Checkpoint 4 — IQ4: Ksp, Qsp & Common Ion | Chem Y12 M5 | HSC Chemistry Year 12 Module 5

What's Covered

L15

Dissolution Equilibria

Lattice & hydration energy
Exo vs endo dissolution
Saturated solutions
ATSI cycad chemistry

L16

Solubility Rules & Ppt

NAGSAG framework
Four-ion method
AgCl, PbI₂, BaSO₄
Net ionic equations

L17

Ksp Expressions

Writing Ksp expressions
Molar solubility from Ksp
Ksp from solubility
Comparison trap (formula type)

L18

Qsp & Common Ion

Qsp vs Ksp decision rule
Critical dilution step
Common ion reduces solubility
Kidney stones & applications

Section A — Multiple Choice (10 questions)

Question 1

A saturated solution of AgCl is described as a dynamic equilibrium: $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$. Which statement is correct?

A In a saturated solution, dissolution has stopped completely

B In a saturated solution, the rate of dissolution equals the rate of crystallisation, so [Ag⁺] and [Cl⁻] remain constant while AgCl crystals continue to dissolve and recrystallise

C An unsaturated solution is at equilibrium because it has dissolved all available AgCl

D More AgCl dissolves at equilibrium if pressure is increased

Correct: B. A saturated solution is a dynamic equilibrium: solid is continuously dissolving (at a rate proportional to surface area) and ions are continuously re-crystallising onto the solid. These two rates are equal, so macroscopic properties (concentrations) remain constant while processes continue at the molecular level. Option A describes a static situation (wrong). Option C is wrong — an unsaturated solution has no undissolved solid present, so there is no equilibrium. Option D is wrong — pressure has negligible effect on solution equilibria.

Question 2

Which set of ions, when mixed, will produce a precipitate? (Use NAGSAG: Nitrates Always, Group 1 Always, Sulfates Always except Ba²⁺/Pb²⁺, Ag⁺ precipitates halides, Carbonates precipitate except Group 1)

A Na⁺(aq) + Cl⁻(aq)

B K⁺(aq) + SO₄²⁻(aq)

C Ba²⁺(aq) + SO₄²⁻(aq)

D NH₄⁺(aq) + NO₃⁻(aq)

Correct: C. BaSO₄ is insoluble — a key exception to the "sulfates are generally soluble" rule. Ba²⁺ and Pb²⁺ form insoluble sulfates. Option A: NaCl is soluble (Group 1 cation and chloride — both generally soluble without a precipitating partner). Option B: K₂SO₄ is soluble (Group 1 cation). Option D: NH₄NO₃ is soluble (ammonium and nitrate are always soluble).

Question 3

Write the correct Ksp expression for lead(II) iodide: $\text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq)$

A $K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2$

B $K_{sp} = [\text{Pb}^{2+}][\text{I}^-]$

C $K_{sp} = \dfrac{[\text{Pb}^{2+}][\text{I}^-]^2}{[\text{PbI}_2]}$

D $K_{sp} = [\text{Pb}^{2+}]^2[\text{I}^-]$

Correct: A. Ksp = [Pb²⁺][I⁻]². The coefficient 2 in front of I⁻ in the dissolution equation becomes the power 2 on [I⁻] in the Ksp expression. PbI₂ is a solid and is excluded from the expression. Option B omits the square. Option C incorrectly includes the solid (which has constant "concentration" and is excluded). Option D reverses the powers.

Question 4

The molar solubility of AgCl is 1.34 × 10⁻⁵ mol/L. What is Ksp?

A 1.34 × 10⁻⁵

B 2.68 × 10⁻⁵

C 1.79 × 10⁻¹⁰ × 2 = 3.58 × 10⁻¹⁰

D $K_{sp} = s^2 = (1.34 \times 10^{-5})^2 = 1.8 \times 10^{-10}$

Correct: D. For AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), if molar solubility = s, then [Ag⁺] = s and [Cl⁻] = s. Ksp = [Ag⁺][Cl⁻] = s × s = s² = (1.34 × 10⁻⁵)² = 1.8 × 10⁻¹⁰. Option A just gives s, not Ksp. Option B gives 2s (the total dissolved ions, not Ksp). Option C incorrectly multiplies by 2 — this isn't a 1:2 type compound.

Question 5

⚠️ TRAP QUESTION: CaF₂ has Ksp = 3.9 × 10⁻¹¹ and AgCl has Ksp = 1.8 × 10⁻¹⁰. Which compound is more soluble?

A AgCl, because it has a larger Ksp

B CaF₂, because despite having a smaller Ksp, its 1:2 formula type means molar solubility s = ∛(Ksp/4) ≈ 2.14 × 10⁻⁴ mol/L, which is greater than AgCl's s = √Ksp ≈ 1.34 × 10⁻⁵ mol/L

C They have equal solubility because both have similar Ksp values

D Neither dissolves appreciably; both have very small Ksp values

Correct: B. This is the most common Ksp misconception. You CANNOT directly compare Ksp values when the compounds have different formula types. CaF₂ (1:2 type): Ksp = 4s³, so s = ∛(3.9×10⁻¹¹/4) = ∛(9.75×10⁻¹²) ≈ 2.14 × 10⁻⁴ mol/L. AgCl (1:1 type): Ksp = s², so s = √(1.8×10⁻¹⁰) ≈ 1.34 × 10⁻⁵ mol/L. CaF₂ is about 16× MORE soluble than AgCl despite having a smaller Ksp. Direct comparison is only valid for compounds of the same formula type.

Question 6

10.0 mL of 0.020 mol/L Pb(NO₃)₂ is mixed with 10.0 mL of 0.020 mol/L Na₂SO₄. Ksp(PbSO₄) = 2.5 × 10⁻⁸. Will a precipitate form?

A No precipitate; Qsp = [Pb²⁺][SO₄²⁻] = (0.020)(0.020) = 4.0 × 10⁻⁴ < Ksp

B Yes; adding the solutions does not change the concentrations, so Qsp = 4.0 × 10⁻⁴ > Ksp

C Yes; after mixing (total volume = 20.0 mL), [Pb²⁺] = [SO₄²⁻] = 0.010 mol/L, so Qsp = (0.010)(0.010) = 1.0 × 10⁻⁴ > Ksp = 2.5 × 10⁻⁸

D Cannot determine without temperature data

Correct: C. Critical step: concentrations are HALVED upon mixing (each solution doubles in volume from 10 mL to 20 mL total). [Pb²⁺] = 0.020 × 10.0/20.0 = 0.010 mol/L. [SO₄²⁻] = 0.020 × 10.0/20.0 = 0.010 mol/L. Qsp = (0.010)(0.010) = 1.0 × 10⁻⁴. Since Qsp = 1.0 × 10⁻⁴ > Ksp = 2.5 × 10⁻⁸, a precipitate forms. Option A uses pre-mixing concentrations (forgetting dilution). Option B also forgets dilution but gets the precipitate conclusion right by accident.

Question 7

The common ion effect is observed when NaCl is added to a saturated AgCl solution. Which correctly describes what happens?

A Adding NaCl increases [Cl⁻], which increases Qsp above Ksp. The equilibrium shifts left, causing more AgCl to precipitate and [Ag⁺] to decrease to a very low value

B Adding NaCl has no effect because NaCl and AgCl are different compounds

C Adding NaCl increases the solubility of AgCl because more ions are present in solution

D Adding NaCl decreases [Cl⁻] in the solution, allowing more AgCl to dissolve

Correct: A. The Cl⁻ ion is common to both NaCl and AgCl. Adding NaCl introduces extra Cl⁻, raising [Cl⁻] so that Qsp = [Ag⁺][Cl⁻] > Ksp. To reduce Qsp back to Ksp, the equilibrium shifts left — more AgCl precipitates and [Ag⁺] decreases dramatically. The solubility of AgCl is greatly reduced. This is the common ion effect — a shared ion suppresses solubility. Options B, C, and D all fail to account for the shared Cl⁻ ion.

Question 8

For the reaction $\text{Ag}_2\text{SO}_4(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{SO}_4^{2-}(aq)$, the molar solubility is s = 1.5 × 10⁻² mol/L. What is Ksp?

A Ksp = s³ = (1.5 × 10⁻²)³ = 3.4 × 10⁻⁶

B Ksp = (1.5 × 10⁻²)² = 2.25 × 10⁻⁴

C Ksp = 2s × s = (3.0 × 10⁻²)(1.5 × 10⁻²) = 4.5 × 10⁻⁴

D $K_{sp} = (2s)^2 \times s = 4s^3 = 4 \times (1.5 \times 10^{-2})^3 = 1.35 \times 10^{-5}$

Correct: D. For Ag₂SO₄: each formula unit dissolves to give 2 Ag⁺ and 1 SO₄²⁻. If molar solubility = s, then [Ag⁺] = 2s and [SO₄²⁻] = s. Ksp = [Ag⁺]²[SO₄²⁻] = (2s)² × s = 4s³ = 4 × (1.5 × 10⁻²)³ = 4 × 3.375 × 10⁻⁶ = 1.35 × 10⁻⁵. Option A uses s³ (misses the coefficient of 4). Option B uses s² (wrong formula). Option C uses 2s × s without squaring the 2s.

Question 9

Traditional cycad seed preparation by Aboriginal and Torres Strait Islander peoples involves leaching seeds in running water over several days. Which chemical principle is best applied to explain why running water is more effective than still water?

A Running water has a higher temperature, increasing the rate of dissolution

B Running water breaks down toxins by chemical reactions with oxygen

C Running water continuously removes dissolved toxins, maintaining a low concentration of toxins in the surrounding water; by Le Chatelier's Principle (LCP), the equilibrium shifts to dissolve more toxin from the seed

D Running water provides more water molecules, which increases the total amount of dissolution possible

Correct: C. The toxins (cycasin, macrozamin) are in equilibrium between the seed and the surrounding water. In still water, the surrounding water eventually reaches equilibrium with the seed — dissolution stops. Running water continuously removes the dissolved toxins (reducing their concentration in the water), which shifts the equilibrium right (LCP) to dissolve more toxin from the seed. This is the LCP-based explanation required by NESA. Option A is incorrect — temperature isn't the key factor. Option D confuses quantity of water with the equilibrium principle.

Question 10

When writing a net ionic equation for the precipitation of lead(II) iodide from Pb(NO₃)₂ and KI solutions, which ionic equation is correct?

A Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)

B Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s)

C Pb²⁺(aq) + I⁻(aq) → PbI(s)

D Pb²⁺(aq) + 2I⁻(aq) + 2K⁺(aq) + 2NO₃⁻(aq) → PbI₂(s) + 2K⁺(aq) + 2NO₃⁻(aq)

Correct: B. The net ionic equation shows only the ions that participate in forming the precipitate — spectator ions (K⁺ and NO₃⁻) are cancelled out. Pb²⁺ combines with 2 I⁻ (stoichiometry from PbI₂) to form the insoluble PbI₂(s). Option A is the molecular equation (not ionic). Option C has wrong stoichiometry (charge doesn't balance: 2+ ≠ 1−). Option D is the full ionic equation — it includes spectator ions and is not yet simplified to the net ionic equation.

Section B — Short Answer

Question 11

The molar solubility of PbSO₄ is 1.5 × 10⁻⁴ mol/L at 25°C. (a) Write the dissolution equation and Ksp expression for PbSO₄. (b) Calculate Ksp. (c) A solution contains Pb²⁺ at 1.0 × 10⁻⁴ mol/L and SO₄²⁻ at 3.0 × 10⁻⁴ mol/L. Will a precipitate form? (4 marks)

4 marks

Model Answer (4 marks):

(a) Equation and Ksp expression (1 mark): $\text{PbSO}_4(s) \rightleftharpoons \text{Pb}^{2+}(aq) + \text{SO}_4^{2-}(aq)$. $K_{sp} = [\text{Pb}^{2+}][\text{SO}_4^{2-}]$

(b) Calculate Ksp (1 mark): For PbSO₄ (1:1 type), s = [Pb²⁺] = [SO₄²⁻] = 1.5 × 10⁻⁴ mol/L. $K_{sp} = s^2 = (1.5 \times 10^{-4})^2 = 2.25 \times 10^{-8} \approx 2.3 \times 10^{-8}$

(c) Precipitate prediction (2 marks): Calculate Qsp: $Q_{sp} = [\text{Pb}^{2+}][\text{SO}_4^{2-}] = (1.0 \times 10^{-4})(3.0 \times 10^{-4}) = 3.0 \times 10^{-8}$ (1 mark). Compare: Qsp = 3.0 × 10⁻⁸ > Ksp = 2.3 × 10⁻⁸. Since Qsp > Ksp, the solution is supersaturated and a precipitate of PbSO₄ WILL form (1 mark).

Question 12

15.0 mL of 0.0400 mol/L AgNO₃ is mixed with 25.0 mL of 0.0200 mol/L NaCl. Ksp(AgCl) = 1.8 × 10⁻¹⁰. (a) Calculate the concentration of each ion after mixing but before any reaction. (b) Calculate Qsp. (c) Predict whether a precipitate forms, and if so, identify it. (5 marks)

5 marks

Model Answer (5 marks):

⚠️ Key step: Always dilute concentrations first using $c_{new} = c_{original} \times \frac{V_{original}}{V_{total}}$

(a) Concentrations after mixing (2 marks): Total volume = 15.0 + 25.0 = 40.0 mL (1 mark for this step).

[Ag⁺] = 0.0400 × 15.0/40.0 = 0.0150 mol/L

[Cl⁻] = 0.0200 × 25.0/40.0 = 0.0125 mol/L (1 mark for both dilutions)

(b) Calculate Qsp (1 mark): $Q_{sp} = [\text{Ag}^+][\text{Cl}^-] = (0.0150)(0.0125) = 1.875 \times 10^{-4}$

(c) Prediction and identification (2 marks): Qsp = 1.875 × 10⁻⁴ >> Ksp = 1.8 × 10⁻¹⁰. Since Qsp > Ksp, the solution is highly supersaturated (1 mark). A white precipitate of AgCl(s) will form as Ag⁺ and Cl⁻ ions combine (1 mark).

Question 13

A saturated solution of CaF₂ (Ksp = 3.9 × 10⁻¹¹) has molar solubility s = 2.14 × 10⁻⁴ mol/L. (a) Show how s was calculated from Ksp. (b) If NaF is added to make [F⁻] = 0.100 mol/L, calculate the new molar solubility of CaF₂. (c) Explain the common ion effect observed. (5 marks)

5 marks

Model Answer (5 marks):

(a) Calculating s (2 marks): $\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq)$. ICE: [Ca²⁺] = s, [F⁻] = 2s (1 mark). $K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = s \times (2s)^2 = 4s^3$. $s = \sqrt[3]{\dfrac{K_{sp}}{4}} = \sqrt[3]{\dfrac{3.9 \times 10^{-11}}{4}} = \sqrt[3]{9.75 \times 10^{-12}} \approx 2.14 \times 10^{-4}$ mol/L (1 mark).

(b) New solubility with [F⁻] = 0.100 mol/L (2 marks): In the presence of excess F⁻, [F⁻] ≈ 0.100 mol/L (the contribution from CaF₂ dissolution is negligible since s will be very small) (1 mark). $K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = s \times (0.100)^2 = s \times 0.0100 = 3.9 \times 10^{-11}$. $s = \dfrac{3.9 \times 10^{-11}}{0.0100} = 3.9 \times 10^{-9}$ mol/L (1 mark). This is ~55,000× less soluble than in pure water.

(c) Common ion explanation (1 mark): The F⁻ ion is common to both NaF and CaF₂. Adding NaF dramatically increases [F⁻], making Qsp > Ksp. The equilibrium shifts left — CaF₂ precipitates and [Ca²⁺] decreases until Qsp = Ksp again. The solubility of CaF₂ is suppressed by the high [F⁻] from the common ion NaF — this is the common ion effect.

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Self-Assessment

Section A — MC (Q1–10) /10

Q11 — Ksp from solubility + Qsp /4

Q12 — Dilution + Qsp + precipitate /5

Q13 — Ksp→s, common ion, explain /5

Total /24—

Checkpoint 4 complete — IQ4 Ksp, Qsp & Common Ion