How to Use This Worksheet
Work in order. Section A checks the foundations: writing Ksp expressions, the bookkeeping that links molar solubility to each ion concentration, and the three Qsp cases. Section B practises the two calculation directions, Ksp from solubility and solubility from Ksp, a fair comparison between two salts, and a precipitation prediction. Section C is exam style: a solubility quoted in grams per litre, a mixing problem where the precipitate does not form, and the four-ion case of Fe(OH)₃.
Reference facts. Ksp is the equilibrium constant for a solid dissolving into its ions, with the solid itself omitted. Each ion concentration is raised to its coefficient in the dissolution equation. Molar solubility s is in mol/L; ion concentrations are s times the ion’s coefficient. The patterns are
Ksp = s² for a 1:1 salt
Ksp = 4s³ for a 1:2 or 2:1 salt
Ksp = 27s⁴ for a 1:3 salt
Qsp less than Ksp is unsaturated, Qsp equal to Ksp is saturated, Qsp greater than Ksp precipitates until Qsp falls back to Ksp. Mixing dilutes: equal volumes halve each ion concentration before Qsp is computed. Ksp ranks solubility directly only between salts of the same stoichiometry; otherwise compare molar solubilities. Values at 25 °C: AgCl Ksp = 1.8 × 10⁻¹⁰, s = 1.3 × 10⁻⁵ M; CaF₂ Ksp = 3.9 × 10⁻¹¹, s = 2.1 × 10⁻⁴ M; BaSO₄ Ksp = 1.1 × 10⁻¹⁰, s = 1.0 × 10⁻⁵ M; Mg(OH)₂ Ksp = 5.6 × 10⁻¹², s = 1.1 × 10⁻⁴ M; PbI₂ Ksp = 9.8 × 10⁻⁹, s = 1.3 × 10⁻³ M; Fe(OH)₃ Ksp = 2.8 × 10⁻³⁹, s = 1.0 × 10⁻¹⁰ M; PbCl₂ Ksp = 1.6 × 10⁻⁵.
Three habits protect every answer. Write the ion concentrations in terms of s, multipliers included, before substituting numbers. Check the stoichiometry before you compare two salts, and never rank solubilities from Ksp across different stoichiometries. And use the concentrations in the mixture, after any dilution, when you compute Qsp.
Do not look at the answer key until you have attempted every question.
Section A: Recall and Concepts
A1. Write the Ksp expression for each dissolution, and in each case state why the solid does not appear. (a) BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq). (b) PbCl₂(s) ⇌ Pb²⁺(aq) + 2Cl⁻(aq). (c) Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq). (d) Fe(OH)₃(s) ⇌ Fe³⁺(aq) + 3OH⁻(aq).
A2. Judge each statement, giving the correction. (a) “Ksp for AgCl is [Ag⁺][Cl⁻] divided by [AgCl], because the solid is a reactant.” (b) “A salt with Ksp = 1.0 × 10⁻¹² must be less soluble than a salt with Ksp = 1.0 × 10⁻¹⁰.” (c) “If twice as much solid is stirred into a saturated solution, the molar solubility doubles.” (d) “Qsp = Ksp means the solution is supersaturated and a precipitate is about to form.”
A3. (a) Mg(OH)₂ dissolves as Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq). If the molar solubility is s, state [Mg²⁺] and [OH⁻] in terms of s, and hence Ksp in terms of s. (b) Do the same for Fe(OH)₃(s) ⇌ Fe³⁺(aq) + 3OH⁻(aq). (c) Do the same for Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq).
A4. A solution of a salt’s ions is tested with Qsp. State what each comparison means and what happens next. (a) Qsp greater than Ksp. (b) Qsp equal to Ksp. (c) Qsp less than Ksp. (d) In the cases where solid forms or dissolves, state the value Qsp moves toward.
A5. Ksp for AgCl is 1.8 × 10⁻¹⁰ and for CaF₂ is 3.9 × 10⁻¹¹. Their molar solubilities are 1.3 × 10⁻⁵ M and 2.1 × 10⁻⁴ M. (a) Which salt has the larger Ksp? (b) Which salt is more soluble? (c) Explain why the two answers differ. (d) Ksp for BaSO₄ is 1.1 × 10⁻¹⁰. State which of AgCl and BaSO₄ is more soluble, and explain why this comparison is safe.
Section B: Core Practice
B1. The molar solubility of AgCl in pure water is 1.3 × 10⁻⁵ M. (a) State [Ag⁺] and [Cl⁻] in the saturated solution. (b) Calculate Ksp for AgCl. (c) Calculate the solubility of AgCl in g/L, given its molar mass is 143.32 g/mol. (d) Calculate the mass of AgCl that dissolves in 500 mL of water at saturation.
B2. The solubility of CaF₂ is 1.64 × 10⁻² g/L. The molar mass of CaF₂ is 78.08 g/mol. (a) Convert the solubility to mol/L. (b) State [Ca²⁺] and [F⁻] in the saturated solution. (c) Calculate Ksp for CaF₂. (d) Check your answer with the 4s³ pattern.
B3. Ksp for Mg(OH)₂ is 5.6 × 10⁻¹². (a) Write Ksp in terms of the molar solubility s. (b) Calculate s. (c) State [Mg²⁺] and [OH⁻] in a saturated solution. (d) Calculate the solubility in g/L, given the molar mass of Mg(OH)₂ is 58.33 g/mol.
B4. Ksp for AgCl is 1.8 × 10⁻¹⁰ and for Ag₂CrO₄ is 1.1 × 10⁻¹². (a) Calculate the molar solubility of each salt. (b) State which salt is more soluble in pure water, and by about what factor. (c) Explain why the Ksp order alone gives the wrong answer here. (d) State [Ag⁺] in a saturated solution of Ag₂CrO₄.
B5. Equal volumes of 0.0010 M CaCl₂ and 0.0010 M NaF are mixed. Ksp for CaF₂ is 3.9 × 10⁻¹¹. (a) State [Ca²⁺] and [F⁻] immediately after mixing. (b) Calculate Qsp. (c) State whether a precipitate of CaF₂ forms, and why. (d) As precipitation proceeds, state what happens to [Ca²⁺], to [F⁻], and to Qsp, and the value Qsp reaches when the changes stop. (e) State which ion concentration falls faster as precipitation proceeds, and why.
B6. Ksp for PbI₂ is 9.8 × 10⁻⁹. (a) Write Ksp in terms of s. (b) Calculate the molar solubility of PbI₂. (c) State [Pb²⁺] and [I⁻] in a saturated solution. (d) Calculate the solubility in g/L, given the molar mass of PbI₂ is 461.0 g/mol. (e) Calculate the mass of PbI₂ that dissolves in 250 mL of water at saturation.
Section C: Exam Style
C1. The solubility of CaCO₃ in water is 6.9 × 10⁻³ g/L. The molar mass of CaCO₃ is 100.09 g/mol. (a) Convert the solubility to a molar solubility. (b) State [Ca²⁺] and [CO₃²⁻] in the saturated solution. (c) Calculate Ksp for CaCO₃. (d) Tables quote Ksp = 4.7 × 10⁻⁹ for CaCO₃. State whether your answer agrees, and explain any difference. (e) More solid CaCO₃ is added to the saturated solution and the mixture is stirred. State what happens to [Ca²⁺], and why.
C2. A 100 mL sample of 0.0020 M Pb(NO₃)₂ is mixed with 100 mL of 0.0040 M KI. Ksp for PbI₂ is 9.8 × 10⁻⁹. (a) State [Pb²⁺] and [I⁻] in the mixture. (b) Calculate Qsp and state whether PbI₂ precipitates. (c) The iodide concentration in the mixture is raised tenfold, to 0.020 M, with [Pb²⁺] unchanged. Calculate the new Qsp and state whether PbI₂ precipitates now. (d) The original mixture from part (a) is diluted with water to twice its volume. Calculate the new Qsp and state whether the conclusion changes. (e) Solid PbI₂ is added to the original mixture from part (a). State what happens to it, and why.
C3. Ksp for Fe(OH)₃ is 2.8 × 10⁻³⁹. (a) Write the Ksp expression and then Ksp in terms of the molar solubility s. (b) Calculate s. (c) State [Fe³⁺] and [OH⁻] in a saturated solution. (d) Calculate the mass of Fe(OH)₃ that dissolves in 1.0 L of water at saturation, given its molar mass is 106.87 g/mol. (e) Calculate the mass that dissolves in 10 L of water at saturation.
Show the answer key and full working (only after you have attempted every question)
A1. (a) Ksp = [Ba²⁺][SO₄²⁻]. (b) Ksp = [Pb²⁺][Cl⁻]², with the chloride concentration squared for its coefficient of 2. (c) Ksp = [Ag⁺]²[CrO₄²⁻], with the silver concentration squared. (d) Ksp = [Fe³⁺][OH⁻]³, with the hydroxide concentration cubed. In every case the solid is omitted because a pure solid’s amount does not enter the equilibrium expression; only dissolved ions appear, each raised to its coefficient.
A2. (a) False. The solid never appears in a Ksp expression: Ksp for AgCl is simply [Ag⁺][Cl⁻]. (b) False as stated. Ksp orders solubility only between salts of the same stoichiometry. A 1:2 salt with Ksp = 1.0 × 10⁻¹² has s = cube root of (Ksp ÷ 4) = 6.3 × 10⁻⁵ M, while a 1:1 salt with Ksp = 1.0 × 10⁻¹⁰ has s = √Ksp = 1.0 × 10⁻⁵ M, so the smaller Ksp belongs to the more soluble salt. (c) False. Molar solubility is fixed by Ksp at that temperature. Extra solid cannot dissolve into a solution that is already saturated, so s is unchanged. (d) False. Qsp = Ksp is the definition of a saturated solution in equilibrium with its solid: stable, with no net precipitation or dissolving. Supersaturated is Qsp greater than Ksp.
A3. (a) Each formula unit that dissolves gives one Mg²⁺ and two OH⁻, so [Mg²⁺] = s and [OH⁻] = 2s. Ksp = [Mg²⁺][OH⁻]² = (s)(2s)² = 4s³. (b) [Fe³⁺] = s and [OH⁻] = 3s. Ksp = [Fe³⁺][OH⁻]³ = (s)(3s)³ = 27s⁴. (c) [Ag⁺] = 2s and [CrO₄²⁻] = s. Ksp = [Ag⁺]²[CrO₄²⁻] = (2s)²(s) = 4s³. The multiplier belongs to the concentration first; the power from the expression is applied to the whole term, multiplier included.
A4. (a) Qsp greater than Ksp: the solution holds more ions than saturation allows (supersaturated), so solid precipitates until Qsp falls to Ksp; the dissolved ion concentrations fall as the amount of solid grows. (b) Qsp equal to Ksp: the solution is exactly saturated; no net change, and any added solid stays undissolved. (c) Qsp less than Ksp: the solution is unsaturated; no precipitate forms, and added solid dissolves, raising Qsp toward Ksp until either saturation is reached or the solid runs out. (d) In both moving cases the destination is the same: Qsp = Ksp, with solid present.
A5. (a) AgCl has the larger Ksp: 1.8 × 10⁻¹⁰ against 3.9 × 10⁻¹¹. (b) CaF₂ is more soluble: 2.1 × 10⁻⁴ M against 1.3 × 10⁻⁵ M, about sixteen times as much. (c) The stoichiometries differ. For AgCl, Ksp = s², so s is a square root of Ksp. For CaF₂, Ksp = 4s³, so s is a cube root of Ksp ÷ 4, and a cube root of a tiny number is far larger than its square root. Ksp values can only be compared directly when the salts share the same ion ratio. (d) AgCl is more soluble: 1.3 × 10⁻⁵ M against 1.0 × 10⁻⁵ M for BaSO₄. The comparison is safe because both salts are 1:1, so both give s = √Ksp and the larger Ksp means the larger solubility.
B1. (a) AgCl dissolves 1:1, so [Ag⁺] = [Cl⁻] = s = 1.3 × 10⁻⁵ M. (b) Ksp = [Ag⁺][Cl⁻] = (1.3 × 10⁻⁵)(1.3 × 10⁻⁵) = 1.7 × 10⁻¹⁰. (c) Solubility = s × molar mass = (1.3 × 10⁻⁵)(143.32) = 1.9 × 10⁻³ g/L. (d) Mass in 500 mL = (1.9 × 10⁻³ g/L)(0.500 L) = 9.3 × 10⁻⁴ g.
B2. (a) s = 1.64 × 10⁻² ÷ 78.08 = 2.1 × 10⁻⁴ M. Dividing by the molar mass converts grams of CaF₂ per litre into moles per litre; every later step needs moles. (b) [Ca²⁺] = s = 2.1 × 10⁻⁴ M and [F⁻] = 2s = 4.2 × 10⁻⁴ M, because each dissolving formula unit releases two fluoride ions. (c) Ksp = [Ca²⁺][F⁻]² = (2.1 × 10⁻⁴)(4.2 × 10⁻⁴)² = (2.1 × 10⁻⁴)(1.76 × 10⁻⁷) = 3.7 × 10⁻¹¹. (d) The pattern agrees: 4s³ = 4(2.1 × 10⁻⁴)³ = 4(9.3 × 10⁻¹²) = 3.7 × 10⁻¹¹.
B3. (a) Ksp = (s)(2s)² = 4s³. (b) s = cube root of (Ksp ÷ 4) = cube root of (5.6 × 10⁻¹² ÷ 4) = cube root of 1.4 × 10⁻¹² = 1.1 × 10⁻⁴ M. (c) [Mg²⁺] = s = 1.1 × 10⁻⁴ M and [OH⁻] = 2s = 2.2 × 10⁻⁴ M. (d) Solubility = s × molar mass = (1.1 × 10⁻⁴)(58.33) = 6.4 × 10⁻³ g/L.
B4. (a) AgCl: s = √Ksp = √(1.8 × 10⁻¹⁰) = 1.3 × 10⁻⁵ M. Ag₂CrO₄: Ksp = 4s³, so s = cube root of (1.1 × 10⁻¹² ÷ 4) = cube root of 2.75 × 10⁻¹³ = 6.5 × 10⁻⁵ M. (b) Ag₂CrO₄ is more soluble, by a factor of 6.5 × 10⁻⁵ ÷ 1.3 × 10⁻⁵ = 5.0. (c) The salts have different stoichiometries, so their Ksp values are built from different powers of s: a square for AgCl and a cube (times 4) for Ag₂CrO₄. A cube root of a very small number exceeds a square root of a somewhat larger one, so the Ksp ranking reverses when both are converted to molar solubility, the quantity that actually measures dissolving. (d) [Ag⁺] = 2s = 2 × 6.5 × 10⁻⁵ = 1.3 × 10⁻⁴ M.
B5. (a) Mixing equal volumes halves each concentration: [Ca²⁺] = 0.00050 M and [F⁻] = 0.00050 M. (b) Qsp = [Ca²⁺][F⁻]² = (5.0 × 10⁻⁴)(5.0 × 10⁻⁴)² = (5.0 × 10⁻⁴)(2.5 × 10⁻⁷) = 1.3 × 10⁻¹⁰. (c) Qsp = 1.3 × 10⁻¹⁰ exceeds Ksp = 3.9 × 10⁻¹¹, so the mixture is supersaturated and CaF₂ precipitates until Qsp falls to Ksp. (d) Removing CaF₂(s) takes one Ca²⁺ and two F⁻ out of solution per formula unit, so both concentrations fall, Qsp falls with them, and precipitation stops when Qsp reaches exactly Ksp = 3.9 × 10⁻¹¹, with solid CaF₂ present: the saturated equilibrium. (e) [F⁻] falls twice as fast as [Ca²⁺], because each formula unit that precipitates removes two fluoride ions for every one calcium ion.
B6. (a) Dissolving s gives [Pb²⁺] = s and [I⁻] = 2s, so Ksp = (s)(2s)² = 4s³. (b) s = cube root of (9.8 × 10⁻⁹ ÷ 4) = cube root of 2.45 × 10⁻⁹ = 1.3 × 10⁻³ M. Check: 4(1.35 × 10⁻³)³ = 9.8 × 10⁻⁹. (c) [Pb²⁺] = 1.3 × 10⁻³ M and [I⁻] = 2s = 2.7 × 10⁻³ M. (d) Solubility = (1.35 × 10⁻³)(461.0) = 0.62 g/L, using the unrounded value of s from part (b). (e) Mass in 250 mL = (0.62 g/L)(0.250 L) = 0.16 g.
C1. (a) s = 6.9 × 10⁻³ ÷ 100.09 = 6.9 × 10⁻⁵ M. (b) CaCO₃ dissolves 1:1, so [Ca²⁺] = [CO₃²⁻] = 6.9 × 10⁻⁵ M. (c) Ksp = [Ca²⁺][CO₃²⁻] = (6.9 × 10⁻⁵)² = 4.8 × 10⁻⁹. (d) Yes, it agrees: the tabulated 4.7 × 10⁻⁹ differs only in the last digit, because the quoted solubility is given to two significant figures. Working backwards, √(4.7 × 10⁻⁹) = 6.9 × 10⁻⁵ M, the value the question started from. (e) Nothing happens to [Ca²⁺]. The solution is already saturated, so Qsp already equals Ksp and the added solid cannot dissolve; [Ca²⁺] stays at 6.9 × 10⁻⁵ M and the extra solid simply rests at the bottom.
C2. (a) Mixing equal volumes halves each concentration: [Pb²⁺] = 0.0010 M and [I⁻] = 0.0020 M. (b) Qsp = [Pb²⁺][I⁻]² = (0.0010)(0.0020)² = (0.0010)(4.0 × 10⁻⁶) = 4.0 × 10⁻⁹. This is below Ksp = 9.8 × 10⁻⁹, so the mixture is unsaturated and no PbI₂ precipitates. (c) Qsp = (0.0010)(0.020)² = (0.0010)(4.0 × 10⁻⁴) = 4.0 × 10⁻⁷, above Ksp = 9.8 × 10⁻⁹, so yellow PbI₂ precipitates. (d) Diluting twofold halves both concentrations again: [Pb²⁺] = 5.0 × 10⁻⁴ M and [I⁻] = 0.0010 M, so Qsp = (5.0 × 10⁻⁴)(0.0010)² = 5.0 × 10⁻¹⁰, smaller still. The conclusion is unchanged: no precipitate. (e) The solid dissolves. Because Qsp is less than Ksp the solution is unsaturated, so PbI₂ dissolves, raising both ion concentrations, until Qsp reaches Ksp (the solution saturates) or the added solid has all dissolved.
C3. (a) Ksp = [Fe³⁺][OH⁻]³. With [Fe³⁺] = s and [OH⁻] = 3s, Ksp = (s)(3s)³ = 27s⁴. (b) s = fourth root of (Ksp ÷ 27) = fourth root of (2.8 × 10⁻³⁹ ÷ 27) = fourth root of 1.04 × 10⁻⁴⁰ = 1.0 × 10⁻¹⁰ M. A fourth root can be taken as two square roots: √(1.04 × 10⁻⁴⁰) = 1.02 × 10⁻²⁰, and √(1.02 × 10⁻²⁰) = 1.0 × 10⁻¹⁰. (c) [Fe³⁺] = s = 1.0 × 10⁻¹⁰ M and [OH⁻] = 3s = 3.0 × 10⁻¹⁰ M. (d) Mass dissolved per litre = s × molar mass = (1.0 × 10⁻¹⁰)(106.87) = 1.1 × 10⁻⁸ g. (e) Mass in 10 L = (1.1 × 10⁻⁸ g/L)(10 L) = 1.1 × 10⁻⁷ g.