Predict, using Le Chatelier's principle, the effect on the solubility of PbI_2 of adding KI to the solution. [2 points]

College Board AP 2021 (style)-style practice: Section I (multiple choice). Adding NaF to a saturated solution of CaF_2 will (A) increase the solubility of CaF_2 (B) decrease the solubility of CaF_2 (C) not affect the solubility (D) increase K_sp. Justify your choice.

A 1-point conceptual MCQ. The answer is (B). F^- from NaF is a common ion; it shifts the CaF_2 dissolution equilibrium toward the solid, decreasing solubility. K_sp itself is unchanged (only temperature changes it). The trap is (D): the common ion changes the solubility, not K_sp.

United StatesChemistrySyllabus dot point

Why does the solubility of a salt decrease when a common ion is already present in solution?

Topic 7.12 Common-Ion Effect: explain and calculate the reduced solubility of a salt in a solution that already contains one of its ions, using Le Chatelier's principle and Ksp.

A focused answer to AP Chemistry Topic 7.12, covering the common-ion effect, why a shared ion lowers solubility, and how to calculate the reduced molar solubility using an ICE table and Ksp, with full worked examples.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The common-ion effect is the decrease in the solubility of a salt when the solution already contains one of the salt's ions. By Le Chatelier's principle, the extra common ion shifts the dissolution equilibrium toward the solid, so less salt dissolves. The solubility product $K_\text{sp}$ is unchanged (only temperature changes it), but the molar solubility falls. To calculate the reduced solubility, set up an ICE table in which the initial concentration of the common ion is the amount already present (from the other source), let the salt dissolve by $s$ , and substitute into $K_\text{sp}$ . Because the common ion is usually present in much larger amount than the salt would supply, you can often approximate its concentration as the amount added, which simplifies the algebra.

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What this topic is asking
What the common-ion effect is
Why solubility decreases (Le Chatelier)
Calculating the reduced solubility
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What this topic is asking

The College Board (Topic 7.12) wants you to explain and calculate the reduced solubility of a salt in a solution that already contains one of its ions, using Le Chatelier's principle and $K_\text{sp}$ . This is a direct application of equilibrium shifts to solubility: a shared ion suppresses dissolving.

What the common-ion effect is

For example, silver chloride dissolves less in a sodium chloride solution than in pure water, because the chloride ion is common to both. The added chloride pushes the equilibrium $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$ back toward the solid.

Why solubility decreases (Le Chatelier)

This is the Q-versus- $K_\text{sp}$ logic in action: adding the common ion raises the ion product $Q$ above $K_\text{sp}$ , so the salt precipitates (the equilibrium shifts left) until $Q = K_\text{sp}$ again, leaving less dissolved. The constant $K_\text{sp}$ is fixed, but the balance of dissolved versus solid shifts.

Calculating the reduced solubility

Set up an ICE table where the initial concentration of the common ion is the amount already present from the other source, not zero. Let the salt dissolve by $s$ , add the contributions to each ion, and substitute the equilibrium concentrations into $K_\text{sp}$ . Because the common ion is usually present in far greater amount than the salt alone would give, its equilibrium concentration is approximately the amount added, so you can treat it as a constant and solve for $s$ directly. Always confirm the salt's own contribution to the common ion is negligible.

Solubility with a common ion

Find the molar solubility of $\text{BaSO}_4$ ( $K_\text{sp} = 1.1 \times 10^{-10}$ ) in $0.020$ M $\text{Na}_2\text{SO}_4$ .

step 1 Write the equilibrium and Ksp

$\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+} + \text{SO}_4^{2-}$ ; $K_\text{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]$ .

step 2 Set up the ICE concentrations

Dissolving $s$ gives $[\text{Ba}^{2+}] = s$ and $[\text{SO}_4^{2-}] = 0.020 + s$ (the $0.020$ M is from the $\text{Na}_2\text{SO}_4$ ).

step 3 Approximate the common ion

Since $s$ is tiny, $0.020 + s \approx 0.020$ M.

step 4 Solve for s

$K_\text{sp} = s(0.020) = 1.1 \times 10^{-10}$ , so $s = \dfrac{1.1 \times 10^{-10}}{0.020} = 5.5 \times 10^{-9}$ M, far less than in pure water ( $\sqrt{1.1 \times 10^{-10}} = 1.0 \times 10^{-5}$ M).

Try this

Q1. Predict, using Le Chatelier's principle, the effect on the solubility of $\text{PbI}_2$ of adding KI to the solution. [2 points]

Cue. $\text{I}^-$ is a common ion; it shifts the equilibrium toward the solid, so the solubility of $\text{PbI}_2$ decreases.

Q2. State whether the common-ion effect changes the value of $K_\text{sp}$ . [1 point]

Cue. No; only temperature changes $K_\text{sp}$ . The common ion changes the solubility, not the constant.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). The solubility product of

\text{AgCl}

K_\text{sp} = 1.8 \times 10^{-10}

. (a) Calculate the molar solubility of AgCl in pure water. (b) Predict, using Le Chatelier's principle, how the solubility changes in

0.10

M NaCl. (c) Calculate the molar solubility of AgCl in

0.10

M NaCl. (d) Justify why the answer in (c) is smaller than in (a).

Show worked answer →

A 4-point quantitative FRQ on the common-ion effect.

(a) In pure water (1 point): $\text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$ ; $K_\text{sp} = s^2 = 1.8 \times 10^{-10}$ , so $s = 1.3 \times 10^{-5}$ M.
(b) Le Chatelier prediction (1 point): the added $\text{Cl}^-$ is a common ion; it shifts the dissolution equilibrium left, so the solubility decreases.
(c) In 0.10 M NaCl (1 point): $[\text{Cl}^-] \approx 0.10$ M (the small amount from AgCl is negligible); $K_\text{sp} = s(0.10) = 1.8 \times 10^{-10}$ , so $s = 1.8 \times 10^{-9}$ M.
(d) Justify (1 point): the high $[\text{Cl}^-]$ from NaCl drives the equilibrium toward the solid (Le Chatelier), so far less AgCl dissolves; the solubility is about $10^{4}$ times smaller.

Markers reward the pure-water solubility, the Le Chatelier prediction, the reduced solubility with the common ion, and the reasoning for the decrease.

AP 2021 (style)1 marksSection I (multiple choice). Adding

\text{NaF}

to a saturated solution of

\text{CaF}_2

will (A) increase the solubility of

\text{CaF}_2

(B) decrease the solubility of

\text{CaF}_2

K_\text{sp}

. Justify your choice.

Show worked answer →

A 1-point conceptual MCQ. The answer is (B).

$\text{F}^-$ from NaF is a common ion; it shifts the $\text{CaF}_2$ dissolution equilibrium toward the solid, decreasing solubility. $K_\text{sp}$ itself is unchanged (only temperature changes it). The trap is (D): the common ion changes the solubility, not $K_\text{sp}$ .

Related dot points

Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)