Chemistry • Year 12 • Module 5 • Lesson 18

Qsp, Precipitation & Common Ion Effect

Lock in core vocabulary, the Qsp vs Ksp decision rule, and the mechanism of the common ion effect before moving to calculation work.

Build · Vocab & Recall

1. Term–definition match

The definitions below are shuffled. Write the matching term from this list in the right-hand column: ion product (Qsp), solubility product (Ksp), common ion effect, molar solubility, supersaturated, unsaturated, Le Chatelier's Principle, selective precipitation, saturation, common ion. 10 marks

#DefinitionMatching term
1.1The equilibrium constant for the dissolution of a sparingly soluble salt; its value changes only with temperature.
1.2The product of ion concentrations raised to stoichiometric powers evaluated at any moment — not necessarily at equilibrium.
1.3The state of a solution in which Qsp exceeds Ksp; excess ions will deposit as solid until equilibrium is restored.
1.4The state of a solution where Qsp is less than Ksp; more solid can dissolve and no precipitation occurs.
1.5The concentration of solute (in mol L−1) that dissolves to give a saturated solution at a given temperature.
1.6An ion that is shared between a dissolving salt and another solute already present in solution.
1.7The decrease in solubility of a sparingly soluble salt caused by the presence of a shared ion in the solution.
1.8When a system at equilibrium is disturbed, it shifts to oppose the change and restore equilibrium.
1.9Using controlled addition of a reagent ion to precipitate one ion from a mixture while leaving a second ion in solution.
1.10The condition where Qsp equals Ksp — the solution is neither gaining nor losing solute at a macroscopic level.
Stuck? Revisit the Key Terms panel and formula panel in Lesson 18.

2. True or false — with correction

Circle T or F. If false, write the corrected statement on the line provided. 10 marks (1 T/F + 1 correction per item)

2.1 Qsp and Ksp use the same mathematical expression, but Qsp is evaluated using current (non-equilibrium) ion concentrations whereas Ksp is the value at equilibrium.    T  /  F

2.2 If Qsp < Ksp, a precipitate will form immediately because the solution is supersaturated.    T  /  F

2.3 When NaCl is dissolved in a saturated AgCl solution, the value of Ksp(AgCl) increases because more Cl− ions are present.    T  /  F

2.4 Adding a common ion to a solution of a sparingly soluble salt shifts the dissolution equilibrium to the left, decreasing solubility without changing Ksp.    T  /  F

2.5 When mixing two solutions to determine Qsp, you must dilute the concentrations using cnew = coriginal × Voriginal / Vtotal before calculating the ion product.    T  /  F

Stuck? Revisit the formula panel and Card 1 (Qsp decision rule), Card 2 (dilution step) and Card 3 (common ion effect) in the lesson.

3. Fill the blank — Qsp decision pathway

Complete the paragraph by filling each blank with the correct term or symbol from the word bank. Use each term once only. 8 marks

Word bank:   Qsp  •  Ksp  •  equilibrium  •  precipitate  •  supersaturated  •  unsaturated  •  left  •  dissolve

When two solutions containing different ions are mixed, the first step is to dilute all ion concentrations for the larger total volume. The ion product __________ is then calculated using those diluted concentrations. This value is compared with __________, the equilibrium constant for the dissolution reaction. If the ion product is greater than the equilibrium constant, the solution is __________ and a __________ forms — ions deposit as solid until the ion product falls back to the equilibrium constant. When the ion product is less than the equilibrium constant, the solution is __________ and more solid can __________ without forming a solid. The system eventually reaches __________ by shifting __________ (precipitation) or right (dissolution) until Qsp = Ksp.

Stuck? Read the formula panel in Lesson 18 and trace the three Qsp conditions.

4. Build a concept map

Draw labelled arrows between the six terms below to show how they are related. Each arrow must carry a short linking phrase (e.g. "shifts equilibrium to", "is compared with", "causes"). Aim for at least six labelled arrows. 6 marks

Supplied terms: QspKspcommon ion effectprecipitationLe Chatelier's Principlemolar solubility

Qsp
Ksp
precipitation
Le Chatelier's Principle
molar solubility
common ion effect
Hint: The chain Qsp > Ksp → precipitation → equilibrium shifts left (Le Chatelier) is one key path. Common ion effect links to solubility and to Qsp > Ksp.

5. Function recall

Answer each question in 1–2 sentences using precise lesson terms. 8 marks (2 each)

5.1 What is the purpose of the dilution step when calculating Qsp after mixing two solutions?

5.2 Why does adding NaCl to a saturated AgCl solution cause some AgCl to precipitate?

5.3 Why does Ksp change when temperature changes but not when a common ion is added?

5.4 What is the significance of Qsp = Ksp for a dissolving ionic compound?

Stuck? Revisit the formula panel, Card 2 (dilution) and Card 3 (common ion effect) in the lesson.
Answers — Do not peek before attempting

Q1 — Term–definition matches

1.1 solubility product (Ksp) • 1.2 ion product (Qsp) • 1.3 supersaturated • 1.4 unsaturated • 1.5 molar solubility • 1.6 common ion • 1.7 common ion effect • 1.8 Le Chatelier's Principle • 1.9 selective precipitation • 1.10 saturation

Q2 — True / false with corrections

2.1 True.

2.2 False. Correction: If Qsp < Ksp, the solution is unsaturated — no precipitate forms. Precipitation only occurs when Qsp > Ksp (supersaturated).

2.3 False. Correction: Ksp is unchanged by adding NaCl — only temperature changes Ksp. Adding NaCl increases [Cl−], raising Qsp above Ksp, which shifts the equilibrium left and causes more AgCl to precipitate (common ion effect), but Ksp itself remains constant.

2.4 True.

2.5 True.

Q3 — Cloze answers (in order)

QspKspsupersaturatedprecipitateunsaturateddissolveequilibriumleft

Q4 — Sample concept map connections

Key arrows (accept any biologically valid linking phrase):

  • Qspis compared withKsp
  • Qsp > Ksptriggersprecipitation
  • Le Chatelier's Principleexplains direction ofprecipitation
  • common ion effectraises Qsp above Ksp, causingprecipitation
  • common ion effectdecreasesmolar solubility
  • Kspdetermines maximummolar solubility

Award 1 mark per correct arrow with a valid linking phrase, up to 6 marks.

Q5.1 — Dilution step purpose (2 marks)

When two solutions are mixed, each ion's concentration decreases because the total volume increases. Using the original (undiluted) concentrations would overestimate Qsp and could falsely indicate precipitation. The dilution step ensures Qsp is calculated from the actual concentrations present in the combined solution. [1 mark for identifying volume increase causes dilution; 1 mark for stating that using undiluted concentrations overestimates Qsp / gives wrong prediction.]

Q5.2 — Why AgCl precipitates when NaCl is added (2 marks)

NaCl fully dissociates to release Cl−, a common ion shared with AgCl's dissolution equilibrium (AgCl(s) ⇌ Ag+(aq) + Cl−(aq)). The increased [Cl−] causes Qsp = [Ag+][Cl−] to exceed Ksp, so the system is now supersaturated. By Le Chatelier's Principle, the equilibrium shifts left — more AgCl solid forms — until Qsp returns to Ksp. [1 mark for identifying Cl− as common ion that raises Qsp above Ksp; 1 mark for left shift via LCP causing precipitation.]

Q5.3 — Why Ksp changes with temperature but not common ion addition (2 marks)

Ksp is a thermodynamic equilibrium constant — it depends only on temperature because temperature changes the position of equilibrium and the thermodynamic favourability of dissolution. Adding a common ion shifts the equilibrium position (changes how much solid dissolves) but does not change the energy difference between dissolved and solid states; therefore Ksp is unchanged. [1 mark for temperature changes thermodynamic stability / equilibrium position; 1 mark for common ion changes position of equilibrium only, not the constant itself.]

Q5.4 — Significance of Qsp = Ksp (2 marks)

When Qsp = Ksp, the solution is exactly at saturation — it is in dynamic equilibrium with any undissolved solid. The rate of dissolution equals the rate of precipitation, so there is no net change in the amount of solid or in ion concentrations. No precipitation occurs and no additional solid dissolves. [1 mark for saturation / dynamic equilibrium; 1 mark for no net change / equal rates of forward and reverse processes.]