These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. All K_sp values are at 25 °C. Salts named with capital letters (E, M, L, G, and A to E in Question 3) are fictional; the ion charges are given or follow from the formula.
Question 1 (multiple choice · foundation)
The fictional salt E₂(SO₄)₃ contains E³⁺ and SO₄²⁻ ions. Which expression is K_sp for E₂(SO₄)₃?
- (A) [E³⁺]²[SO₄²⁻]³
- (B) [E³⁺][SO₄²⁻]
- (C) (2[E³⁺])(3[SO₄²⁻])
- (D) [E³⁺]²[SO₄²⁻]³ / [E₂(SO₄)₃]
Answer and explanation
Answer: (A). The dissolving equation is E₂(SO₄)₃(s) ⇌ 2 E³⁺(aq) + 3 SO₄²⁻(aq). Each ion concentration is raised to the power of its coefficient, and the solid is left out.
- (B) ignores the coefficients.
- (C) multiplies by the coefficients instead of using them as powers.
- (D) puts the solid in a denominator. A pure solid never appears in an equilibrium expression.
Question 2 (multiple choice · core)
A fictional salt M₂X₃ dissolves as M₂X₃(s) ⇌ 2 M³⁺(aq) + 3 X²⁻(aq). If its molar solubility is s, which expression equals K_sp?
- (A) s⁵
- (B) 6s⁵
- (C) 72s⁵
- (D) 108s⁵
Answer and explanation
Answer: (D). [M³⁺] = 2s and [X²⁻] = 3s. K_sp = (2s)²(3s)³ = 4s² × 27s³ = 108s⁵.
- (A) forgets that each formula unit releases 2 M³⁺ and 3 X²⁻ ions.
- (B) multiplies the coefficients (2 × 3) but does not raise them to powers.
- (C) swaps the powers: 2³ × 3² = 72. The coefficient 2 belongs with the power 2, and 3 with the power 3.
Question 3 (multiple choice · core)
Four fictional salts have these K_sp values. Which salt has the greatest molar solubility in water?
| Salt | Ion ratio | K_sp |
|---|---|---|
| AB | 1:1 | 1.0 × 10⁻⁸ |
| AC₂ | 1:2 | 4.0 × 10⁻⁹ |
| A₂D | 2:1 | 3.2 × 10⁻¹¹ |
| AE₃ | 1:3 | 2.7 × 10⁻¹⁵ |
- (A) AB
- (B) AC₂
- (C) A₂D
- (D) AE₃
Answer and explanation
Answer: (B). Calculate s for each salt.
| Salt | K_sp in terms of s | s (mol L⁻¹) |
|---|---|---|
| AB | s² | √(1.0 × 10⁻⁸) = 1.0 × 10⁻⁴ |
| AC₂ | 4s³ | ∛(1.0 × 10⁻⁹) = 1.0 × 10⁻³ |
| A₂D | 4s³ | ∛(8.0 × 10⁻¹²) = 2.0 × 10⁻⁴ |
| AE₃ | 27s⁴ | ⁴√(1.0 × 10⁻¹⁶) = 1.0 × 10⁻⁴ |
- (A) has the largest K_sp, which is the trap: K_sp values cannot be compared directly when the ion ratios differ.
- (C) is more soluble than AB but five times less soluble than AC₂.
- (D) has by far the smallest K_sp but the same solubility as AB, because its expression has the power 4.
Question 4 (multiple choice · foundation)
Which of these salts would you expect to have K_sp greater than 1 at 25 °C?
- (A) AgBr
- (B) BaSO₄
- (C) CaCO₃
- (D) NH₄NO₃
Answer and explanation
Answer: (D). All ammonium salts and all nitrate salts are soluble. A soluble salt gives large ion concentrations in a saturated solution, so its K_sp is greater than 1.
- (A), (B) and (C) contain none of the always-soluble ions (Na⁺, K⁺, NH₄⁺, NO₃⁻). Each is a familiar precipitate, so its K_sp is much less than 1. You do not need to memorise extra solubility rules: the ammonium and nitrate rule is enough to pick (D).
Question 5 (calculation · core)
The fictional salt LD₂ (molar mass 210.0 g mol⁻¹) dissolves as LD₂(s) ⇌ L²⁺(aq) + 2 D⁻(aq). Its K_sp is 5.0 × 10⁻⁷.
(a) Calculate the molar solubility of LD₂. (b) State the concentration of D⁻ in the saturated solution. (c) Calculate the maximum mass of LD₂ that can dissolve in 500.0 mL of water.
Worked solution and suggested Marlbridge rubric
(a) K_sp = [L²⁺][D⁻]² = (s)(2s)² = 4s³. So s³ = 5.0 × 10⁻⁷ ÷ 4 = 1.25 × 10⁻⁷ and s = ∛(1.25 × 10⁻⁷) = 5.0 × 10⁻³ mol L⁻¹.
(b) [D⁻] = 2s = 1.0 × 10⁻² mol L⁻¹.
(c) Moles in 0.5000 L = 5.0 × 10⁻³ × 0.5000 = 2.5 × 10⁻³ mol. Mass = 2.5 × 10⁻³ mol × 210.0 g mol⁻¹ = 0.53 g (0.525 g before rounding to 2 significant figures).
| Point | What earns it |
|---|---|
| 1 | K_sp = 4s³ (or s(2s)²) set up correctly |
| 1 | s = 5.0 × 10⁻³ mol L⁻¹ |
| 1 | [D⁻] = 1.0 × 10⁻² mol L⁻¹ (carry forward 2 × their s) |
| 1 | Mass 0.53 g, using the volume in litres and the molar mass |
Common errors: s = √K_sp = 7.1 × 10⁻⁴ M (treats the salt as 1:1); s = ∛K_sp = 7.9 × 10⁻³ M (forgets the 4).
Question 6 (calculation · core)
A saturated solution of the fictional salt M₂X is made by stirring excess solid with water. The salt dissolves as M₂X(s) ⇌ 2 M⁺(aq) + X²⁻(aq). Analysis shows [M⁺] = 2.6 × 10⁻⁴ M in the saturated solution.
(a) Calculate the molar solubility of M₂X and [X²⁻]. (b) Calculate K_sp for M₂X. (c) A student mixes 25.0 mL of 1.0 × 10⁻³ M MNO₃ with 25.0 mL of 1.0 × 10⁻³ M Na₂X. Predict, with a calculation, whether M₂X precipitates.
Worked solution and suggested Marlbridge rubric
(a) Each formula unit releases 2 M⁺, so s = [M⁺] ÷ 2 = 1.3 × 10⁻⁴ mol L⁻¹, and [X²⁻] = s = 1.3 × 10⁻⁴ M.
(b) K_sp = [M⁺]²[X²⁻] = (2.6 × 10⁻⁴)²(1.3 × 10⁻⁴) = 8.8 × 10⁻¹².
(c) Mixing doubles the volume, so each concentration halves: [M⁺] = 5.0 × 10⁻⁴ M and [X²⁻] = 5.0 × 10⁻⁴ M. Q = [M⁺]²[X²⁻] = (5.0 × 10⁻⁴)²(5.0 × 10⁻⁴) = 1.25 × 10⁻¹⁰. Q > K_sp (about 14 times larger), so the solution holds more ions than equilibrium allows: M₂X precipitates until Q falls to K_sp.
| Point | What earns it |
|---|---|
| 1 | s = [M⁺] ÷ 2 = 1.3 × 10⁻⁴ M, with [X²⁻] equal to s |
| 1 | K_sp = 8.8 × 10⁻¹² from [M⁺]²[X²⁻] |
| 1 | Concentrations halved on mixing and Q = 1.25 × 10⁻¹⁰ (or 1.3 × 10⁻¹⁰) |
| 1 | Compares Q with K_sp and concludes a precipitate forms |
Common errors: using 2.6 × 10⁻⁴ M for both ions gives K_sp = 1.8 × 10⁻¹¹ (the X²⁻ value is double what it should be). Forgetting dilution gives Q = 1.0 × 10⁻⁹; the conclusion is the same here, but the method point is lost.
Question 7 (constructed response · stretch)
Two fictional salts have the same K_sp at 25 °C, 1.0 × 10⁻¹². Salt GH dissolves as GH(s) ⇌ G⁺(aq) + H⁻(aq); salt GJ₂ dissolves as GJ₂(s) ⇌ G²⁺(aq) + 2 J⁻(aq).
(a) A student claims: “The two salts have the same K_sp, so they are equally soluble.” Use calculations to evaluate this claim. (b) A second student adds another 1 g of solid GH to a beaker of saturated GH solution (with solid already present) and stirs at 25 °C. She predicts that [G⁺] will increase. Is she right? Explain.
Worked solution and suggested Marlbridge rubric
(a) For GH: K_sp = s², so s = √(1.0 × 10⁻¹²) = 1.0 × 10⁻⁶ mol L⁻¹. For GJ₂: K_sp = 4s³, so s = ∛(2.5 × 10⁻¹³) = 6.3 × 10⁻⁵ mol L⁻¹. The claim is incorrect: GJ₂ is about 63 times more soluble. The K_sp expressions have different forms (s² against 4s³), so equal K_sp values do not mean equal solubilities. K_sp values compare directly only for salts with the same ion ratio.
(b) She is not right. The solution is already saturated, so dissolving and precipitating are happening at equal rates. K_sp = [G⁺][H⁻] does not include the solid, so adding more solid does not change Q or K_sp; nothing shifts. [G⁺] stays at 1.0 × 10⁻⁶ M, and the extra solid simply stays undissolved. Only a change in temperature would change K_sp and the ion concentrations.
| Point | What earns it |
|---|---|
| 1 | Both molar solubilities calculated correctly |
| 1 | Claim rejected with the reason that the expressions differ (different ion ratios or powers) |
| 1 | States [G⁺] does not change |
| 1 | Explains that the solid is not in the K_sp expression (or that the solution is already at equilibrium), so Q still equals K_sp |
How did you do?
- Q1 or Q2 wrong: rebuild K_sp from the dissolving equation, as in “The solubility-product constant” in the study guide.
- Q3 or Q7(a) wrong: go back to Worked example 3: calculate s before you rank.
- Q4 wrong: re-read “Connecting K_sp to the solubility rules”.
- Q5 or Q6 wrong: check the coefficient inside each bracket (2s, not s) and the power outside it. Use Figure 2 for unit changes.
- Q6(c) wrong: remember to dilute before calculating Q, then compare Q with K_sp.
Then tick off the topic checklist.