11.1
Solubility Product Ksp
Solubility Product KspFor a sparingly soluble salt MᵋXᵘ ⇋ a Mⁿ⁺ + b Xᵗ⁻:
Kᵘᵔ = [Mⁿ⁺]ᵃ [Xᵗ⁻]ᵇ (units: molⁿ⁺ᵇ/Lⁿ⁺ᵇ)
Kᵘᵔ is constant at fixed temperature.
Common Ksp Values (25°C)
| Salt | Equilibrium | Kᵘᵔ |
|---|---|---|
| AgCl | Ag⁺ + Cl⁻ | 1.8 × 10⁻¹⁰ |
| AgBr | Ag⁺ + Br⁻ | 5.0 × 10⁻¹³ |
| AgI | Ag⁺ + I⁻ | 8.5 × 10⁻¹⁷ |
| CaCO₃ | Ca²⁺ + CO₃²⁻ | 3.4 × 10⁻⁹ |
| BaSO₄ | Ba²⁺ + SO₄²⁻ | 1.1 × 10⁻¹⁰ |
| PbSO₄ | Pb²⁺ + SO₄²⁻ | 2.5 × 10⁻⁸ |
| Ca₃(PO₄)₂ | 3Ca²⁺ + 2PO₄³⁻ | 2.1 × 10⁻³³ |
| Fe(OH)₃ | Fe³⁺ + 3OH⁻ | 2.8 × 10⁻³¹ |
Calculating Solubility from Kᵘᵔ
Example 1 — AgCl (1:1):
Kᵘᵔ = [Ag⁺][Cl⁻] = s² = 1.8×10⁻¹⁰
s = √(1.8×10⁻¹⁰) = 1.34×10⁻⁵ mol/L = 1.93×10⁻⁵ g/L (M=143.3)
Example 2 — Ca₃(PO₄)₂ (3:2):
Kᵘᵔ = [Ca²⁺]³[PO₄³⁻]² = (3s)³(2s)² = 108s⁵
s = (Kᵘᵔ/108)^(1/5)
11.2
Ion Product & Precipitation Criteria
Reaction Quotient Qᵘ
Qᵘ = ion product (calculated from actual concentrations, before equilibrium)
If Qᵘ < Kᵘᵔ → unsaturated solution (more solid can dissolve, no ppt)
If Qᵘ = Kᵘᵔ → saturated solution (equilibrium)
If Qᵘ > Kᵘᵔ → supersaturated → PRECIPITATION occurs
Example: Mix 50 mL of 2.0×10⁻⁴ mol/L AgNO₃ with 50 mL 2.0×10⁻⁴ mol/L NaCl.
After mixing: [Ag⁺] = [Cl⁻] = 1.0×10⁻⁴ mol/L (halved by dilution)
Q = [Ag⁺][Cl⁻] = (1.0×10⁻⁴)² = 1.0×10⁻⁸
Kᵘᵔ(AgCl) = 1.8×10⁻¹⁰
Q > Kᵘᵔ → AgCl PRECIPITATES (white ppt forms)
11.3
Common Ion Effect
Effect on Solubility
Adding a common ion suppresses solubility (Le Chatelier principle).
AgCl solubility in 0.1 mol/L NaCl:
Kᵘᵔ = [Ag⁺][Cl⁻] = s(s + 0.1) ≈ 0.1s (since s << 0.1)
s = Kᵘᵔ/0.1 = 1.8×10⁻¹⁰/0.1 = 1.8×10⁻⁹ mol/L
Compare: pure water s = 1.34×10⁻⁵ mol/L
Solubility reduced by factor ~10⁴ — dramatic suppression!
Applications:
• Washing precipitates with salt of common ion reduces peptisation
• Fractional precipitation exploits differences in Kᵘᵔ
11.4
Fractional Precipitation
Selective Precipitation
Different salts have different Kᵘᵔ → precipitate at different [anion].
To separate Cl⁻ and I⁻ with Ag⁺:
Kᵘᵔ(AgCl) = 1.8×10⁻¹⁰; Kᵘᵔ(AgI) = 8.5×10⁻¹⁷
AgI starts to ppt when: [Ag⁺] > Kᵘᵔ(AgI)/[I⁻] — at much lower [Ag⁺]
AgCl starts to ppt when [Ag⁺] is higher.
∴ Add Ag⁺ slowly → AgI ppts first (less soluble), then AgCl.
Good separation when Kᵘᵔ values differ by >10⁶.
Application in qualitative analysis: Group I precipitates (AgCl, AgBr, AgI, PbCl₂)
with dilute HCl — selectively precipitates Pb²⁺, Ag⁺, Hg₂²⁺
11.5
Effect of pH on Solubility
Solubility and pH
Salts of weak acids dissolve more readily in acid (H⁺ consumes the anion):
CaCO₃(s) ⇋ Ca²⁺ + CO₃²⁻
CO₃²⁻ + H⁺ → HCO₃⁻ (then HCO₃⁻ + H⁺ → H₂O + CO₂)
Le Chatelier: anion consumed → equilibrium shifts right → more dissolves
Metal hydroxides: solubility decreases as pH rises (more OH⁻ → common ion effect)
Fe(OH)₃ ⇋ Fe³⁺ + 3OH⁻
Adding NaOH (high pH): [OH⁻] increases → Q > Kᵘᵔ → precipitates more
Amphoteric hydroxides (Al(OH)₃, Zn(OH)₂):
Dissolve in acid: Al(OH)₃ + 3H⁺ → Al³⁺ + 3H₂O
Dissolve in base: Al(OH)₃ + OH⁻ → Al(OH)₄⁻ (aluminate)
Therefore maximum precipitation at intermediate pH (isoelectric point)
11.6
Complex Ion Formation and Solubility
Dissolution by Complexation
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AgCl is insoluble in water, but dissolves in NH₃(aq):
AgCl(s) ⇋ Ag⁺ + Cl⁻ Kᵘᵔ = 1.8×10⁻¹⁰
Ag⁺ + 2NH₃ ⇋ [Ag(NH₃)₂]⁺ Kᵮ = 1.6×10⁷ (stability constant)
Net: AgCl + 2NH₃ ⇋ [Ag(NH₃)₂]⁺ + Cl⁻ K = Kᵘᵔ×Kᵮ = 2.9×10⁻³
AgBr dissolves in conc. NH₃; AgI does NOT dissolve in NH₃ (Kᵘᵔ too small)
Applications:
• Qualitative analysis: AgCl dissolves in dilute NH₃ (distinguishes from AgBr/AgI)
• Photography: thiosulfate Na₂S₂O₃ dissolves unexposed AgBr:
AgBr + 2S₂O₃²⁻ → [Ag(S₂O₃)₂]³⁻ + Br⁻ (fixing)
✏️
Exercises
- Calculate the solubility of AgCl (Kᵘᵔ = 1.8×10⁻¹⁰) in (a) pure water; (b) 0.10 mol/L NaCl.(a) Kᵘᵔ = s² = 1.8×10⁻¹⁰. s = 1.34×10⁻⁵ mol/L. Mass solubility = 1.34×10⁻⁵ × 143.3 = 1.92×10⁻⁷ g/L. (b) Kᵘᵔ = s(0.10+s) ≈ 0.10s (s<<0.10). s = 1.8×10⁻¹⁰/0.10 = 1.8×10⁻⁹ mol/L. Factor of ~74,000 less soluble!
- Determine whether a precipitate forms when 100 mL of 2×10⁻³ mol/L BaCl₂ is mixed with 100 mL of 2×10⁻³ mol/L Na₂SO₄. Kᵘᵔ(BaSO₄) = 1.1×10⁻¹⁰.After mixing: [Ba²⁺] = [SO₄²⁻] = 10⁻³ mol/L (each halved). Q = (10⁻³)² = 10⁻⁶. Kᵘᵔ = 1.1×10⁻¹⁰. Q = 10⁻⁶ >> 1.1×10⁻¹⁰ → Q > Kᵘᵔ → BaSO₄ PRECIPITATES.
- A mixture contains 0.10 mol/L Cl⁻ and 0.10 mol/L I⁻. AgNO₃ is added slowly. Which precipitates first and why?AgI (Kᵘᵔ = 8.5×10⁻¹⁷) is much less soluble than AgCl (Kᵘᵔ = 1.8×10⁻¹⁰). AgI begins to ppt when [Ag⁺] > Kᵘᵔ(AgI)/[I⁻] = 8.5×10⁻¹⁷/0.10 = 8.5×10⁻¹⁶ mol/L. AgCl ppts when [Ag⁺] > 1.8×10⁻¹⁰/0.10 = 1.8×10⁻⁹ mol/L. AgI ppts first (at much lower [Ag⁺]), allowing separation.
- Explain why CaCO₃ scale in kettles dissolves when treated with vinegar (dilute CH₃COOH).CaCO₃(s) ⇋ Ca²⁺ + CO₃²⁻. Acetic acid protonates CO₃²⁻: CO₃²⁻ + 2H⁺ → H₂O + CO₂ (gas, escapes). Removal of CO₃²⁻ (Le Chatelier) shifts equilibrium right → more CaCO₃ dissolves. The acid protonation of the weak-acid anion effectively increases the solubility of the salt. Effervescence (CO₂) is observed.
- Calculate Kᵘᵔ of Pb(OH)₂ if its solubility in water is 6.7×10⁻⁶ mol/L.Pb(OH)₂ ⇋ Pb²⁺ + 2OH⁻. s = 6.7×10⁻⁶ mol/L; [Pb²⁺] = s; [OH⁻] = 2s. Kᵘᵔ = [Pb²⁺][OH⁻]² = s(2s)² = 4s³ = 4×(6.7×10⁻⁶)³ = 4×3.00×10⁻¹⁷ = 1.2×10⁻¹⁶.
- AgCl dissolves in NH₃ but AgI does not. Explain using equilibrium constants.AgCl + 2NH₃ ⇋ [Ag(NH₃)₂]⁺ + Cl⁻. K = Kᵘᵔ(AgCl) × Kf([Ag(NH₃)₂]⁺) = 1.8×10⁻¹⁰ × 1.6×10⁷ = 2.9×10⁻³ (significant → dissolves). For AgI: K = Kᵘᵔ(AgI) × Kf = 8.5×10⁻¹⁷ × 1.6×10⁷ = 1.4×10⁻¹⁰ (too small → AgI does not dissolve in NH₃). Kᵘᵔ(AgI) is so much smaller that even the very large stability constant Kf cannot compensate.
🧠
Quiz — 25 Questions
Unit 11: Solubility & Solubility Product
25 QsQ1
Kᵘᵔ for CaF₂ is 3.45×10⁻¹¹. Its molar solubility is:
CaF₂ ⇋ Ca²⁺ + 2F⁻. [Ca²⁺]=s, [F⁻]=2s. Kᵘᵔ=s(2s)²=4s³=3.45×10⁻¹¹. s=(3.45×10⁻¹¹/4)^(1/3)=2.05×10⁻⁴ mol/L.
Q2
Qᵘ > Kᵘᵔ means:
If Q > Kᵘᵔ: the ion product exceeds equilibrium → reaction proceeds left → precipitation occurs until equilibrium is re-established. Q = Kᵘᵔ = equilibrium (saturated). Q < Kᵘᵔ = unsaturated (more solid can dissolve).
Q3
The common ion effect reduces solubility because:
Adding common ion (e.g. Cl⁻ to AgCl solution): immediately Q > Kᵘᵔ → AgCl precipitates until Q = Kᵘᵔ again. The equilibrium [Ag⁺] (= s) is much lower when [Cl⁻] is already high: s = Kᵘᵔ/[Cl⁻]added. Much less AgCl dissolves.
Q4
BaSO₄ (Kᵘᵔ=1.1×10⁻¹⁰) is used as a medical contrast agent (barium meal) because:
Ba²⁺ is toxic, but BaSO₄ Kᵘᵔ = 1.1×10⁻¹⁰ → [Ba²⁺] in saturated solution ≈ 10⁻⁵ mol/L — far below toxic levels. BaSO₄ is opaque to X-rays, making the GI tract visible. Safe because essentially insoluble.
Q5
To separate Ag⁺ and Pb²⁺ from solution, you would add:
Group I qualitative analysis: dilute HCl precipitates Ag⁺ (as AgCl, white), Pb²⁺ (as PbCl₂, white, soluble in hot water), and Hg₂²⁺ (as Hg₂Cl₂, white). These are the only cations that precipitate with dilute HCl at room temperature.
Q6
Increasing pH (adding NaOH) decreases solubility of Fe(OH)₃ because:
Fe(OH)₃ ⇋ Fe³⁺ + 3OH⁻. Adding OH⁻ (high pH): [OH⁻] increases → Q > Kᵘᵔ → precipitates Fe(OH)₃. Solubility = s = Kᵘᵔ/[OH⁻]³ — decreases rapidly as pH increases. At pH 3: s is significant; at pH 7: nearly completely precipitated.
Q7
AgI does not dissolve in NH₃ but AgCl does. The reason is:
Net K = Kᵘᵔ × Kf. For AgCl: K=1.8×10⁻¹⁰×1.6×10⁷=2.9×10⁻³ (reaction goes forward). For AgI: K=8.5×10⁻¹⁷×1.6×10⁷=1.4×10⁻¹⁰ (too small, reaction barely proceeds). Kᵘᵔ(AgI) is 10⁷ smaller than Kᵘᵔ(AgCl) — complexation cannot overcome this.
Q8
CaCO₃ dissolves in acid because:
CO₃²⁻ + 2H⁺ → H₂O + CO₂ (g, escapes). Removing CO₃²⁻ shifts CaCO₃ ⇋ Ca²⁺ + CO₃²⁻ to the right → more dissolves. Effervescence (CO₂) observed. This is why limestone (CaCO₃) dissolves in acidic rain (acid rain) forming karst landscapes.
Q9
The Kᵘᵔ of Mg(OH)₂ = 5.6×10⁻¹². The solubility in pure water is:
Mg(OH)₂ ⇋ Mg²⁺ + 2OH⁻. [Mg²⁺]=s; [OH⁻]=2s. Kᵘᵔ=4s³=5.6×10⁻¹². s=(5.6×10⁻¹²/4)^(1/3) = (1.4×10⁻¹²)^(1/3) = 1.12×10⁻⁴ mol/L.
Q10
In photography, sodium thiosulfate (‘hypo’) fixes film by:
Na₂S₂O₃ + AgBr: Ag⁺ + 2S₂O₃²⁻ → [Ag(S₂O₃)₂]³⁻. This water-soluble complex dissolves unexposed silver halide, removing it from the film/print and making the image permanent. Without fixing, unexposed AgBr would slowly darken in light.
Q11
Fractional precipitation is most effective when:
Large difference in Kᵘᵔ → large difference in [anion] at which each precipitates → good separation. If Kᵘᵔ values are similar, both start precipitating at similar [anion] → co-precipitation → poor separation.
Q12
A solution of Kᵘᵔ(BaSO₄)=1.1×10⁻¹⁰ means that at equilibrium, [Ba²⁺][SO₄²⁻]:
Kᵘᵔ applies at equilibrium (saturated solution in contact with excess solid at constant T). In unsaturated solutions: Q < Kᵘᵔ. Adding other ions (common ion effect) changes individual concentrations but the product of ion concentrations still equals Kᵘᵔ at equilibrium.
Q13
Which precipitate is most suitable for gravimetric analysis of Ba²⁺?
BaSO₄ ideal for gravimetric analysis: (1) Kᵘᵔ very small → >99.999% precipitation; (2) stable, non-hygroscopic; (3) not reduced by heat; (4) pure white. Compare: BaCO₃ dissolves in acid; Ba(OH)₂ is too soluble; BaCl₂ is highly soluble.
Q14
Amphoteric hydroxides like Al(OH)₃ are precipitated at:
Al(OH)₃ dissolves in acid (Al(OH)₃+3H⁺→Al³⁺+3H₂O) and in excess NaOH (Al(OH)₃+OH⁻→Al(OH)₄⁻). Maximum precipitation at intermediate pH ~5–7. This amphoteric behaviour is used in water treatment to coagulate particles.
Q15
Increasing temperature generally:
Dissolution of most ionic salts is endothermic. Higher T favours the endothermic process → Kᵘᵔ increases → more soluble. Exceptions: some salts (e.g. Li₂SO₄, Ce₂(SO₄)₃) have exothermic dissolution → less soluble at higher T (Kᵘᵔ decreases).
Q16
Which statement about Kᵘᵔ is CORRECT?
Kᵘᵔ is a thermodynamic equilibrium constant — depends only on temperature, not on the presence of other ions or common ions. It does not include [solid] (activity of pure solid = 1). Units vary: AgCl has units (mol/L)²; Ca₃(PO₄)₂ has units (mol/L)⁵.
Q17
If s is the molar solubility of Ag₂CrO₄ (Kᵘᵔ=1.12×10⁻¹²):
Ag₂CrO₄ ⇋ 2Ag⁺ + CrO₄²⁻. [Ag⁺]=2s; [CrO₄²⁻]=s. Kᵘᵔ=(2s)²(s)=4s³=1.12×10⁻¹². s=(1.12×10⁻¹²/4)^(1/3)=(2.8×10⁻¹³)^(1/3)=0.014 mol/L.
Q18
When qualitative analysis uses Group I: adding dilute HCl precipitates:
Group I cations: Ag⁺ (AgCl, white), Pb²⁺ (PbCl₂, white), Hg₂²⁺ (Hg₂Cl₂, white). All other common metal chlorides are soluble. The group separation uses dilute HCl to precipitate only these three.
Q19
Solubility of Ag₂S is extremely low (Kᵘᵔ~6×10⁻⁵⁰). This is used in:
Group II/III in qualitative analysis: H₂S at controlled pH precipitates metal sulfides selectively. Ag₂S (black) precipitates at any pH. pH controls [S²⁻] (H₂S ⇋ 2H⁺ + S²⁻) → selective precipitation of CuS, PbS (low Kᵘᵔ) before ZnS, FeS (higher Kᵘᵔ).
Q20
The ion product Q is:
Q can be calculated at any instant using actual concentrations (even non-equilibrium). Comparing Q with Kᵘᵔ predicts direction: Q < Kᵘᵔ (dissolves more); Q = Kᵘᵔ (equilibrium); Q > Kᵘᵔ (precipitates).
Q21
The molar solubility of PbCl₂ (Kᵘᵔ=1.2×10⁻⁵) in 0.20 mol/L NaCl is:
PbCl₂ ⇋ Pb²⁺ + 2Cl⁻. [Cl⁻] ≈ 0.20 (common ion, s<<0.20). [Pb²⁺]=s. Kᵘᵔ = s(0.20)² = 0.040s. s = 1.2×10⁻⁵/0.040 = 3.0×10⁻⁴ mol/L. Compare: in pure water s=(Kᵘᵔ/4)^(1/3)=(3×10⁻⁶)^(1/3)=6.7×10⁻² mol/L — 220× less soluble in 0.20 mol/L NaCl.
Q22
To dissolve CaF₂ in the laboratory, you would use:
F⁻ is the conjugate base of HF (pKa=3.2). In acid: H⁺ + F⁻ → HF. Removing F⁻ drives CaF₂ ⇋ Ca²⁺ + 2F⁻ to the right. However: HF is very corrosive — handle carefully. Same principle as CaCO₃ + acid, but forming HF instead of CO₂.
Q23
Which of these is MOST soluble?
For 1:1 salts: s=√Kᵘᵔ. Largest Kᵘᵔ = most soluble. AgCl (Kᵘᵔ = 1.8×10⁻¹⁰) is most soluble. Order: AgCl >> AgBr >> AgI >> Ag₂S. Reflected in colour: AgCl=white, AgBr=pale yellow, AgI=yellow (darker = less soluble).
Q24
At constant temperature, adding more solid CaCO₃ to a saturated CaCO₃ solution:
At equilibrium, adding more pure solid does not change concentrations. The solid activity = 1 (pure solid, fixed). Ion concentrations stay at their equilibrium values. Kᵘᵔ is temperature-dependent only. No change in [Ca²⁺] or [CO₃²⁻].
Unit 11 Quiz — Solubility Product (25 Questions)
Select one answer eachQ1
The solubility product (Ksp) of AgCl is defined as:
For AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq): Ksp = [Ag⁺][Cl⁻]. Solid AgCl is not included (activity = 1).
Q2
If the solubility of BaSO₄ is s mol/L, then Ksp(BaSO₄) =
BaSO₄ ⇌ Ba²⁺ + SO₄²⁻. [Ba²⁺] = [SO₄²⁻] = s. Ksp = s × s = s².
Q3
Precipitation occurs when:
If Q > Ksp, the solution is supersaturated and a precipitate forms until Q = Ksp (equilibrium).
Q4
The common ion effect on solubility means:
Adding Cl⁻ to AgCl solution shifts AgCl ⇌ Ag⁺ + Cl⁻ left → more AgCl precipitates → lower solubility of AgCl.
Q5
Ksp of Ca₃(PO₄)₂ (which gives 3 Ca²⁺ and 2 PO₄³⁻) is:
Ca₃(PO₄)₂ ⇌ 3Ca²⁺ + 2PO₄³⁻. Ksp = [Ca²⁺]³[PO₄³⁻]².
Q6
If Ksp(AgCl) = 1.8×10⁻¹⁰, the molar solubility of AgCl is:
s² = 1.8×10⁻¹⁰ → s = √(1.8×10⁻¹⁰) ≈ 1.3×10⁻⁵ mol/L.
Q7
Fractional precipitation is used to:
When two ions have very different Ksp values for their precipitates, one can be selectively removed by adding just enough precipitating agent.
Q8
CaCO₃ dissolves in excess HCl because:
CO₃²⁻ + 2H⁺ → CO₂ + H₂O. This removes CO₃²⁻ from solution, dissolving the precipitate (Le Chatelier).
Q9
Kidney stones (calcium oxalate) form when:
High [Ca²⁺] and [C₂O₄²⁻] in urine → Q exceeds Ksp → CaC₂O₄ precipitates → kidney stones form.
Q10
Qualitative analysis uses selective precipitation to identify:
H₂S in acid precipitates Cu²⁺, Pb²⁺, etc. (group II); in alkali precipitates Zn²⁺, Ni²⁺ etc. — group separations.
Q11
Amphoteric hydroxides like Al(OH)₃ dissolve in:
Al(OH)₃ + 3H⁺ → Al³⁺ + 3H₂O (in acid); Al(OH)₃ + OH⁻ → Al(OH)₄⁻ (aluminate, in excess base).
Q12
The solubility of CaCO₃ increases in acidic solution because:
Lower [CO₃²⁻] means Q falls below Ksp → more CaCO₃ dissolves. Acidic caves form stalactites/stalagmites this way.
Q13
The solubility product of AgBr is 5×10⁻¹³. If [Ag⁺] = 0.1 mol/L, what is the maximum [Br⁻] before precipitation?
[Br⁻] < Ksp/[Ag⁺] = 5×10⁻¹³/0.1 = 5×10⁻¹² mol/L. Any higher causes precipitation.
Q14
Formation of a complex ion can dissolve a precipitate by:
AgCl + 2NH₃ → [Ag(NH₃)₂]⁺ + Cl⁻. Complexation removes Ag⁺ → Q falls below Ksp → AgCl dissolves.
Q15
An insoluble salt with a larger Ksp is:
For salts with the same ion ratio (e.g. both 1:1), larger Ksp → larger s. Be careful comparing salts with different stoichiometry.
Q16
The pH at which metal hydroxide M(OH)₂ begins to precipitate depends on:
M(OH)₂ ⇌ M²⁺ + 2OH⁻. Ksp = [M²⁺][OH⁻]². Precipitation starts when [OH⁻]² > Ksp/[M²⁺].
Q17
Supersaturation refers to:
A supersaturated solution has more dissolved solute than the equilibrium value. It is metastable — precipitation can occur with a nucleation site.
Q18
Which compound has the lowest molar solubility given these Ksp values: AgCl=1.8×10⁻¹⁰, AgBr=5×10⁻¹³, AgI=8.5×10⁻¹⁷?
For 1:1 salts, s = √Ksp. AgI has the smallest Ksp → smallest s → least soluble.
Q19
Toothpaste uses fluoride to protect teeth because:
Ca₁₀(PO₄)₆(OH)₂ (hydroxyapatite) → Ca₁₀(PO₄)₆F₂ (fluorapatite). Fluorapatite is less soluble in acid → resists decay.
Q20
The ion product Q is calculated the same way as Ksp but:
Q uses actual [ions] at any point. Comparing Q to Ksp predicts direction: Q < Ksp (unsaturated, dissolve), Q > Ksp (precipitate).
Q21
Stalagmites and stalactites form by:
Ca(HCO₃)₂ → CaCO₃↓ + CO₂ + H₂O as CO₂ pressure drops in cave air. Slow precipitation builds limestone formations.
Q22
Selective precipitation of Cl⁻ in the presence of Br⁻ uses AgNO₃ because:
Ksp(AgCl)=1.8×10⁻¹⁰ > Ksp(AgBr)=5×10⁻¹³. AgBr precipitates at lower [Ag⁺]. So adding slow AgNO₃ precipitates Br⁻ first.
Q23
Hard water deposits (scale) in kettles form by:
Solubility of CaCO₃ decreases with temperature. Heating hard water → Q > Ksp → CaCO₃ scale deposits.
Q24
Gravimetric analysis relies on:
Analyte precipitated as insoluble compound → filtered, dried, weighed. Mass of precipitate → moles of analyte.
Q25
The solubility of Mg(OH)₂ in a buffer at pH 10 ([OH⁻] = 10⁻⁴) given Ksp = 5.6×10⁻¹²:
Ksp = [Mg²⁺][OH⁻]². [Mg²⁺] = Ksp/[OH⁻]² = 5.6×10⁻¹²/(10⁻⁴)² = 5.6×10⁻¹²/10⁻⁸ = 5.6×10⁻⁴ mol/L.
📝
Unit Test — 50 marks
Section A
30 marksQ1 [5]
Write Kᵘᵔ expressions for: (a) Ag₂SO₄; (b) Fe(OH)₃; (c) Ca₃(PO₄)₂. Calculate the molar solubility of Ag₂SO₄ if Kᵘᵔ = 1.2×10⁻⁵.
(a) Kᵘᵔ=[Ag⁺]²[SO₄²⁻]; (b) Kᵘᵔ=[Fe³⁺][OH⁻]³; (c) Kᵘᵔ=[Ca²⁺]³[PO₄³⁻]². Ag₂SO₄⇋2Ag⁺+SO₄²⁻: [Ag⁺]=2s, [SO₄²⁻]=s. Kᵘᵔ=(2s)²s=4s³=1.2×10⁻⁵. s=(3×10⁻¹⁰)^(1/3)=6.7×10⁻⁴ mol/L.
Q2 [5]
Will a precipitate form when 50 mL 0.010 mol/L CaCl₂ is mixed with 50 mL 0.010 mol/L Na₂SO₄? Kᵘᵔ(CaSO₄)=4.93×10⁻⁵.
After mixing: [Ca²⁺]=[SO₄²⁻]=0.010/2=0.005 mol/L. Q=(0.005)²=2.5×10⁻⁵. Kᵘᵔ=4.93×10⁻⁵. Q=2.5×10⁻⁵>4.93×10⁻⁵... actually Q
Q3 [5]
A mixture contains 0.050 mol/L Cl⁻ and 0.050 mol/L Br⁻. AgNO₃ is added drop by drop. Calculate [Ag⁺] at which each ion begins to precipitate. What is [Cl⁻] remaining when AgBr just begins to precipitate? Kᵘᵔ(AgCl)=1.8×10⁻¹⁰; Kᵘᵔ(AgBr)=5.0×10⁻¹³.
AgBr ppts when [Ag⁺]>Kᵘᵔ(AgBr)/[Br⁻]=5.0×10⁻¹³/0.050=1.0×10⁻¹¹ mol/L. AgCl ppts when [Ag⁺]>1.8×10⁻¹⁰/0.050=3.6×10⁻⁹ mol/L. AgBr ppts FIRST (at much lower [Ag⁺]). When AgBr starts pptting: [Ag⁺]=1.0×10⁻¹¹. [Cl⁻]remaining=Kᵘᵔ(AgCl)/[Ag⁺]=1.8×10⁻¹⁰/1.0×10⁻¹¹=0.018 mol/L (96.4% Cl⁻ still in solution). Excellent separation!
Q4 [5]
Explain how pH control is used in qualitative analysis to separate metal ions as hydroxides. Include Al³⁺ and Fe³⁺. Kᵘᵔ[Al(OH)₃]=1.3×10⁻³²; Kᵘᵔ[Fe(OH)₃]=2.8×10⁻³¹.
Fe(OH)₃ ppts at lower pH than Al(OH)₃ (Kᵘᵔ smaller). To ppt Fe(OH)₃: [OH⁻]=( Kᵘᵔ/[Fe³⁺])^(1/3)=(2.8×10⁻³¹/0.01)^(1/3)=6.5×10⁻¹⁰ mol/L → pH=4.8. Al(OH)₃ begins pptting at [OH⁻]=(1.3×10⁻³²/0.01)^(1/3)=5×10⁻¹¹ → pH=3+3=... at pH~4.0 for Al³⁺. By carefully controlling pH to ~4–6, Fe(OH)₃ (Kᵘᵔ smaller) ppts selectively while Al³⁺ remains in solution, allowing separation.
Q5 [5]
The solubility of BaSO₄ increases slightly in the presence of EDTA (ethylenediaminetetraacetate). Explain this observation and write a general equation to show the principle.
EDTA is a strong chelating agent (hexadentate ligand) that forms extremely stable complexes with metal ions, including Ba²⁺: Ba²⁺ + EDTA⁴⁻ → [Ba-EDTA]²⁻ (stable complex, Kf~large). This removes Ba²⁺ from solution, reducing [Ba²⁺]. By Le Chatelier, BaSO₄ ⇋ Ba²⁺ + SO₄²⁻ shifts right → more BaSO₄ dissolves. Net: BaSO₄ + EDTA⁴⁻ ⇋ [Ba-EDTA]²⁻ + SO₄²⁻. K=Kᵘᵔ×Kf. Same principle as NH₃ dissolving AgCl by forming [Ag(NH₃)₂]⁺.
Q6 [5]
Derive an expression for the solubility of a 2:3 salt M₂X₃ in pure water, where M has charge +3 and X has charge -2. Calculate s if Kᵘᵔ = 1.0×10⁻³⁰.
M₂X₃ ⇋ 2M³⁺ + 3X²⁻. [M³⁺]=2s; [X²⁻]=3s. Kᵘᵔ=(2s)²(3s)³=4s²×27s³=108s⁵=1.0×10⁻³⁰. s=(1.0×10⁻³⁰/108)^(1/5)=(9.26×10⁻³³)^(1/5)=(9.26)^(1/5)×10⁻⁶·⁶=1.56×10⁻⁷ mol/L. [Compare Ca₃(PO₄)₂ type.]
Section B
20 marksQ7 [10]
(a) Explain the concept of selective/fractional precipitation with reference to AgCl and AgI. [4] (b) A solution contains 0.10 mol/L Ag⁺ and 0.10 mol/L Pb²⁺. Is it possible to selectively precipitate PbSO₄ without significantly precipitating Ag₂SO₄? Kᵘᵔ(PbSO₄)=2.5×10⁻⁸; Kᵘᵔ(Ag₂SO₄)=1.2×10⁻⁵. [6]
(a) Selective precipitation exploits large differences in Kᵘᵔ. AgI (Kᵘᵔ=8.5×10⁻¹⁷) is much less soluble than AgCl (Kᵘᵔ=1.8×10⁻¹⁰). Adding Ag⁺: AgI starts pptting when [Ag⁺]>Kᵘᵔ(AgI)/[I⁻]=8.5×10⁻¹⁷/0.1=8.5×10⁻¹⁶ mol/L. AgCl ppts when [Ag⁺]>1.8×10⁻¹⁰/0.1=1.8×10⁻⁹ mol/L. Ratio = 1.8×10⁻⁹/8.5×10⁻¹⁶=210,000. When Ag⁺ reaches 1.8×10⁻⁹ mol/L (AgCl starts pptting), [I⁻]=Kᵘᵔ(AgI)/1.8×10⁻⁹=4.7×10⁻⁵ mol/L — 99.995% of I⁻ already precipitated as AgI. Excellent separation possible.
(b) PbSO₄ ppts when [SO₄²⁻]>Kᵘᵔ/[Pb²⁺]=2.5×10⁻⁸/0.10=2.5×10⁻⁷ mol/L. Ag₂SO₄ ppts when [SO₄²⁻]>Kᵘᵔ/[Ag⁺]²=1.2×10⁻⁵/(0.10)²=1.2×10⁻⁷ mol/L. Ratio: 1.2×10⁻⁷/2.5×10⁻⁷=4.8. When [SO₄²⁻]=2.5×10⁻⁷, PbSO₄ just starts pptting. At this point: [Ag⁺]²=Kᵘᵔ(Ag₂SO₄)/[SO₄²⁻]=1.2×10⁻⁵/2.5×10⁻⁷=4.8×10⁻³; [Ag⁺]=0.069 mol/L. So when PbSO₄ begins pptting, [Ag⁺]=0.069 vs initial 0.10 — 31% of Ag⁺ has already precipitated! Separation is NOT clean enough (30% co-precipitation). The Kᵘᵔ ratio is only 4.8 — too small for good separation.
Q8 [10]
Discuss the importance of solubility product in three areas: (a) water treatment; (b) kidney stones; (c) qualitative analysis. For each, include relevant equilibria and practical implications. [10]
(a) Water treatment: hardness removal (Ca²⁺ and Mg²⁺): CaCO₃ (Kᵘᵔ=3.4×10⁻⁹) and Mg(OH)₂ (Kᵘᵔ=5.6×10⁻¹²) are precipitated by adding Ca(OH)₂ (lime) and Na₂CO₃ (soda). CaO/Na₂CO₃ process: Ca²⁺+CO₃²⁻→CaCO₃↓; Mg²⁺+2OH⁻→Mg(OH)₂↓. Precipitates removed by settling/filtration. Fluoridation: NaF added to achieve [F⁻]=1 ppm (prevents tooth decay); CaF₂ (Kᵘᵔ=3.45×10⁻¹¹) precipitation on tooth enamel strengthens it.
(b) Kidney stones: most are calcium oxalate (CaC₂O₄, Kᵘᵔ=2.7×10⁻⁹). Formed when urinary [Ca²⁺][C₂O₄²⁻] > Kᵘᵔ. Risk factors: dehydration (increases concentrations), excess dietary Ca or oxalate, low citrate (citrate complexes Ca²⁺, raising Kᵘᵔeff). Prevention: drink 2+ L water/day (dilutes ions, keeps Q
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