How to Use This Worksheet
Work in order. Section A checks the ideas: spotting a common ion, what adding one does to Qsp and Ksp, and how the ICE table is built when one ion starts with a head start. Section B practises the standard calculation in six versions: a 1:1 salt, a squared common ion, an exact quadratic solve, a hydroxide, a common cation in a different place, and a case where the approximation fails. Section C is exam style: a mixing problem, an approximation error judged honestly, and a comparison of how strongly two salts are suppressed.
Reference facts. A common ion is an ion shared between the sparingly soluble salt and another solute. Solubility with a common ion present is always lower than in pure water, and falls further as the common-ion concentration rises. Adding a common ion does not change Ksp. With the common ion at concentration C, solve Ksp = s × Cⁿ, where n is the exponent on the common ion. The common ion’s equilibrium concentration is C plus its multiple of s. For a 1:1 salt, doubling C halves s; where the common ion is squared, doubling C cuts s by a factor of four. Ksp and pure-water s at 25 °C: AgCl 1.8 × 10⁻¹⁰ and 1.3 × 10⁻⁵ M; CaF₂ 4.0 × 10⁻¹¹ and 2.2 × 10⁻⁴ M; Ag₂CrO₄ 1.1 × 10⁻¹² and 6.5 × 10⁻⁵ M; Ca(OH)₂ 5.5 × 10⁻⁶ and 1.1 × 10⁻² M; PbI₂ 9.8 × 10⁻⁹ and 1.3 × 10⁻³ M; BaSO₄ 1.1 × 10⁻¹⁰ and 1.0 × 10⁻⁵ M; Mg(OH)₂ 5.6 × 10⁻¹² and 1.1 × 10⁻⁴ M.
Three habits protect every answer. Write both ion concentrations in terms of s, the head start included, before substituting numbers. Run the 5 percent check on any approximation. Pause on every added species, name its ions, and check each against the dissolution equation before doing any arithmetic.
Do not look at the answer key until you have attempted every question.
Section A: Recall and Concepts
A1. For each addition to a saturated solution, state the common ion, or state that there is none. (a) NaCl(aq) added to saturated AgBr(aq). (b) Na₂SO₄(aq) added to saturated BaSO₄(aq). (c) KNO₃(aq) added to saturated PbI₂(aq). (d) AgNO₃(aq) added to saturated Ag₂CrO₄(aq). (e) NaOH(aq) added to saturated Mg(OH)₂(aq).
A2. Judge each statement, giving the correction. (a) “Adding NaCl(aq) to a saturated AgCl solution lowers the Ksp of AgCl.” (b) “Once the salt starts dissolving, the common ion’s concentration stays exactly at its starting value C.” (c) “A common ion at 0.10 M always cuts a salt’s solubility by the same factor.” (d) “Adding more solid to a common-ion solution raises the solubility back toward the pure-water value.”
A3. PbI₂ is placed in KI(aq) in which [I⁻] starts at C. If the molar solubility of PbI₂ in this solution is s, state (a) [Pb²⁺] at equilibrium, (b) [I⁻] at equilibrium, (c) Ksp written in terms of s and C. (d) The same salt is placed instead in Pb(NO₃)₂(aq) in which [Pb²⁺] starts at C. State [Pb²⁺] and [I⁻] at equilibrium, and Ksp in terms of s and C.
A4. Solid NaCl is stirred into a saturated AgCl solution that already has solid AgCl present. (a) State what happens to Qsp at the moment of addition, and why. (b) State what the system then does. (c) State what happens to [Ag⁺], and what [Cl⁻] ends at compared with its value before the addition. (d) State the value Qsp has when the changes stop.
A5. The molar solubility of AgCl is 1.3 × 10⁻⁵ M in pure water, 1.8 × 10⁻⁸ M in 0.010 M NaCl, and 1.8 × 10⁻⁹ M in 0.10 M NaCl. (a) Rank the three solutions by solubility. (b) For this 1:1 salt, state the factor by which s falls when [Cl⁻] rises by a factor of ten, and connect it to the form of the approximate equation s ≈ Ksp ÷ C. (c) Predict the molar solubility of AgCl in 0.20 M NaCl. (d) State what happens to the solubility of AgCl in 0.10 M NaCl if the solution is diluted with an equal volume of pure water.
Section B: Core Practice
B1. Ksp for AgCl is 1.8 × 10⁻¹⁰. (a) Write [Ag⁺] and [Cl⁻] in terms of s for AgCl dissolving in 0.050 M NaCl, and hence Ksp in terms of s. (b) Calculate s using the small-s approximation. (c) Run the check on the approximation. (d) By what factor is the solubility below its pure-water value of 1.3 × 10⁻⁵ M? (e) State [Cl⁻] at equilibrium.
B2. Ksp for PbI₂ is 9.8 × 10⁻⁹. (a) Write Ksp in terms of s for PbI₂ dissolving in 0.050 M KI. (b) Calculate s using the small-s approximation. (c) Run the check on the approximation. (d) State [Pb²⁺] and [I⁻] at equilibrium. (e) State the factor by which the solubility falls short of the pure-water value of 1.3 × 10⁻³ M.
B3. Ksp for BaSO₄ is 1.1 × 10⁻¹⁰. Calculate the molar solubility of BaSO₄ in 1.0 × 10⁻³ M Na₂SO₄ (a) by the small-s approximation, (b) exactly, by solving the quadratic, and (c) state the factor by which it falls short of the pure-water value of 1.0 × 10⁻⁵ M. (d) Run the check on the approximation. (e) State [SO₄²⁻] at equilibrium.
B4. Ksp for Mg(OH)₂ is 5.6 × 10⁻¹². (a) Write Ksp in terms of s for Mg(OH)₂ dissolving in 0.10 M NaOH. (b) Calculate s using the small-s approximation. (c) Run the check on the approximation. (d) State the factor by which the solubility falls short of the pure-water value of 1.1 × 10⁻⁴ M. (e) State [OH⁻] at equilibrium.
B5. Ksp for Ag₂CrO₄ is 1.1 × 10⁻¹². Ag₂CrO₄ is dissolved in 0.010 M K₂CrO₄, so this time the common ion is CrO₄²⁻. (a) Write [Ag⁺] and [CrO₄²⁻] in terms of s, and hence Ksp in terms of s. (b) Calculate s using the small-s approximation. (c) Run the check on the approximation. (d) State the factor by which the solubility falls short of the pure-water value of 6.5 × 10⁻⁵ M. (e) For comparison, Ag₂CrO₄ in 0.010 M AgNO₃ has s = 1.1 × 10⁻⁸ M. State which common ion suppresses the solubility more strongly at the same concentration, and explain the difference using the powers in the two Ksp expressions.
B6. Ksp for PbCl₂ is 1.6 × 10⁻⁵, and its molar solubility in pure water is 1.6 × 10⁻² M. PbCl₂ is placed in 0.010 M NaCl. (a) Write Ksp in terms of s and calculate the value the small-s approximation returns. (b) Show two separate reasons why that value cannot be trusted. (c) The exact root is s = 1.27 × 10⁻² M. Verify it by back-substitution. (d) State the factor by which the common ion suppresses the solubility. (e) Explain why the suppression is so weak here.
Section C: Exam Style
C1. A saturated AgCl solution, with [Ag⁺] = [Cl⁻] = 1.3 × 10⁻⁵ M and solid present, is mixed with an equal volume of 0.20 M NaCl. Ksp for AgCl is 1.8 × 10⁻¹⁰. (a) State [Ag⁺] and [Cl⁻] immediately after mixing, before any shift. (b) Calculate Qsp at that moment and state what happens next. (c) State [Ag⁺] and [Cl⁻] once equilibrium is reestablished. (d) Calculate how many moles of AgCl precipitate per litre of the mixture. (e) Calculate the mass of AgCl that precipitates per litre of the mixture, given its molar mass is 143.32 g/mol.
C2. Ksp for Ca(OH)₂ is 5.5 × 10⁻⁶. Ca(OH)₂ is placed in 0.10 M NaOH. (a) Write Ksp in terms of s and calculate s using the small-s approximation. (b) Run the 5 percent check. (c) The exact root is s = 5.4 × 10⁻⁴ M. State the percentage error of the approximation and whether its direction matches the known bias of the method. (d) The solubility of Ca(OH)₂ in pure water is 1.1 × 10⁻² M. State the suppression factor. (e) State [OH⁻] at equilibrium, using the exact root.
C3. AgCl in 0.10 M NaCl falls from 1.3 × 10⁻⁵ M to 1.8 × 10⁻⁹ M, a factor of about 7,000; CaF₂ in 0.010 M NaF falls from 2.2 × 10⁻⁴ M to 4.0 × 10⁻⁷ M, a factor of about 550. In pure saturated AgCl, [Cl⁻] = 1.3 × 10⁻⁵ M; in pure saturated CaF₂, [F⁻] = 4.4 × 10⁻⁴ M. (a) Confirm both factors from the solubilities given. (b) Show that each factor is close to (C divided by the common ion’s concentration in the pure saturated solution) raised to the power of that ion in the Ksp expression. (c) Explain why AgCl is suppressed more strongly in the comparison that matters, even though its C is larger in absolute terms. (d) Predict the molar solubility of CaF₂ in 0.10 M NaF, and the suppression factor against pure water. (e) State which salt’s suppression grows faster as C rises, and why.
Show the answer key and full working (only after you have attempted every question)
A1. (a) None: NaCl supplies Na⁺ and Cl⁻, while the AgBr equilibrium involves Ag⁺ and Br⁻. (b) SO₄²⁻, present in both the added Na₂SO₄ and the BaSO₄ dissolution. (c) None: K⁺ and NO₃⁻ appear nowhere in the PbI₂ equilibrium. (d) Ag⁺, supplied by AgNO₃ and released by Ag₂CrO₄. (e) OH⁻, supplied by NaOH and released by Mg(OH)₂.
A2. (a) False. Ksp is a constant at a fixed temperature and does not respond to added solute. The addition raises Qsp above Ksp; the solubility falls because the concentrations change. (b) False. The common ion’s equilibrium concentration is C plus whatever the dissolving salt adds: C + s for Cl⁻ from AgCl, C + 2s for I⁻ from PbI₂. (c) False. The factor depends on C relative to the ion levels the salt produces alone, and on the power of the common ion: 0.10 M of Cl⁻ cuts AgCl’s solubility about 7,000-fold, while 0.010 M of F⁻ cuts CaF₂’s only about 550-fold. (d) False. A saturated solution cannot dissolve more solid at the same temperature; s is fixed by Ksp and C.
A3. (a) [Pb²⁺] = s: lead ion comes only from the dissolving salt, one per formula unit. (b) [I⁻] = C + 2s: the solution starts with C, and each formula unit adds two more. (c) Ksp = [Pb²⁺][I⁻]² = s(C + 2s)². The multiplier of 2 belongs to s before the square is applied. (d) [Pb²⁺] = C + s and [I⁻] = 2s, so Ksp = (C + s)(2s)² = 4s²(C + s). The common ion sits in a different place, so the math changes with it.
A4. (a) Qsp jumps above Ksp: the added NaCl raises [Cl⁻] sharply while [Ag⁺] is momentarily unchanged, so [Ag⁺][Cl⁻] exceeds 1.8 × 10⁻¹⁰. (b) The system precipitates AgCl: Ag⁺ and Cl⁻ leave the solution together as solid, in a 1:1 ratio. (c) [Ag⁺] falls steeply, by orders of magnitude. [Cl⁻] falls slightly but ends far above its value before the addition. (d) Precipitation stops when Qsp has fallen back to exactly Ksp, with solid AgCl present, but at a much lower [Ag⁺].
A5. (a) Pure water (1.3 × 10⁻⁵ M), then 0.010 M NaCl (1.8 × 10⁻⁸ M), then 0.10 M NaCl (1.8 × 10⁻⁹ M). (b) Each tenfold rise in [Cl⁻] cuts s tenfold: with s ≈ Ksp ÷ C, multiplying C by 10 divides s by 10. That clean inverse pattern is special to the 1:1 case; where the common ion is squared, as for F⁻ in CaF₂, s falls with 1 ÷ C² instead. (c) s = Ksp ÷ C = (1.8 × 10⁻¹⁰) ÷ 0.20 = 9.0 × 10⁻¹⁰ M. (d) Dilution halves [Cl⁻] to 0.050 M, so s rises to (1.8 × 10⁻¹⁰) ÷ 0.050 = 3.6 × 10⁻⁹ M: more solid dissolves.
B1. (a) [Ag⁺] = s and [Cl⁻] = 0.050 + s, so Ksp = s(0.050 + s). (b) Neglecting s beside 0.050: s ≈ (1.8 × 10⁻¹⁰) ÷ 0.050 = 3.6 × 10⁻⁹ M. (c) The neglected term is s itself, 3.6 × 10⁻⁹ against C = 0.050: far under 5 percent. (d) Factor: (1.3 × 10⁻⁵) ÷ (3.6 × 10⁻⁹) = 3,600. (e) [Cl⁻] = 0.050 + s ≈ 0.050 M.
B2. (a) [Pb²⁺] = s and [I⁻] = 0.050 + 2s, so Ksp = s(0.050 + 2s)². (b) Neglecting 2s beside 0.050: s ≈ (9.8 × 10⁻⁹) ÷ (0.050)² = (9.8 × 10⁻⁹) ÷ (2.5 × 10⁻³) = 3.9 × 10⁻⁶ M. (c) 2s = 7.8 × 10⁻⁶, 0.016 percent of 0.050: the check passes. (d) [Pb²⁺] = 3.9 × 10⁻⁶ M and [I⁻] = 0.050 + 2s ≈ 0.050 M. (e) Factor: (1.3 × 10⁻³) ÷ (3.9 × 10⁻⁶) = 330.
B3. (a) Approximately: Ksp = s(1.0 × 10⁻³ + s); neglecting s gives s ≈ (1.1 × 10⁻¹⁰) ÷ (1.0 × 10⁻³) = 1.1 × 10⁻⁷ M. (b) Exactly: s² + (1.0 × 10⁻³)s − 1.1 × 10⁻¹⁰ = 0, whose positive root is s = 1.1 × 10⁻⁷ M. (c) Factor: (1.0 × 10⁻⁵) ÷ (1.1 × 10⁻⁷) = 91. (d) The neglected term s = 1.1 × 10⁻⁷ M is 0.011 percent of C, far under the 5 percent limit. (e) [SO₄²⁻] at equilibrium is 1.0 × 10⁻³ + s ≈ 1.0 × 10⁻³ M.
B4. (a) [Mg²⁺] = s and [OH⁻] = 0.10 + 2s, so Ksp = s(0.10 + 2s)². (b) Neglecting 2s beside 0.10: s ≈ (5.6 × 10⁻¹²) ÷ (0.10)² = (5.6 × 10⁻¹²) ÷ 0.010 = 5.6 × 10⁻¹⁰ M. (c) 2s = 1.1 × 10⁻⁹, negligible beside 0.10. (d) Factor: (1.1 × 10⁻⁴) ÷ (5.6 × 10⁻¹⁰) = 2.0 × 10⁵. The tiny Ksp is the cause: in pure water Mg(OH)₂ produces only about 10⁻⁴ M of ions. (e) [OH⁻] = 0.10 + 2s ≈ 0.10 M.
B5. (a) The chromate starts with the head start: [Ag⁺] = 2s and [CrO₄²⁻] = 0.010 + s, so Ksp = (2s)²(0.010 + s) = 4s²(0.010 + s). (b) Neglecting s beside 0.010: 4s²(0.010) = 1.1 × 10⁻¹², so s² = (1.1 × 10⁻¹²) ÷ 0.040 = 2.75 × 10⁻¹¹ and s = 5.2 × 10⁻⁶ M. (c) The neglected term is s = 5.2 × 10⁻⁶ against C = 0.010, which is 0.052 percent: the check passes. (d) Factor: (6.5 × 10⁻⁵) ÷ (5.2 × 10⁻⁶) = 13. (e) The Ag⁺ common ion suppresses far more strongly: s = 1.1 × 10⁻⁸ M against 5.2 × 10⁻⁶ M here. With Ag⁺, Ksp ≈ C²s and s shrinks with 1 ÷ C². With CrO₄²⁻, Ksp ≈ 4s²C and s shrinks with 1 ÷ √C.
B6. (a) [Pb²⁺] = s and [Cl⁻] = 0.010 + 2s, so Ksp = s(0.010 + 2s)². The approximation replaces (0.010 + 2s) by 0.010: s ≈ (1.6 × 10⁻⁵) ÷ (0.010)² = (1.6 × 10⁻⁵) ÷ (1.0 × 10⁻⁴) = 0.16 M. (b) First, 2s would be 0.32 M, thirty-two times the C it was supposed to be negligible beside. Second, 0.16 M exceeds PbCl₂’s solubility in pure water, 1.6 × 10⁻² M. (c) With s = 1.27 × 10⁻²: [Cl⁻] = 0.010 + 2(1.27 × 10⁻²) = 0.0354 M, and Ksp = (1.27 × 10⁻²)(0.0354)² = 1.6 × 10⁻⁵. Confirmed. (d) Factor: (1.6 × 10⁻²) ÷ (1.27 × 10⁻²) = 1.3. (e) The suppression is feeble because PbCl₂’s own dissolving supplies 0.025 M of Cl⁻, two and a half times the added head start.
C1. (a) Mixing equal volumes halves each concentration: [Ag⁺] = 6.5 × 10⁻⁶ M and [Cl⁻] = 0.10 M. (b) Qsp = (6.5 × 10⁻⁶)(0.10) = 6.5 × 10⁻⁷, far above Ksp = 1.8 × 10⁻¹⁰, so AgCl precipitates. (c) Precipitation removes Ag⁺ and Cl⁻ in equal amounts; [Cl⁻] barely moves, so [Ag⁺] settles at (1.8 × 10⁻¹⁰) ÷ 0.10 = 1.8 × 10⁻⁹ M, with [Cl⁻] ≈ 0.10 M. (d) Silver precipitated per litre = 6.5 × 10⁻⁶ − 1.8 × 10⁻⁹ ≈ 6.5 × 10⁻⁶ mol. (e) Mass per litre = (6.5 × 10⁻⁶ mol)(143.32 g/mol) = 9.3 × 10⁻⁴ g.
C2. (a) [Ca²⁺] = s and [OH⁻] = 0.10 + 2s, so Ksp = s(0.10 + 2s)². Neglecting 2s: s ≈ (5.5 × 10⁻⁶) ÷ (0.10)² = (5.5 × 10⁻⁶) ÷ 0.010 = 5.5 × 10⁻⁴ M. (b) 2s = 1.1 × 10⁻³, 1.1 percent of 0.10: inside the 5 percent limit. (c) Error: (5.5 × 10⁻⁴ − 5.4 × 10⁻⁴) ÷ (5.4 × 10⁻⁴) = 1.9 percent, about 2 percent, and the approximation is the larger value, exactly the known bias. (d) Factor: (1.1 × 10⁻²) ÷ (5.4 × 10⁻⁴) = 20. (e) [OH⁻] = 0.10 + 2(5.4 × 10⁻⁴) = 0.10 + 1.1 × 10⁻³ = 0.101 M.
C3. (a) AgCl: (1.3 × 10⁻⁵) ÷ (1.8 × 10⁻⁹) = 7,200. CaF₂: (2.2 × 10⁻⁴) ÷ (4.0 × 10⁻⁷) = 550. (b) For AgCl: C ÷ [Cl⁻] in pure solution = 0.10 ÷ (1.3 × 10⁻⁵) = 7,700, to the power 1, giving about 7,700, close to 7,200. For CaF₂: C ÷ [F⁻] in pure solution = 0.010 ÷ (4.4 × 10⁻⁴) = 23, squared, giving about 520, close to 550. (c) What matters is not C itself but C measured against the ion level the salt reaches on its own, raised to the ion’s power. AgCl’s head start is thousands of times the level AgCl produces alone; CaF₂’s smaller head start is only about twenty-three times its own fluoride level, and even squared it suppresses less. (d) In 0.10 M NaF, s = Ksp ÷ C² = (4.0 × 10⁻¹¹) ÷ (0.10)² = (4.0 × 10⁻¹¹) ÷ 0.010 = 4.0 × 10⁻⁹ M. The suppression factor is (2.2 × 10⁻⁴) ÷ (4.0 × 10⁻⁹) = 55,000. (e) CaF₂’s suppression grows faster as C rises, because its common ion is squared: s falls with 1 ÷ C², while AgCl’s s falls only with 1 ÷ C.