Q: What does H2 Chemistry Notes: Topic 10 - Chemistry of Aqueous Solutions cover? A: Understand ionic equilibria, solubility products, hydrolysis, complex ion formation, and qualitative analysis for the extension topic Chemistry of Aqueous Solutions.
TL;DR Aqueous solutions is about ions competing in water: some precipitate, some hydrolyse, and some form complexes. The key skill is deciding which equilibrium controls the observation.
Ions in water can precipitate, react, or stay dissolved: Identify the ions present.
Compare ionic product with Ksp: Decide whether a precipitate forms.
Hydrolysis and complex formation can shift the visible result: Write the relevant equilibrium before explaining the observation.
Concrete example: Adding ammonia to CuX2+
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can first form a precipitate, then dissolve it in excess ammonia to give a deep blue complex. That is not a contradiction; it is two equilibria in sequence.
This extension topic examines how ions behave in water: precipitation, hydrolysis, and complex formation underpin analytical chemistry questions in Papers 1-3 (and show up in Paper 4 through qualitative analysis and complex-ion work).
Status: SEAB's current H2 Chemistry (9476) syllabus PDF is labelled for 2026, and the current Chemistry Data Booklet is labelled 8873/9476/9813 for use from 2026 in non-practical papers. Extension Topic 10 is assessed across Papers 1-4; the data booklet is used in Papers 1-3, not Paper 4 practical.
Quick revision box
What this topic tests: Ionic equilibria, Ksp, hydrolysis, complex ion formation, and QA reasoning.
Top mistakes to avoid: Mixing solubility and concentration terms; incomplete ionic equations; weak hydrolysis/complex-ion justification.
20-minute sprint plan: 5 min Ksp/hydrolysis formula recap; 10 min precipitation/complex-ion problems; 5 min QA decision chain.
1 Solubility Product Ksp
For sparingly soluble salt AB(s)⇋AX+(aq)+BX−(aq):
Ksp=[AX+][BX−]
More generally, for an empirical formula with p cations and q anions in the formula unit, the powers in Ksp come from the dissolution equation: Ksp=[AXq+]p[BXp−]q.
1.1 Ionic Product and Precipitation
If ionic product Qsp<Ksp: unsaturated, no precipitate.
If Qsp=Ksp: saturated solution, equilibrium.
If Qsp>Ksp: precipitation occurs until Qsp=Ksp
Use stoichiometric coefficients carefully when raising concentrations to powers.
Ksp expression checkpoint
Before calculating solubility or ionic product, write the dissolution equation and let the coefficients set the powers. This prevents the common mistake of copying the formula subscripts into the wrong concentration term.
AgCl. Dissolution: AgCl(s)⇌AgX+(aq)+ClX−(aq). Expression: Ksp=[AgX+][ClX−]. Trap: squaring a concentration just because two ions are present.
CaFX2. Dissolution: CaFX2(s)⇌CaX2+(aq)+2FX−(aq). Expression: Ksp=[CaX2+][FX−]X2. Trap: forgetting that fluoride concentration is 2S in pure water.
AgX2CrOX4. Dissolution: AgX2CrOX4(s)⇌2AgX+(aq)+CrOX4X2−(aq). Expression: Ksp=[AgX+]X2[CrOX4X2−]. Trap: putting the square on chromate instead of silver.
Al(OH)X3. Dissolution: Al(OH)X3(s)⇌AlX3+(aq)+3OHX−(aq). Expression: Ksp=[AlX3+][OHX−]X3. Trap: treating hydroxide as one group and missing the cube.
Worked check: for AgX2CrOX4 in pure water, let the molar solubility be S. Then [CrOX4X2−]=S and [AgX+]=2S, so Ksp=(2S)2(S)=4S3. The coefficient changes the concentration first; the power then comes from the same coefficient in the Ksp expression.
Misconception check: Ksp powers come from the balanced dissolution equation, not from the charges on the ions. Charge balance tells you the formula; the dissolution equation tells you the expression.
1.2 Precipitation Decision Flow
Most Ksp mistakes happen before the comparison step. Use this order whenever two aqueous solutions are mixed.
Step
What to write
Why it matters
1
List the possible precipitate from the ions present.
Ksp only applies after you know the sparingly soluble salt.
2
Recalculate ion concentrations after dilution.
Mixing volumes changes concentration before precipitation starts.
3
Build Qsp with the same powers as Ksp.
The powers come from the dissolution equation, not from the amount mixed.
4
Compare Qsp with Ksp.
Only Qsp>Ksp
For example, mix 25.0mL of 2.0⋅10−3mol⋅L−1AgNOX3 with 25.0mL of 2.0⋅10−3mol⋅L−1NaCl. The possible precipitate is AgCl.
After mixing, the total volume is 50.0mL, so both ion concentrations are halved:
[AgX+]=[ClX−]=1.0⋅10−3mol⋅L−1.
Then
Qsp=[AgX+][ClX−]=(1.0×10−3)(1.0×10−3)=1.0×10−6.
If the question gives Ksp(AgCl)=1.8×10−10, then Qsp>Ksp, so AgCl(s) precipitates.
Trap check: do not compare the original 2.0⋅10−3mol⋅L−1 concentrations directly. Dilution comes before the Qsp calculation.
2 Common Ion Effect
Adding a common ion reduces solubility (Le Chatelier). Show algebra by substituting initial concentrations and solving for equilibrium using ICE tables. Example:
Ksp(AgCl)=[AgX+][ClX−]
In a 0.10mol⋅L−1NaCl solution, approximate [ClX−]≈0.10, so [AgX+]=Ksp/0.10.
Common-ion approximation checkpoint
Before using the shortcut, decide whether the added common ion is large enough to dominate the ion made by dissolving the sparingly soluble salt.
Situation
Set-up move
Approximation allowed?
Common trap
Pure water only
Let molar solubility be S.
No. Both ion concentrations come from the salt.
Treating one ion as a fixed number when no common ion was added.
Large common-ion concentration given
Use the given common-ion concentration as nearly constant.
Usually yes, if it is much larger than S.
Forgetting to include the ion power from the dissolution equation.
Small common-ion concentration given
Compare it with the expected S.
Not automatically. You may need the full expression.
Assuming every common ion justifies dropping the added S.
Mixture of two solutions
Recalculate concentrations after dilution first.
Only after dilution has been accounted for.
Using the stock concentration before mixing volumes.
Worked check: for AgCl(s)⇋AgX+(aq)+ClX−(aq) in 0.10mol⋅L−1NaCl,
[ClX−] is dominated by the added chloride. If Ksp is very small, the extra chloride from dissolving AgCl is tiny compared with 0.10mol⋅L−1, so [AgX+]≈Ksp/0.10.
Misconception check: the common-ion effect lowers solubility, but it does not change Ksp at the same temperature. The equilibrium concentration changes; the equilibrium constant does not.
3 Hydrolysis of Salts
Salts of weak acids or bases hydrolyse to affect pH.
Example: NHX4Cl in water produces an acidic solution because NHX4X++HX2O⇋NHX3+HX3OX+.
Use Ka or Kb of conjugates to calculate pH (values are provided in the question when needed). For anions of weak acids (e.g. CHX3COOX−), hydrolysis yields a basic solution.
Salt hydrolysis classification checkpoint
Before calculating pH, classify which ion can react with water. Spectator ions from strong acids or strong bases usually do not control the pH.
Salt type
Ion that hydrolyses
Direction to expect
What to write first
Strong acid + strong base
Neither ion meaningfully hydrolyses.
Approximately neutral.
State that both ions are spectators for acid-base hydrolysis.
Strong acid + weak base
Conjugate acid of the weak base, such as NHX4X+.
Acidic.
Write the ion donating a proton to water to form HX3OX+.
Weak acid + strong base
Conjugate base of the weak acid, such as CHX3COOX−.
Basic.
Weak acid + weak base
Both ions can hydrolyse.
Compare the given Ka and Kb.
Do not guess from the formula alone; use the data provided.
Worked check: NHX4NOX3 contains NHX4X+, the conjugate acid of weak base NHX3, and NOX3X−, a spectator ion from a strong acid. The hydrolysis equation is NHX4X++HX2O⇋NHX3+HX3OX+, so the solution is acidic.
Misconception check: a salt is not automatically neutral because it has no visible HX+ or OHX− in the formula. Look at what each ion becomes when it reacts with water.
Ethanoate (conjugate base): classic weak-acid anion for hydrolysis discussions.
4 Complex Ion Formation
Complexes increase solubility when ligands stabilise ions. Stepwise formation constants:
Large Kf values drive dissolution (basis of ammonia acting as complexing agent in qualitative analysis). Combine Ksp and Kf when solving simultaneous equilibria.
Ammonia: ligand used in qualitative-analysis complex formation.
Ethane-1,2-diamine (en): another chelating ligand reference.
5 Qualitative Analysis Links
Before explaining an observation, decide which equilibrium has changed. The same visible sequence can involve precipitation first, then either amphoteric dissolution or ligand complex formation.
Observation pattern
Likely process
What to write in the explanation
Precipitate forms after adding a few drops of reagent.
Qsp exceeds Ksp.
An insoluble hydroxide or salt has formed because the ionic product is too high.
White precipitate dissolves in excess NaOH.
Amphoteric hydroxide dissolves in excess hydroxide.
A soluble hydroxo complex forms, so the solid precipitate is removed.
Coloured precipitate dissolves in excess NHX3.
Ligand substitution forms a soluble ammine complex.
Ammonia ligands stabilise the metal ion in solution.
Precipitate remains insoluble in excess reagent.
No sufficiently stable soluble complex forms under the stated conditions.
The solid remains because the equilibrium still favours the precipitate.
Misconception check: do not use "complex ion forms" for every precipitate that disappears. Excess NaOH often points to hydroxo complexes from amphoteric behaviour, while excess NHX3 points to ammine complexes.
Ion
Reagent
Observation
Explanation
FeX3+
NaOH
Brown precipitate Fe(OH)X3, insoluble in excess.
Precipitation due to low Ksp; no ligand to redissolve.
CuX2+
NHX3
AlX3+
NaOH
White ppt dissolves in excess.
Amphoteric hydroxide forms [Al(OH)X4]X−
Remember to describe colours precisely; these observations are heavily tested.
Observations here mirror the qualitative analysis tables in the SEAB Chemistry Data Booklet for exams from 2026.
6 Worked Example
Question:
Calculate the solubility of CaFX2 in pure water and in 0.10mol⋅L−1NaF, using the reference value Ksp(CaFX2)=3.9×10−11 at 298 K (always use the value supplied in your paper).
Solution:
In pure water:
Let molar solubility be Smol⋅L−1:
[CaX2+]=S
[FX−]=2S
Ksp=S(2S)2=4S3
S=((3.9×10−11)/4)1/3=2.1×10−4mol⋅L−1
In 0.10mol⋅L−1NaF:
Assume added fluoride dominates ([FX−]=0.10).
Ksp=[CaX2+](0.10)2
[CaX2+]=(3.9×10−11)/(1.0×10−2)=3.9×10−9mol⋅L−1
Solubility drops dramatically due to common ion effect.
7 Laboratory Considerations
Paper 4 planning: When proposing qualitative analysis steps, specify sequence (carbonate test before acidifying, then sulfate, then halides), justify reagent volumes, and note need for acidifying with HNOX3 to avoid introducing interfering ions.
Error control: Mention rinsing precipitates, avoiding contamination of dropper bottles, and using freshly prepared solutions for unstable ions (e.g. FeX2+).
8 Typical Misconceptions
Forgetting stoichiometric factors when computing Ksp (e.g. AgX2CrOX4 yields [AgX+]2).
Assuming amphoteric behaviour applies to all metal hydroxides-limit to specified ions (AlX3+,ZnX2+,PbX2+
Using Ksp values without considering simultaneous equilibria (e.g. pH-dependent solubility).
Describing colours vaguely; examiners expect precise descriptors like “deep blue solution” rather than “bluish”.
9 Quick Drills
Determine whether a precipitate forms when 50.0mL of 1.0×10−3mol⋅L−1Pb(NOX3)X2 is mixed with 50.0mL of 2.0×10−3mol⋅L−1NaX2SOX4; Ksp(PbSOX4)=1.6×10−8.
Write ionic equations for observations when aqueous ammonia is added dropwise then in excess to ZnX2+ solution, explaining amphoteric behaviour.
Explain how addition of NHX4Cl affects the solubility of Mg(OH)X2
Common exam mistakes
Mistake: Forgetting to square (or cube) the ion concentration when deriving Ksp - for AgX2CrOX4, the expression is Ksp=[AgX+]2[CrOX4X2−], not [AgX+][CrOX4X2−].
Mistake: Comparing the ionic product Qsp to Ksp without accounting for dilution when two solutions are mixed - always recalculate concentrations using the combined volume first.
Mistake: Applying the common ion effect approximation without checking whether the simplification is valid - if the common ion concentration is not much larger than S, the approximation breaks down.
Mistake: Extending amphoteric hydroxide behaviour to all metal hydroxides - only AlX3+, ZnX2+, and PbX2+
Mistake: Describing QA observations vaguely - examiners expect precise colour terms such as "pale blue precipitate", "deep blue solution", or "red-brown precipitate", not "blue solid" or "coloured ppt".
Mistake: Confusing hydrolysis direction - salts of weak acids give basic solutions (anion hydrolyses to produce OHX−), while salts of weak bases give acidic solutions (cation hydrolyses to produce HX3OX+
Frequently asked questions
What is the difference between solubility and solubility product? Solubility is the maximum amount of solute that dissolves per litre (in mol⋅L−1). The solubility product Ksp is an equilibrium constant for the dissolution equilibrium and depends only on temperature. Solubility can change with pH or the presence of common ions even at constant temperature; Ksp cannot.
Which metal hydroxides are amphoteric at H2 level? The SEAB syllabus covers AlX3+, ZnX2+, and PbX2+ as amphoteric - their hydroxide precipitates dissolve in excess NaOH to form aluminate, zincate, and plumbate ions respectively. This behaviour does not apply to FeX3+, CuX2+, or MnX2+.
Why does adding a common ion reduce solubility? Adding a common ion increases the concentration of one of the product ions in the dissolution equilibrium, pushing the ionic product Qsp above Ksp. The system responds by shifting left - more solid precipitates - until equilibrium is re-established at a lower solubility.
How does complex ion formation increase solubility? When a ligand (such as excess NHX3) forms a stable complex with the dissolved metal ion, it effectively removes that ion from the equilibrium, decreasing the ionic product below Ksp. More solid dissolves to restore equilibrium, so overall solubility increases.
