The shape of the lesson
By the end of the lesson, learners can
- Writing and using Ksp: Compare sparingly soluble salts in the simulator, then add a common ion.
- Common ions and precipitation: Use K sp the other way round: fix one ion concentration and find the most of the other ion that can stay in solution.
- AB: Ksp = s2, mol2 dm−6.
- AB2: Ksp = 4s3, mol3 dm−9.
- Common ion: lowers solubility, Ksp unchanged.
- Mixing: dilute first; ionic product > Ksp means precipitate.
One picture every idea returns to
In a saturated solution of a sparingly soluble salt, the solid is in dynamic equilibrium with its ions: AgCl(s) ⇌ Ag+(aq) + Cl−(aq). The solubility product is the product of the ion concentrations, each raised to the power of its coefficient: Ksp = [Ag+][Cl−] (mol2 dm−6), Ksp = [Pb2+][I−]2 (mol3 dm−9). The solid does not appear. From a solubility s: AgCl has s = 1.34 × 10−5 mol dm−3, so Ksp = s2 = 1.80 × 10−10 mol2 dm−6; for PbI2, Ksp = 4s3. Adding a common ion shifts the equilibrium to the left and lowers the solubility. If the ionic product of mixed solutions exceeds Ksp, a precipitate forms.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, How much of a sparingly soluble salt dissolves.. Even 'insoluble' salts dissolve a tiny amount, and the saturated solution is an equilibrium. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Writing and using Ksp | Compare sparingly soluble salts in the simulator, then add a common ion. Project tab 5 of the lab (Acid-base equilibria lab, Ksp and common ions) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Common ions and precipitation | Use K sp the other way round: fix one ion concentration and find the most of the other ion that can stay in solution. Project tab 5 of the lab (Acid-base equilibria lab, Ksp and common ions) and let a student drive it while the class predicts. | Numbered Heads Together |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 15 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Common ions and precipitation, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Square or cube the right terms: for PbI 2, [I − ] = 2s and K sp = 4s 3, not s 3 or s 2. | When two solutions are mixed, each concentration is diluted (halved for equal volumes) before you work out the ionic product. The common ion lowers the solubility, not K sp. Units of K sp depend on the formula: mol 2 dm −6 for AB, mol 3 dm −9 for AB 2. |
Solubility product and the common ion effect: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Write Ksp expressions, with units, for (a) BaSO4, (b) PbI2, (c) Ag2CrO4.
The solubility of calcium fluoride is 2.05 × 10−4 mol dm−3. Calculate Ksp.
Ksp for BaSO4 is 1.08 × 10−10 mol2 dm−6. Calculate its solubility in mol dm−3 and in g dm−3 (Mr 233.4).
Ksp for PbI2 is 8.49 × 10−9 mol3 dm−9. Calculate its solubility in (a) water and (b) 0.100 mol dm−3 KI.
Equal volumes of 2.00 × 10−4 mol dm−3 AgNO3 and 2.00 × 10−4 mol dm−3 NaCl are mixed. Ksp(AgCl) = 1.80 × 10−10 mol2 dm−6. Will a precipitate form?
Explain why silver chloride is less soluble in sodium chloride solution than in pure water.
Answer key and marking notes
- (a) [Ba2+][SO42−], mol2 dm−6.
- (b) [Pb2+][I−]2, mol3 dm−9.
- (c) [Ag+]2[CrO42−], mol3 dm−9.
- [Ca2+] = 2.05 × 10−4, [F−] = 4.10 × 10−4 mol dm−3.
- Ksp = 2.05 × 10−4 × (4.10 × 10−4)2 = 3.45 × 10−11 mol3 dm−9.
- s = √(1.08 × 10−10) = 1.04 × 10−5 mol dm−3.
- Mass: 1.04 × 10−5 × 233.4 = 2.43 × 10−3 g dm−3.
- (a) 4s3 = 8.49 × 10−9, so s = 1.29 × 10−3 mol dm−3.
- (b) [I−] = 0.100 (the dissolved PbI2 adds almost nothing), so [Pb2+] = 8.49 × 10−9 ÷ 0.1002.
- s = 8.49 × 10−7 mol dm−3, about 1500 times less: the common ion effect.
- After mixing, [Ag+] = [Cl−] = 1.00 × 10−4 mol dm−3.
- Ionic product = 1.00 × 10−4 × 1.00 × 10−4 = 1.00 × 10−8 mol2 dm−6.
- This is greater than Ksp, so AgCl precipitates.
- AgCl(s) ⇌ Ag+(aq) + Cl−(aq); NaCl adds a common ion, Cl−.
- The equilibrium shifts left to keep [Ag+][Cl−] equal to Ksp, so less AgCl can dissolve.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The diagram shows a saturated solution of lead(II) iodide in contact with the solid. What is the Ksp expression for lead(II) iodide?
- AKsp = [Pb2+][I−]2
- BKsp = [Pb2+][I−]
- CKsp = [Pb2+][I−]2 ÷ [PbI2]
- DKsp = [Pb2+][2I−]
What are the units of Ksp for silver chromate, Ag2CrO4?
- Amol3 dm−9
- Bmol2 dm−6
- Cmol dm−3
- Dno units
The solubility of silver chloride in water at 298 K is 1.34 × 10−5 mol dm−3. What is Ksp?
- A1.80 × 10−10 mol2 dm−6
- B3.66 × 10−3 mol2 dm−6
- C2.68 × 10−5 mol2 dm−6
- D7.18 × 10−10 mol2 dm−6
Solid sodium chloride is dissolved in a saturated solution of silver chloride at constant temperature. What happens?
- Asome AgCl precipitates; its solubility falls but Ksp is unchanged
- Bmore AgCl dissolves, because the solution now contains more ions
- Cnothing happens, because Ksp is constant
- Dsome AgCl precipitates, because Ksp decreases
Equal volumes of 2.00 × 10−4 mol dm−3 AgNO3 and 2.00 × 10−4 mol dm−3 NaCl are mixed. Ksp(AgCl) = 1.80 × 10−10 mol2 dm−6. What happens?
- Aa precipitate forms, because the ionic product, 1.00 × 10−8, is greater than Ksp
- Bno precipitate forms, because the ionic product, 1.00 × 10−8, is less than Ksp
- Cno precipitate forms, because both solutions are very dilute
- Da precipitate forms only if the ionic product equals 1.80 × 10−10 exactly
Ksp for lead(II) iodide is 8.49 × 10−9 mol3 dm−9 at 298 K. What is its solubility?
- A1.29 × 10−3 mol dm−3
- B9.21 × 10−5 mol dm−3
- C2.04 × 10−3 mol dm−3
- D2.12 × 10−9 mol dm−3
The solubility of calcium fluoride, CaF2, is 2.05 × 10−4 mol dm−3 at 298 K. What is Ksp?
- A3.45 × 10−11 mol3 dm−9
- B8.62 × 10−12 mol3 dm−9
- C1.72 × 10−11 mol3 dm−9
- D8.41 × 10−8 mol2 dm−6
Equal volumes of 2.40 × 10−3 mol dm−3 Pb(NO3)2 and 2.40 × 10−3 mol dm−3 KI are mixed. Ksp(PbI2) = 8.49 × 10−9 mol3 dm−9. What happens?
- Ano precipitate: the ionic product is 1.73 × 10−9, below Ksp
- Ba precipitate: the ionic product is 1.38 × 10−8, above Ksp
- Ca precipitate: the ionic product is 5.76 × 10−6, above Ksp
- Da precipitate, because lead(II) iodide is insoluble in water
Quiz key
PbI2(s) ⇌ Pb2+(aq) + 2I−(aq), so Ksp = [Pb2+][I−]2, with units mol3 dm−9. Leaving out the square ignores the 2 in front of I−. The solid PbI2 has a constant concentration and never appears in a Ksp expression. Writing [2I−] puts the coefficient inside the bracket instead of using it as a power; it should be [I−]2.
Ksp = [Ag+]2[CrO42−] multiplies three concentrations, so the units are (mol dm−3)3 = mol3 dm−9. mol2 dm−6 is correct only for a salt with one cation and one anion, such as AgCl or BaSO4. mol dm−3 is the unit of a single concentration, such as the solubility. Ksp always has units, because the concentrations do not cancel as they do in a partition coefficient.
[Ag+] = [Cl−] = s = 1.34 × 10−5 mol dm−3, so Ksp = s2 = 1.80 × 10−10 mol2 dm−6. 3.66 × 10−3 takes the square root instead of squaring. 2.68 × 10−5 adds the two ion concentrations instead of multiplying them. 7.18 × 10−10 is 4s2, wrongly applying the factor used for a salt such as PbI2.
Adding Cl− raises the ionic product [Ag+][Cl−] above Ksp, so the equilibrium shifts to the left and AgCl precipitates until the product falls back to Ksp. This is the common ion effect: the solubility falls, but Ksp itself only changes with temperature. More ions do not make AgCl more soluble when one of them is the same ion. Saying nothing happens confuses a constant Ksp with constant concentrations. Ksp does not decrease; it is the solubility that decreases.
After mixing, [Ag+] = [Cl−] = 1.00 × 10−4 mol dm−3, so the ionic product is 1.00 × 10−8 mol2 dm−6. This is larger than 1.80 × 10−10, so AgCl precipitates until the product falls to Ksp. 1.00 × 10−8 is larger, not smaller, than 1.80 × 10−10: compare the powers of ten carefully. Even very dilute solutions precipitate a salt with a tiny Ksp. A precipitate forms whenever the ionic product exceeds Ksp; equality is the saturated limit, not the trigger.
Ksp = s × (2s)2 = 4s3 = 8.49 × 10−9, so s3 = 2.12 × 10−9 and s = 1.29 × 10−3 mol dm−3 (0.592 g dm−3). 9.21 × 10−5 is √Ksp, treating PbI2 as a 1 : 1 salt like AgCl. 2.04 × 10−3 takes the cube root of Ksp but forgets to divide by 4 (it treats [I−] as s). 2.12 × 10−9 divides by 4 but forgets the cube root.
[Ca2+] = 2.05 × 10−4 and [F−] = 4.10 × 10−4 mol dm−3, so Ksp = 2.05 × 10−4 × (4.10 × 10−4)2 = 4s3 = 3.45 × 10−11 mol3 dm−9. 8.62 × 10−12 is s3, forgetting that [F−] is 2s. 1.72 × 10−11 is 2s3, doubling [F−] but not squaring the 2. 8.41 × 10−8 is s × 2s, forgetting to square [F−], and the units are wrong too.
Mixing equal volumes halves each concentration to 1.20 × 10−3 mol dm−3. Ionic product = (1.20 × 10−3)(1.20 × 10−3)2 = 1.73 × 10−9 mol3 dm−9, which is below 8.49 × 10−9, so no precipitate forms. 1.38 × 10−8 uses the undiluted 2.40 × 10−3 values: forgetting the dilution changes the prediction. 5.76 × 10−6 multiplies the two undiluted concentrations without squaring [I−]. Calling PbI2 insoluble is the misconception: it dissolves to about 1.29 × 10−3 mol dm−3, so whether it precipitates depends on comparing the ionic product with Ksp.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.