Topic 7.12: Common-Ion Effect Quiz

Master AP Chemistry Topic 7.12 with this free practice quiz. Test your skills on common ion effect, decreased solubility, ice tables with initial concentrations, and shifts in solubility to prepare for your AP exam.

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Topic 7.12: Common-Ion Effect Quiz

18 MCQs

Master the common-ion effect with these 18 MCQs on AP Chemistry Topic 7.12: Common-Ion Effect.

Covered Subtopics: Common-ion effect, solubility calculations, identifying common ions

1 / 18

1. Solid NaCl is added to a saturated AgCl solution. In terms of Qsp and Ksp, why does some AgCl precipitate?

Added Na⁺ makes Qsp exceed Ksp, so solid forms until Qsp exceeds Ksp

Added Cl⁻ makes Qsp fall below Ksp, so solid forms until Qsp equals Ksp

Added Cl⁻ makes Ksp fall below Qsp, so solid forms until Ksp equals Qsp

Added Cl⁻ makes Qsp exceed Ksp, so solid forms until Qsp equals Ksp

Extra chloride raises Qsp above Ksp, and precipitation of AgCl continues until the ion product falls back to Ksp.

Extra chloride raises Qsp above Ksp, and precipitation of AgCl continues until the ion product falls back to Ksp.

Extra chloride raises Qsp above Ksp, and precipitation of AgCl continues until the ion product falls back to Ksp.

2 / 18

2. A saturated AgCl solution also contains 0.10 M NaCl(aq). Compared with pure water, the solubility of AgCl is:

Lower, because the added Cl⁻ is a common ion

Higher, because the added Cl⁻ is a common ion

Lower, because NaCl reacts directly with solid AgCl

Unchanged, because NaCl does not affect the equilibrium

Chloride is a common ion, so its higher concentration shifts the dissolution equilibrium left and less AgCl dissolves.

Chloride is a common ion, so its higher concentration shifts the dissolution equilibrium left and less AgCl dissolves.

Chloride is a common ion, so its higher concentration shifts the dissolution equilibrium left and less AgCl dissolves.

3 / 18

3. A saturated AgCl solution (Ksp = 1.8 × 10⁻¹⁰) is mixed with an equal volume of 0.20 M NaCl. What is [Ag⁺] after equilibrium is reestablished?

9.0 × 10⁻¹⁰ M

3.6 × 10⁻⁹ M

1.8 × 10⁻⁹ M

1.3 × 10⁻⁵ M

After mixing, [Cl⁻] is 0.10 M, so [Ag⁺] = Ksp/0.10 = 1.8 × 10⁻⁹ M once equilibrium returns.

After mixing, [Cl⁻] is 0.10 M, so [Ag⁺] = Ksp/0.10 = 1.8 × 10⁻⁹ M once equilibrium returns.

After mixing, [Cl⁻] is 0.10 M, so [Ag⁺] = Ksp/0.10 = 1.8 × 10⁻⁹ M once equilibrium returns.

4 / 18

4. HCl(aq) is added to a saturated AgCl solution, and separately HNO₃(aq) is added to an identical one. Which comparison is correct?

Both acids lower solubility equally through the common ion

HNO₃ lowers solubility through the common ion, while HCl has no effect

HCl lowers solubility through the common ion, while HNO₃ has no effect

Both acids raise solubility by reacting with the solid

HCl supplies chloride, a common ion, but nitric acid supplies neither Ag⁺ nor Cl⁻, leaving the equilibrium untouched.

HCl supplies chloride, a common ion, but nitric acid supplies neither Ag⁺ nor Cl⁻, leaving the equilibrium untouched.

HCl supplies chloride, a common ion, but nitric acid supplies neither Ag⁺ nor Cl⁻, leaving the equilibrium untouched.

5 / 18

5. Adding which solution to saturated PbI₂(aq) does NOT change its solubility?

KNO₃(aq), which provides no common ion

Pb(NO₃)₂(aq), which provides the common ion Pb²⁺

NaI(aq), which provides the common ion I⁻

KI(aq), which provides the common ion I⁻

Potassium nitrate supplies neither Pb²⁺ nor I⁻, so Qsp is unaffected and the solubility stays the same.

Potassium nitrate supplies neither Pb²⁺ nor I⁻, so Qsp is unaffected and the solubility stays the same.

Potassium nitrate supplies neither Pb²⁺ nor I⁻, so Qsp is unaffected and the solubility stays the same.

6 / 18

6. For AgCl(s) ⇌ Ag+(aq) + Cl-(aq), adding NaCl(aq) shifts the equilibrium left. Which statement explains this with Q?

Higher [Cl⁻] lowers K itself, so the system shifts left

Higher [Cl⁻] raises Q above K, so the system shifts left

Higher [Na⁺] raises Q above K, so the system shifts right

Higher [Cl⁻] lowers Q below K, so the system shifts left

The added chloride raises the ion product Q above Ksp, so the reverse reaction runs and solid AgCl forms.

The added chloride raises the ion product Q above Ksp, so the reverse reaction runs and solid AgCl forms.

The added chloride raises the ion product Q above Ksp, so the reverse reaction runs and solid AgCl forms.

7 / 18

7. Which pair demonstrates the common-ion effect?

NaCl(aq) added to saturated AgBr(aq)

NaOH(aq) added to saturated AgCl(aq)

KNO₃(aq) added to saturated PbI₂(aq)

Na₂SO₄(aq) added to saturated BaSO₄(aq)

Sulfate appears in both the added salt and the dissolution equilibrium, so it is the common ion that lowers solubility.

Sulfate appears in both the added salt and the dissolution equilibrium, so it is the common ion that lowers solubility.

Sulfate appears in both the added salt and the dissolution equilibrium, so it is the common ion that lowers solubility.

8 / 18

8. Ksp for CaF₂ is 4.0 × 10⁻¹¹. What is the molar solubility of CaF₂ in 0.010 M NaF(aq)?

4.0 × 10⁻⁷ M

4.0 × 10⁻⁹ M

6.3 × 10⁻⁶ M

4.0 × 10⁻¹¹ M

Here Ksp = s(0.010)², so s = (4.0 × 10⁻¹¹)/(1.0 × 10⁻⁴) = 4.0 × 10⁻⁷ M.

Here Ksp = s(0.010)², so s = (4.0 × 10⁻¹¹)/(1.0 × 10⁻⁴) = 4.0 × 10⁻⁷ M.

Here Ksp = s(0.010)², so s = (4.0 × 10⁻¹¹)/(1.0 × 10⁻⁴) = 4.0 × 10⁻⁷ M.

9 / 18

9. AgCl is more soluble in pure water than in 0.010 M NaCl, which in turn is more soluble than in 0.10 M NaCl. What pattern does this show?

Solubility decreases as common-ion concentration increases

Solubility increases as common-ion concentration increases

Solubility is unrelated to common-ion concentration

Solubility first rises and then falls with common-ion concentration

Each increase in chloride concentration pushes the dissolution equilibrium further left, lowering the solubility step by step.

Each increase in chloride concentration pushes the dissolution equilibrium further left, lowering the solubility step by step.

Each increase in chloride concentration pushes the dissolution equilibrium further left, lowering the solubility step by step.

10 / 18

10. Ksp for PbI₂ is 9.8 × 10⁻⁹. What is the molar solubility of PbI₂ in 0.050 M KI(aq)?

1.4 × 10⁻⁴ M

2.0 × 10⁻⁷ M

9.8 × 10⁻⁹ M

3.9 × 10⁻⁶ M

Ksp = s(0.050)² here, so s = (9.8 × 10⁻⁹)/(2.5 × 10⁻³) = 3.9 × 10⁻⁶ M.

Ksp = s(0.050)² here, so s = (9.8 × 10⁻⁹)/(2.5 × 10⁻³) = 3.9 × 10⁻⁶ M.

Ksp = s(0.050)² here, so s = (9.8 × 10⁻⁹)/(2.5 × 10⁻³) = 3.9 × 10⁻⁶ M.

11 / 18

11. Ksp for AgCl is 1.8 × 10⁻¹⁰. What is the molar solubility of AgCl in 0.10 M NaCl(aq)?

1.8 × 10⁻¹⁰ M

1.8 × 10⁻⁹ M

1.3 × 10⁻⁵ M

1.8 × 10⁻¹¹ M

With [Cl⁻] fixed at 0.10 M, Ksp = s(0.10), so s = (1.8 × 10⁻¹⁰)/0.10 = 1.8 × 10⁻⁹ M.

With [Cl⁻] fixed at 0.10 M, Ksp = s(0.10), so s = (1.8 × 10⁻¹⁰)/0.10 = 1.8 × 10⁻⁹ M.

With [Cl⁻] fixed at 0.10 M, Ksp = s(0.10), so s = (1.8 × 10⁻¹⁰)/0.10 = 1.8 × 10⁻⁹ M.

12 / 18

12. Ksp for Ag₂CrO₄ is 1.1 × 10⁻¹². What is its molar solubility in 0.020 M AgNO₃(aq)?

2.8 × 10⁻⁹ M

1.1 × 10⁻¹² M

5.5 × 10⁻¹¹ M

2.8 × 10⁻¹⁰ M

Ksp = (0.020)²(s) approximately, so s = (1.1 × 10⁻¹²)/(4.0 × 10⁻⁴) = 2.8 × 10⁻⁹ M.

Ksp = (0.020)²(s) approximately, so s = (1.1 × 10⁻¹²)/(4.0 × 10⁻⁴) = 2.8 × 10⁻⁹ M.

Ksp = (0.020)²(s) approximately, so s = (1.1 × 10⁻¹²)/(4.0 × 10⁻⁴) = 2.8 × 10⁻⁹ M.

13 / 18

13. Why does the common-ion effect lower solubility, in Le Chatelier terms?

The added ion raises Ksp, so less solid can dissolve

The added ion is a product of dissolution, so equilibrium shifts left

The added ion is a reactant of dissolution, so equilibrium shifts right

The added ion lowers Qsp, so more solid must dissolve

Dissolved ions are products of the dissolution reaction, so adding one drives the equilibrium back toward the solid.

Dissolved ions are products of the dissolution reaction, so adding one drives the equilibrium back toward the solid.

Dissolved ions are products of the dissolution reaction, so adding one drives the equilibrium back toward the solid.

14 / 18

14. Ksp for BaSO₄ is 1.1 × 10⁻¹⁰. What is the molar solubility of BaSO₄ in 1.0 × 10⁻³ M Na₂SO₄(aq)?

1.1 × 10⁻¹⁰ M

1.1 × 10⁻⁷ M

1.1 × 10⁻¹³ M

1.0 × 10⁻⁵ M

With [SO₄²⁻] fixed at 1.0 × 10⁻³ M, s = Ksp/(1.0 × 10⁻³) = 1.1 × 10⁻⁷ M.

With [SO₄²⁻] fixed at 1.0 × 10⁻³ M, s = Ksp/(1.0 × 10⁻³) = 1.1 × 10⁻⁷ M.

With [SO₄²⁻] fixed at 1.0 × 10⁻³ M, s = Ksp/(1.0 × 10⁻³) = 1.1 × 10⁻⁷ M.

15 / 18

15. Which change increases the solubility of AgCl(s) in a solution already containing NaCl(aq)?

Diluting the solution with pure water

Adding more solid AgCl to the solution

Adding AgNO₃ to the solution

Adding more NaCl to the solution

Dilution lowers the chloride concentration, easing the common-ion effect so more AgCl can dissolve.

Dilution lowers the chloride concentration, easing the common-ion effect so more AgCl can dissolve.

Dilution lowers the chloride concentration, easing the common-ion effect so more AgCl can dissolve.

16 / 18

16. Ksp for Ca(OH)₂ is 5.5 × 10⁻⁶. What is the molar solubility of Ca(OH)₂ in 0.10 M NaOH(aq)?

5.5 × 10⁻⁶ M

2.4 × 10⁻² M

5.5 × 10⁻⁵ M

5.5 × 10⁻⁴ M

Ksp = s(0.10)² approximately, so s = (5.5 × 10⁻⁶)/0.010 = 5.5 × 10⁻⁴ M.

Ksp = s(0.10)² approximately, so s = (5.5 × 10⁻⁶)/0.010 = 5.5 × 10⁻⁴ M.

Ksp = s(0.10)² approximately, so s = (5.5 × 10⁻⁶)/0.010 = 5.5 × 10⁻⁴ M.

17 / 18

17. Which added solution most reduces the solubility of AgCl(s)?

0.10 M NaCl, which supplies the common ion Cl⁻

0.20 M NaCl, which supplies the common ion Cl⁻

0.10 M AgNO₃, which supplies the common ion Ag⁺

0.10 M NaNO₃, which supplies no common ion

Solubility falls as common-ion concentration rises, so the most concentrated common-ion solution, 0.20 M NaCl, suppresses solubility the most.

Solubility falls as common-ion concentration rises, so the most concentrated common-ion solution, 0.20 M NaCl, suppresses solubility the most.

Solubility falls as common-ion concentration rises, so the most concentrated common-ion solution, 0.20 M NaCl, suppresses solubility the most.

18 / 18

18. Mg(OH)₂(s) is placed in 0.10 M NaOH(aq) instead of pure water. What happens to its solubility?

It decreases, because OH⁻ is a common ion

It decreases, because Na⁺ reacts with Mg(OH)₂

It stays the same, because NaOH is a strong base

It increases, because OH⁻ is a common ion

Hydroxide from NaOH is a common ion, so the dissolution equilibrium shifts left and less Mg(OH)₂ dissolves.

Hydroxide from NaOH is a common ion, so the dissolution equilibrium shifts left and less Mg(OH)₂ dissolves.

Hydroxide from NaOH is a common ion, so the dissolution equilibrium shifts left and less Mg(OH)₂ dissolves.

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What the Common-Ion Effect Is

The common-ion effect is the drop in solubility of a sparingly soluble salt when the solution already contains one of the salt’s ions. Dissolving AgCl in 0.10 M NaCl gives far less dissolved AgCl than dissolving it in pure water, because chloride is already present. The added ion is called the common ion.

This is a direct application of Le Chatelier’s principle and of the Qsp versus Ksp comparison.

KNO₃ added to saturated PbI₂ changes nothing, because neither K⁺ nor NO₃⁻ appears in the dissolution equilibrium. Only a shared ion matters.

Why Solubility Falls: Two Views of One Event

The Le Chatelier view. Write the dissolution with the ions as products:

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

Adding NaCl pours extra Cl⁻, a product, into the system. The equilibrium shifts left toward the solid, so less AgCl dissolves. The same logic works for any salt: Mg(OH)₂ shifts left when NaOH supplies extra OH⁻.

The Qsp view. The moment the common ion is added, Qsp jumps above Ksp, and solid precipitates until Qsp falls back to Ksp, at a lower dissolved concentration of the salt’s other ion.

Calculating Solubility with a Common Ion

Let s be the molar solubility of the salt in the common-ion solution, and let C be the given concentration of the common ion. Because s is tiny compared with C, the common ion concentration is treated as fixed at C. Then solve Ksp = s × Cⁿ, where n is the exponent on the common ion.

A 1:1 salt. Ksp for AgCl is 1.8 × 10⁻¹⁰ in 0.10 M NaCl. Approximating 0.10 + s as 0.10, s = Ksp ÷ 0.10 = 1.8 × 10⁻⁹ M, against 1.3 × 10⁻⁵ M in pure water: a factor of about 7,000.

A squared common ion. Ksp for CaF₂ is 4.0 × 10⁻¹¹ in 0.010 M NaF. Here Ksp = [Ca²⁺][F⁻]² = (s)(0.010)², so s = (4.0 × 10⁻¹¹) ÷ (1.0 × 10⁻⁴) = 4.0 × 10⁻⁷ M. The fluoride concentration is squared because its coefficient is 2. Forgetting that exponent is the most common calculation error in this topic.

A common cation. Ksp for Ag₂CrO₄ is 1.1 × 10⁻¹² in 0.020 M AgNO₃. Then Ksp = [Ag⁺]²[CrO₄²⁻] = (0.020)²(s), so s = (1.1 × 10⁻¹²) ÷ (4.0 × 10⁻⁴) = 2.8 × 10⁻⁹ M.

A 1:2 hydroxide. Ksp for Ca(OH)₂ is 5.5 × 10⁻⁶ in 0.10 M NaOH. Then Ksp = (s)(0.10)², giving s = (5.5 × 10⁻⁶) ÷ 0.010 = 5.5 × 10⁻⁴ M.

Checking the Approximation

The shortcut is valid when s is much smaller than C. After solving, glance at the numbers: in the AgCl example s = 1.8 × 10⁻⁹ M against C = 0.10 M, so the approximation is excellent. Treating the common-ion concentration as C alone understates it slightly, so the approximate s comes out slightly too large.

When the shortcut breaks. If a question gives a very dilute common-ion solution, such as 1.0 × 10⁻⁶ M NaCl with AgCl, s = Ksp ÷ C would give 1.8 × 10⁻⁴ M, not much smaller than C, so the full expression Ksp = s(C + s) would be needed.

How Much the Common Ion Matters

Solubility falls as the common-ion concentration rises, and the relationship is inverse. AgCl is more soluble in 0.010 M NaCl than in 0.10 M NaCl, because s = Ksp ÷ C. In 0.010 M NaCl, s = 1.8 × 10⁻⁸ M; in 0.10 M NaCl, s = 1.8 × 10⁻⁹ M. Doubling the common-ion concentration roughly halves the solubility for a 1:1 salt. For salts where the common ion is squared, the effect is stronger: doubling C cuts s by a factor of four. Dilution works the other way: adding pure water lowers C and lets more solid dissolve.

The power of the common ion decides how hard it bites. PbI₂ in 0.010 M NaI is suppressed through iodide: s = Ksp ÷ (0.010)² = 9.8 × 10⁻⁵ M. The same salt in 0.010 M Pb(NO₃)₂ is suppressed through lead: Ksp = (0.010)(4s²), so s = √(Ksp ÷ (4 × 0.010)) = 4.9 × 10⁻⁴ M. The same salt, the same added concentration, but different math because the common ion sits in a different place in the Ksp expression. Always write the full expression first, then decide which terms are negligible.

Distinguishing Real Common Ions

HCl lowers the solubility of AgCl because it supplies Cl⁻, but HNO₃ does not, because it supplies neither Ag⁺ nor Cl⁻. Na₂SO₄ lowers the solubility of BaSO₄ through sulfate, while NaCl leaves BaSO₄ alone.

Watch for polyatomic ions that look similar but are not identical. Adding Na₂SO₄ to saturated BaSO₃ would do nothing, because sulfite and sulfate are different ions.

Common Ion from Two Sources at Once

Sometimes the common ion comes from more than one added salt: AgCl in a solution 0.050 M in NaCl and 0.050 M in KCl has total [Cl⁻] = 0.10 M, so s = Ksp ÷ 0.10 = 1.8 × 10⁻⁹ M. Always add up every source of the common ion before solving. The same care applies when a common-ion solution is prepared by mixing: a saturated AgCl solution mixed with an equal volume of 0.20 M NaCl ends with [Cl⁻] = 0.10 M, so [Ag⁺] = 1.8 × 10⁻⁹ M once equilibrium is reestablished.

Worked Examples

Example 1: AgCl in 0.10 M NaCl. Ksp = [Ag⁺][Cl⁻] = (s)(0.10) = 1.8 × 10⁻¹⁰, so s = (1.8 × 10⁻¹⁰) ÷ 0.10 = 1.8 × 10⁻⁹ M. In pure water s = 1.3 × 10⁻⁵ M, a factor of about 7,000 higher.

Example 2: CaF₂ in 0.010 M NaF. Ksp = [Ca²⁺][F⁻]² = (s)(0.010)² = 4.0 × 10⁻¹¹, so s = (4.0 × 10⁻¹¹) ÷ (1.0 × 10⁻⁴) = 4.0 × 10⁻⁷ M. In pure water s = 2.2 × 10⁻⁴ M, a factor of about 550 higher.

Example 3: Ag₂CrO₄ in 0.020 M AgNO₃. Ksp = [Ag⁺]²[CrO₄²⁻] = (0.020)²(s) = 1.1 × 10⁻¹², so s = (1.1 × 10⁻¹²) ÷ (4.0 × 10⁻⁴) = 2.8 × 10⁻⁹ M. In pure water s = 6.5 × 10⁻⁵ M, a factor of about 2.3 × 10⁴ higher.

Example 4: Ca(OH)₂ in 0.10 M NaOH. Ksp = [Ca²⁺][OH⁻]² = (s)(0.10)² = 5.5 × 10⁻⁶, so s = (5.5 × 10⁻⁶) ÷ 0.010 = 5.5 × 10⁻⁴ M. In pure water s = 1.1 × 10⁻² M, a factor of about 20 higher.

Traps That Cost Marks in Topic 7.12

“Adding a common ion lowers Ksp.” Ksp is constant at a given temperature; only the solubility changes.

“For CaF₂ in NaF, Ksp = s(C + s)².” The fluoride released by the salt is 2s, not s: Ksp = s(C + 2s)².

“The small-s approximation is always safe.” It fails whenever C is not far above the dissolved amount. Run the check every time.

“Use the pure-water solubility out of habit.” In a common-ion solution the solubility must be solved fresh from Ksp = s(C)ⁿ.

“Any added salt is a common ion.” Only a salt that shares an ion with the dissolving solid counts.

“Adding more solid raises the solubility back toward pure water.” A saturated solution cannot dissolve more solid at the same temperature however much is added.

Key Relationships

• A common ion is an ion shared between the sparingly soluble salt and another solute in the same solution
• 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; it changes the ion concentrations at which the Ksp product is reached
• 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 stoichiometric 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
• If an addition pushes Qsp above Ksp, solid precipitates until Qsp returns to Ksp
• Solubility changes caused by pH belong to Topic 7.13; in this topic only the shared ion changes the solubility