Q: What does H2 Chemistry Notes: EXTENSION TOPICS, 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.
This extension topic examines how ions behave in water-precipitation, hydrolysis, and complex formation underpin analytical chemistry questions in Papers 2-4.
More generally, AXpBXq(s)-using the empirical formula with p and q in lowest terms-yields 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.
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.
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. For anions of weak acids (e.g. CHX3COOX−), hydrolysis yields a basic solution.
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.
5 Qualitative Analysis Links
Ion
Reagent
Observation
Explanation
\(\ce{Fe^{3+}}\)
\(\ce{NaOH}\)
Brown precipitate \(\ce{Fe(OH)3}\), insoluble in excess.
Precipitation due to low \(K_{sp}\); no ligand to redissolve.
\(\ce{Cu^{2+}}\)
\(\ce{NH3}\)
Light blue ppt dissolves in excess to give deep blue solution.
Formation of \(\ce{[Cu(NH3)4]^{2+}}\).
\(\ce{Al^{3+}}\)
\(\ce{NaOH}\)
White ppt dissolves in excess.
Amphoteric hydroxide forms \(\ce{[Al(OH)4]^-}\).
Remember to describe colours precisely; these observations are heavily tested.
6 Worked Example
Question:
Calculate the solubility of CaFX2 in pure water and in 0.10mol⋅L−1NaF, given Ksp(CaFX2)=3.9×10−11.
Solution:
In pure water:
Let molar solubility be SmolL−1:
[CaX2+]=S
[F−]=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 Zn2+ solution, explaining amphoteric behaviour.
Explain how addition of NHX4Cl affects the solubility of Mg(OH)X2
The aqueous solutions topic rewards systematic use of equilibria ideas. Consolidate these notes with structured lab practice, and revisit https://eclatinstitute.sg/blog/h2-chemistry-notes for supplementary drills.