules — no ions
Ethanol solution	Covalent (molecular)	No	Dissolves as neutral molecules — no ions
HCl(aq)	Covalent, but ionises	Yes	Reacts with water to form H⁺(aq) and Cl⁻(aq)

Solubility Rules

Not all ionic compounds dissolve in water. Chemists use a set of general solubility rules to predict whether a compound will be soluble (dissolves, forming a clear solution) or insoluble (remains a solid, forming a precipitate).

Ion / compound family	Solubility rule
Group 1 (Na⁺, K⁺, Li⁺...) and NH₄⁺	Almost all compounds are soluble
Nitrates (NO₃⁻)	All nitrates are soluble
Chlorides (Cl⁻)	Soluble, EXCEPT AgCl, PbCl₂ (and Hg₂Cl₂)
Sulfates (SO₄²⁻)	Soluble, EXCEPT BaSO₄, PbSO₄ (and CaSO₄ is only slightly soluble)
Carbonates (CO₃²⁻)	Generally insoluble, EXCEPT Group 1 carbonates and (NH₄)₂CO₃
Hydroxides (OH⁻)	Generally insoluble, EXCEPT Group 1 hydroxides and Ba(OH)₂ (slightly soluble)

Learn these rules well — they let you predict, without a solubility chart, whether mixing two solutions will produce a precipitate.

Precipitation Reactions

A precipitate is an insoluble solid that forms and separates out when two aqueous solutions are mixed. A precipitation reaction occurs when the cations from one solution combine with the anions from another solution to form a new, insoluble ionic compound.

This is a type of double replacement (metathesis) reaction: the positive and negative ions "swap partners." If the new combination is insoluble (by the solubility rules), a precipitate forms and can be observed as cloudiness, a colour change, or a solid settling out. If both possible new combinations are soluble, no visible reaction occurs — all four ion types simply remain dissolved together in solution.

Full Ionic Equations and Net Ionic Equations

A chemical equation can be written at three levels of detail:

Molecular (formula) equation: shows complete compound formulae, as if no dissociation occurs.
Full ionic equation: shows every soluble ionic compound broken into its free (aq) ions; insoluble products and pure liquids are kept as whole formulae.
Net ionic equation: the full ionic equation with spectator ions removed — spectator ions are ions that appear unchanged, in the same form, on both sides of the equation. They take no real part in the reaction.

Worked Example — Mixing silver nitrate and sodium chloride solutions:

Step 1 — Molecular equation:
AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

Step 2 — Full ionic equation (dissociate every soluble compound; AgCl is insoluble so it stays as a solid formula):
Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + Cl⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)

Step 3 — Identify spectator ions: Na⁺(aq) and NO₃⁻(aq) appear unchanged on both sides — they are spectators and can be cancelled.

Step 4 — Net ionic equation:
Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

This net ionic equation captures the true chemistry of the reaction: silver ions and chloride ions combine to form solid silver chloride, regardless of which soluble silver salt or which soluble chloride salt you started with.

Method summary: (1) Write the balanced molecular equation and predict the precipitate using solubility rules. (2) Rewrite all soluble (aq) ionic compounds as separate ions; keep the precipitate, water, and any gas as whole formulae. (3) Cancel any ion that appears identically on both sides — these are the spectators. (4) What remains is the net ionic equation.

IEB Extension — Applying Net Ionic Equations to New Combinations

The real value of learning net ionic equations is being able to predict the outcome of unfamiliar combinations, not just memorise one example. Consider mixing barium chloride and sodium sulfate solutions.

Step 1 — Molecular equation:
BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)

Step 2 — Full ionic equation:
Ba²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) + 2Na⁺(aq) + 2Cl⁻(aq)

Step 3 — Cancel spectators (Na⁺ and Cl⁻ appear unchanged on both sides):
Net ionic equation: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

Notice that barium sulfate forming from Ba²⁺ and SO₄²⁻ is the same net reaction no matter which soluble barium salt or soluble sulfate salt you use — this is why the net ionic equation is considered the most chemically meaningful way to represent a precipitation reaction. This idea of a common underlying reaction, independent of the spectator ions present, is a key conceptual bridge to acid–base neutralisation reactions (H⁺(aq) + OH⁻(aq) → H₂O(l)) studied later.

Precipitation Reaction Simulator

Solution A

Solution B

How to use

Pick two aqueous solutions to mix. Watch the ions disperse in the beaker — if an insoluble combination forms, a precipitate settles out and the net ionic equation is shown below. If not, the ions simply stay mixed in solution.

Question 1 (1 mark)

When solid NaCl dissolves in water, this process is best described as:

Dissociation is the separation of an ionic compound into its already-existing ions when it dissolves: NaCl(s) → Na⁺(aq) + Cl⁻(aq). Ionisation refers to covalent molecules forming ions (e.g. HCl reacting with water).

Question 2 (1 mark)

Which solution will NOT conduct electricity?

Sugar (a covalent/molecular compound) dissolves as neutral molecules — no free-moving ions are formed, so the solution cannot conduct electricity. NaCl, CuSO₄, and KI are all ionic and dissociate into free ions.

Question 3 (1 mark)

According to the solubility rules, which of the following compounds is INSOLUBLE in water?

BaSO₄ (barium sulfate) is one of the key exceptions to the "sulfates are soluble" rule, along with PbSO₄. NaCl, KNO₃, and NH₄Cl are all soluble (Group 1 / ammonium salts and nitrates are always soluble).

Question 4 (2 marks)

In the reaction Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq), the spectator ions are:

K⁺ and NO₃⁻ appear unchanged, still dissolved as (aq) ions, on both sides of the equation — they are the spectator ions. Pb²⁺ and I⁻ actually combine to form the solid precipitate PbI₂, so they are NOT spectators.

Question 5 (2 marks)

What is the correct net ionic equation for mixing BaCl₂(aq) with Na₂SO₄(aq)?

The precipitate BaSO₄ forms directly from Ba²⁺ and SO₄²⁻ ions combining. Na⁺ and Cl⁻ remain as spectator ions in solution and are cancelled out, leaving Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) as the net ionic equation.

Question 6 (1 mark)

Mixing NaCl(aq) with KNO₃(aq) produces:

Both possible new combinations, NaNO₃ and KCl, are soluble according to the solubility rules (all nitrates soluble; most chlorides soluble). Since no insoluble product can form, all four ion types — Na⁺, Cl⁻, K⁺, NO₃⁻ — simply remain dissolved together with no visible change.

Question 7 (Analysis)

A learner mixes equal volumes of three solutions together in one beaker: silver nitrate (AgNO₃), barium chloride (BaCl₂), and potassium sulfate (K₂SO₄). Using the solubility rules, which precipitate(s) will form?

With three solutions mixed, you must check every possible new ion pairing, not just one. Ag⁺ (from AgNO₃) pairs with Cl⁻ (from BaCl₂): most chlorides are soluble EXCEPT those of Ag⁺, Pb²⁺, and Hg₂²⁺ — so AgCl is insoluble and precipitates. Separately, Ba²⁺ (from BaCl₂) pairs with SO₄²⁻ (from K₂SO₄): most sulfates are soluble EXCEPT those of Ba²⁺, Pb²⁺, and Ca²⁺ — so BaSO₄ is also insoluble and precipitates. K⁺ and NO₃⁻ remain as spectator ions throughout. Because two separate insoluble combinations exist among the six possible ion pairings, two different precipitates form at once.

IEB Question 1 IEB (2 marks)

Silver nitrate solution is mixed with potassium iodide solution, forming a yellow precipitate. What is the net ionic equation for this reaction?

AgI is insoluble, so it precipitates. K⁺ and NO₃⁻ remain dissolved as spectator ions on both sides and cancel out, leaving the net ionic equation Ag⁺(aq) + I⁻(aq) → AgI(s).

IEB Question 2 IEB (1 mark)

Which underlying idea links precipitation net ionic equations (e.g. Ba²⁺ + SO₄²⁻ → BaSO₄) to acid–base neutralisation (H⁺ + OH⁻ → H₂O)?

In both cases, the net ionic equation strips away spectator ions to reveal the actual reacting particles — the reaction is the same no matter which soluble salts or acids/bases supplied those ions.

Show ALL working. Use the solubility rules table from the lesson where needed.

Question 1

(a) Write a balanced equation, with state symbols, for the dissociation of magnesium chloride (MgCl₂) in water.
(b) Explain, in terms of particle behaviour, why molten (melted) MgCl₂ conducts electricity but solid MgCl₂ does not.
(c) Explain why a sugar solution does not conduct electricity even though sugar dissolves readily in water.

Question 2

Using the solubility rules, predict whether each of the following compounds is soluble or insoluble in water. Give a reason in each case.
(a) KNO₃ (b) PbCl₂ (c) Na₂CO₃ (d) CaSO₄ (e) (NH₄)₂SO₄ (f) Fe(OH)₃

Question 3

Aqueous solutions of lead(II) nitrate and potassium iodide are mixed.
(a) Write the balanced molecular equation for this reaction.
(b) Write the full ionic equation.
(c) Identify the spectator ions.
(d) Write the net ionic equation.
(e) State one visible observation that would confirm a reaction has taken place.

Question 4

A learner mixes barium chloride solution with sodium sulfate solution and observes a white precipitate.
(a) Identify the precipitate formed and explain, using the solubility rules, why it is insoluble.
(b) Write the full ionic equation and the net ionic equation for this reaction.
(c) The learner then mixes barium chloride solution with sodium nitrate solution. Predict, with a reason, whether a precipitate will form.

Question 5

Two unlabelled beakers, X and Y, each contain a colourless aqueous solution. A learner mixes a small sample of each together and a white precipitate forms.
(a) Suggest a possible pair of soluble ionic compounds (one in X, one in Y) that could produce this observation. Justify your choice using the solubility rules.
(b) Write the net ionic equation for your suggested reaction.
(c) Explain why writing a net ionic equation is more useful to a chemist than only writing the full molecular equation.

Question 6 — Interpreting a Results Table

A learner mixes pairs of solutions in four test tubes and records the results:

Test tube	Solutions mixed	Observation
1	AgNO₃(aq) + NaCl(aq)	White precipitate forms
2	AgNO₃(aq) + NaBr(aq)	Pale yellow precipitate forms
3	BaCl₂(aq) + Na₂SO₄(aq)	White precipitate forms
4	BaCl₂(aq) + NaNO₃(aq)	No visible change

(a) Name the precipitate formed in test tubes 1, 2, and 3.
(b) Write the net ionic equation for the reaction in test tube 3.
(c) Using the solubility rules, explain why test tube 4 shows no visible change.
(d) Identify the spectator ion(s) present in test tube 1.
(e) Based on the pattern across test tubes 1–3, what general rule can you state about compounds containing Ag⁺ or Ba²⁺ ions and certain negative ions?