Solubility Equilibrium: Ksp, Common-Ion Effect & Precipitation

Calculate Ksp and predict precipitation

Written and reviewed by the Study Mondo Education TeamLast updated April 28, 2026

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Solubility Equilibrium: Ksp, Common-Ion Effect & Precipitation

A "sparingly soluble" ionic solid placed in water reaches a dynamic equilibrium between dissolution and precipitation. We describe that equilibrium with a special equilibrium constant called $K_{sp}$ (the solubility product). Mastering $K_{sp}$ lets you (1) calculate molar solubility, (2) predict whether a precipitate will form when two solutions are mixed, (3) explain why solubility drops in the presence of a common ion, (4) explain why pH affects the solubility of certain salts, and (5) design selective precipitation schemes — all skills tested directly on the AP Chemistry exam (Unit 7 equilibrium and Unit 8 acid–base extensions).

1. $K_{sp}$ Expressions — the foundation

For a generic salt $A_p B_q(s) \rightleftharpoons p\,A^{m+}(aq) + q\,B^{n-}(aq)$ ,

$K_{sp} = [A^{m+}]^{p}\,[B^{n-}]^{q}.$

Pure solids and pure liquids do not appear in the expression. Examples:

$\text{AgCl}(s) \rightleftharpoons \text{Ag}^{+} + \text{Cl}^{-}$ : $K_{sp} = [\text{Ag}^{+}][\text{Cl}^{-}]$
$\text{PbCl}_{2}(s) \rightleftharpoons \text{Pb}^{2+} + 2\,\text{Cl}^{-}$ : $K_{sp} = [\text{Pb}^{2+}][\text{Cl}^{-}]^{2}$
$\text{Ca}_{3}(\text{PO}_{4})_{2}(s) \rightleftharpoons 3\,\text{Ca}^{2+} + 2\,\text{PO}_{4}^{3-}$ : $K_{sp} = [\text{Ca}^{2+}]^{3}[\text{PO}_{4}^{3-}]^{2}$

The smaller the $K_{sp}$ , the less soluble the salt. But you can only directly compare solubilities of two salts by $K_{sp}$ if they have the same stoichiometry — otherwise compute molar solubility first.

2. Molar Solubility from $K_{sp}$

Let $s$ = molar solubility (mol/L of salt that dissolves to saturate the solution).

For $\text{AgCl}$ : $[\text{Ag}^{+}] = s$ , $[\text{Cl}^{-}] = s$ → $K_{sp} = s^{2}$ , so $s = \sqrt{K_{sp}}$ .

For $\text{PbCl}_{2}$ : $[\text{Pb}^{2+}] = s$ , $[\text{Cl}^{-}] = 2s$ → $K_{sp} = s(2s)^{2} = 4s^{3}$ , so $s = \sqrt[3]{K_{sp}/4}$ .

For $\text{Ca}_{3}(\text{PO}_{4})_{2}$ : $K_{sp} = (3s)^{3}(2s)^{2} = 108\,s^{5}$ .

AP tip. Always start with the dissolution equation, then read off the stoichiometric coefficients to set the ion concentrations in terms of $s$ . Most " $K_{sp}$ vs. solubility" wrong answers come from forgetting the coefficient ( $2s$ , not $s$ ) or failing to raise to the right power.

3. The Common-Ion Effect

By Le Châtelier's principle, adding an ion already present in the dissolution equilibrium shifts the equilibrium back toward the solid, decreasing solubility.

Example. $K_{sp}(\text{AgCl}) = 1.8 \times 10^{-10}$ . In pure water, $s = \sqrt{K_{sp}} = 1.3 \times 10^{-5}$ M.

In 0.10 M NaCl, $[\text{Cl}^{-}]$ is essentially fixed at 0.10 M (the trace from AgCl is negligible). So $[\text{Ag}^{+}] = K_{sp}/[\text{Cl}^{-}] = (1.8 \times 10^{-10})/(0.10) = 1.8 \times 10^{-9}\,\text{M}.$ The molar solubility crashes from $1.3 \times 10^{-5}$ M to $1.8 \times 10^{-9}$ M — about four orders of magnitude lower.

The common-ion effect underlies why "salting out" works in many separations.

4. $Q$ vs. $K_{sp}$ — Will a Precipitate Form?

The reaction quotient for a dissolution uses the same expression as $K_{sp}$ but plugged with current (not equilibrium) ion concentrations.

Comparison	Meaning
$Q < K_{sp}$	Unsaturated. More solid can dissolve. No precipitate.
$Q = K_{sp}$	Saturated equilibrium. Solid and dissolved ions in balance.
$Q > K_{sp}$	Supersaturated. Precipitate forms until $Q$ falls back to $K_{sp}$ .

Worked example. Mix 50 mL of 0.0010 M $\text{Pb(NO}_{3})_{2}$ with 50 mL of 0.020 M NaI. $K_{sp}(\text{PbI}_{2}) = 9.8 \times 10^{-9}$ .

After mixing (volumes add → each is diluted by ½):

$[\text{Pb}^{2+}] = 5.0 \times 10^{-4}$ M
$[\text{I}^{-}] = 0.010$ M
$Q = (5.0 \times 10^{-4})(0.010)^{2} = 5.0 \times 10^{-8}$

Since $Q\;(5.0 \times 10^{-8}) > K_{sp}\;(9.8 \times 10^{-9})$ , PbI₂ precipitates.

5. pH Effects on Solubility

A salt becomes more soluble in acidic solution if its anion is a conjugate base of a weak acid (or if its cation reacts with OH⁻ in basic solution).

Examples.

$\text{CaF}_{2}$ is more soluble in acidic solution. The $\text{F}^{-}$ ion reacts with $\text{H}^{+}$ to form HF (a weak acid), removing $\text{F}^{-}$ from solution and pulling the dissolution equilibrium right.
$\text{Mg(OH)}_{2}$ is more soluble in acidic solution (acid neutralizes OH⁻) and less soluble in basic solution (added OH⁻ is a common ion).
$\text{AgCl}$ solubility is not noticeably affected by pH because $\text{Cl}^{-}$ is the conjugate base of a strong acid (HCl) and doesn't react with $\text{H}^{+}$ .

Rule of thumb. Salts of weak-acid anions (F⁻, OH⁻, S²⁻, $\text{CO}_{3}^{2-}$ , $\text{PO}_{4}^{3-}$ , …) become more soluble at low pH. Salts of strong-acid anions (Cl⁻, Br⁻, I⁻, $\text{NO}_{3}^{-}$ , $\text{ClO}_{4}^{-}$ ) are essentially pH-independent.

6. Selective Precipitation

When a solution contains two ions that both form an insoluble salt with a common reagent, the one with the smaller $K_{sp}$ precipitates first (assuming similar stoichiometry).

Example. A solution is 0.10 M in both $\text{Cl}^{-}$ and $\text{I}^{-}$ . Slowly add AgNO₃. Use $K_{sp}(\text{AgCl}) = 1.8 \times 10^{-10}$ and $K_{sp}(\text{AgI}) = 8.5 \times 10^{-17}$ .

AgI starts to precipitate when $[\text{Ag}^{+}] = K_{sp}/[\text{I}^{-}] = 8.5 \times 10^{-17}/0.10 = 8.5 \times 10^{-16}$ M.
AgCl starts when $[\text{Ag}^{+}] = 1.8 \times 10^{-10}/0.10 = 1.8 \times 10^{-9}$ M.

AgI begins to precipitate at a much lower [Ag⁺]. By the time AgCl just begins to precipitate, $[\text{I}^{-}]$ has been reduced to $K_{sp}(\text{AgI})/(1.8 \times 10^{-9}) = 4.7 \times 10^{-8}$ M — i.e., over 99.99999% of the iodide has been removed before any chloride precipitates. This is the basis of qualitative-analysis cation/anion separation schemes.

7. Synthesis & AP Review

Big ideas to leave with:

$K_{sp}$ is just an equilibrium constant for the dissolution of a sparingly soluble salt. Solid omitted, ions to their stoichiometric powers.
Always relate molar solubility $s$ to $K_{sp}$ via the dissolution stoichiometry (e.g., $K_{sp} = 4s^{3}$ for $A B_{2}$ ).
The common-ion effect dramatically lowers solubility (Le Châtelier on the dissolution equilibrium).
Compare $Q$ to $K_{sp}$ to decide if a precipitate forms. Don't forget to dilute ion concentrations after mixing.
Solubility of salts with weak-acid anions increases at low pH; salts of strong-acid anions are essentially pH-independent.
Selective precipitation exploits differences in $K_{sp}$ to separate ions.

Common mistakes

Forgetting to raise ion concentrations to the stoichiometric power in $K_{sp}$ ( $[\text{Cl}^{-}]^{2}$ , not $[\text{Cl}^{-}]$ , for $\text{PbCl}_{2}$ ).
Using molar solubility $s$ as a $K_{sp}$ value, or vice versa.
Forgetting to dilute when mixing two solutions before computing $Q$ .
Comparing $K_{sp}$ values of salts with different stoichiometries to rank solubilities. Compute $s$ first.
Saying that pH affects all salts. It only matters when the cation reacts with OH⁻ or the anion reacts with H⁺.

Quick reference card

$K_{sp}$ for $A B$ : $K_{sp} = s^{2}$
$K_{sp}$ for $A B_{2}$ or $A_{2} B$ : $K_{sp} = 4 s^{3}$
$K_{sp}$ for $A B_{3}$ or $A_{3} B$ : $K_{sp} = 27 s^{4}$
$K_{sp}$ for $A_{3} B_{2}$ : $K_{sp} = 108 s^{5}$
Common ion → solubility ↓ (Le Châtelier)
$Q > K_{sp}$ → precipitate forms
Weak-acid anion (F⁻, OH⁻, $\text{CO}_{3}^{2-}$ , …) → solubility ↑ at low pH

📚 Practice Problems

1Problem 1easy

❓ Question:

Write the dissolution equation and the $K_{sp}$ expression for silver chloride, $\text{AgCl}(s)$ .

💡 Show Solution

Dissolution: $\text{AgCl}(s) \rightleftharpoons \text{Ag}^{+}(aq) + \text{Cl}^{-}(aq)$ .

The pure solid is omitted from the equilibrium expression:

$\boxed{K_{sp} = [\text{Ag}^{+}][\text{Cl}^{-}]}.$

At 25 °C, $K_{sp}(\text{AgCl}) \approx 1.8 \times 10^{-10}$ .

2Problem 2easy

❓ Question:

Write the $K_{sp}$ expression for calcium phosphate, $\text{Ca}_{3}(\text{PO}_{4})_{2}(s)$ .

💡 Show Solution

Dissolution: $\text{Ca}_{3}(\text{PO}_{4})_{2}(s) \rightleftharpoons 3\,\text{Ca}^{2+}(aq) + 2\,\text{PO}_{4}^{3-}(aq)$ .

$\boxed{K_{sp} = [\text{Ca}^{2+}]^{3}[\text{PO}_{4}^{3-}]^{2}}.$

Note both the coefficients (3 cations, 2 anions per formula unit) and the corresponding exponents.

3Problem 3easy

❓ Question:

For $\text{BaSO}_{4}$ , $K_{sp} = 1.1 \times 10^{-10}$ at 25 °C. Calculate the molar solubility in pure water.

💡 Show Solution

$\text{BaSO}_{4}(s) \rightleftharpoons \text{Ba}^{2+} + \text{SO}_{4}^{2-}$ .

Let $s$ = molar solubility. Then $[\text{Ba}^{2+}] = s$ and $[\text{SO}_{4}^{2-}] = s$ :

$K_{sp} = s^{2} = 1.1 \times 10^{-10}$ → $s = \sqrt{1.1 \times 10^{-10}} = \boxed{1.05 \times 10^{-5}\,\text{M}}$ .

4Problem 4medium

❓ Question:

The $K_{sp}$ of $\text{PbCl}_{2}$ is $1.7 \times 10^{-5}$ . Calculate its molar solubility in pure water.

💡 Show Solution

$\text{PbCl}_{2}(s) \rightleftharpoons \text{Pb}^{2+} + 2\,\text{Cl}^{-}$ .

ICE: $[\text{Pb}^{2+}] = s$ , $[\text{Cl}^{-}] = 2s$ .

$K_{sp} = s(2s)^{2} = 4s^{3} = 1.7 \times 10^{-5}$

$s^{3} = 4.25 \times 10^{-6}$ → $s = \boxed{1.6 \times 10^{-2}\,\text{M}}$ .

A notably more-soluble salt than the silver halides above — consistent with $K_{sp}$ being roughly $10^{5}$ times larger.

5Problem 5medium

❓ Question:

Explain the common-ion effect. Quantitatively, by what factor does the molar solubility of $\text{AgCl}$ ( $K_{sp} = 1.8 \times 10^{-10}$ ) drop on going from pure water to 0.10 M NaCl?

💡 Show Solution

Concept. Adding an ion already in the dissolution equilibrium shifts the equilibrium back to solid (Le Châtelier), reducing solubility.

In pure water: $s = \sqrt{K_{sp}} = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5}$ M.

In 0.10 M NaCl: $[\text{Cl}^{-}] \approx 0.10$ M (the trace from AgCl is negligible). Then

$[\text{Ag}^{+}] = K_{sp}/[\text{Cl}^{-}] = (1.8 \times 10^{-10})/(0.10) = 1.8 \times 10^{-9}$ M.

$[\text{Ag}^{+}]$ now equals the molar solubility, so $s' = 1.8 \times 10^{-9}$ M.

Factor: $s/s' = 1.34 \times 10^{-5}/1.8 \times 10^{-9} = \boxed{\sim 7400 \times \text{ less soluble}}$ .

6Problem 6medium

❓ Question:

100 mL of $2.0 \times 10^{-3}$ M $\text{AgNO}_{3}$ is mixed with 100 mL of $2.0 \times 10^{-3}$ M NaCl. $K_{sp}(\text{AgCl}) = 1.8 \times 10^{-10}$ . Will a precipitate form?

💡 Show Solution

Step 1. After mixing, total volume = 200 mL, so each species is diluted by a factor of 2:

$[\text{Ag}^{+}] = 1.0 \times 10^{-3}$ M, $[\text{Cl}^{-}] = 1.0 \times 10^{-3}$ M.

Step 2. $Q = [\text{Ag}^{+}][\text{Cl}^{-}] = (1.0 \times 10^{-3})(1.0 \times 10^{-3}) = 1.0 \times 10^{-6}$ .

Step 3. $Q\;(1.0 \times 10^{-6}) \gg K_{sp}\;(1.8 \times 10^{-10})$ , so $\boxed{\text{yes, AgCl precipitates}}$ until $Q$ falls back to $K_{sp}$ .

7Problem 7medium

❓ Question:

Why is $\text{Mg(OH)}_{2}$ much more soluble in 0.10 M HCl than in pure water, but essentially insoluble in 0.10 M NaOH?

💡 Show Solution

Dissolution: $\text{Mg(OH)}_{2}(s) \rightleftharpoons \text{Mg}^{2+} + 2\,\text{OH}^{-}$ .

In 0.10 M HCl (acid): $\text{H}^{+}$ neutralizes $\text{OH}^{-}$ ( $\text{H}^{+} + \text{OH}^{-} \to \text{H}_{2}\text{O}$ ), removing a product. Le Châtelier shifts dissolution right, dramatically increasing solubility.

In 0.10 M NaOH (base): $\text{OH}^{-}$ is a common ion at 0.10 M. The equilibrium shifts left, suppressing dissolution. Solubility crashes.

Quantitatively, in 0.10 M NaOH, $[\text{Mg}^{2+}] = K_{sp}(\text{Mg(OH)}_{2})/[\text{OH}^{-}]^{2} = (5.6 \times 10^{-12})/(0.10)^{2} = 5.6 \times 10^{-10}$ M (essentially nothing dissolves).

8Problem 8hard

❓ Question:

A solution is 0.010 M in both $\text{Cl}^{-}$ and $\text{I}^{-}$ . Solid $\text{AgNO}_{3}$ is added slowly with stirring. ( $K_{sp}(\text{AgCl}) = 1.8 \times 10^{-10}$ ; $K_{sp}(\text{AgI}) = 8.5 \times 10^{-17}$ .) (a) Which precipitates first? (b) When the second salt just begins to precipitate, what is $[\text{I}^{-}]$ ? Has separation been effective?

💡 Show Solution

(a) AgI starts when $[\text{Ag}^{+}] = K_{sp}/[\text{I}^{-}] = (8.5 \times 10^{-17})/(0.010) = 8.5 \times 10^{-15}$ M.

AgCl starts when $[\text{Ag}^{+}] = (1.8 \times 10^{-10})/(0.010) = 1.8 \times 10^{-8}$ M.

AgI requires far less $[\text{Ag}^{+}]$ , so $\boxed{\text{AgI precipitates first}}$ .

(b) When AgCl just begins to precipitate, $[\text{Ag}^{+}] = 1.8 \times 10^{-8}$ M. At that point

$[\text{I}^{-}] = K_{sp}(\text{AgI})/[\text{Ag}^{+}] = (8.5 \times 10^{-17})/(1.8 \times 10^{-8}) = \boxed{4.7 \times 10^{-9}\,\text{M}}$ .

Fraction remaining: $4.7 \times 10^{-9}/0.010 = 4.7 \times 10^{-7}$ , i.e. >99.99995% of the iodide has been removed before any chloride precipitates. Excellent separation.

9Problem 9hard

❓ Question:

Calculate the molar solubility of $\text{Ag}_{2}\text{CrO}_{4}$ ( $K_{sp} = 1.1 \times 10^{-12}$ ) in (a) pure water, and (b) 0.010 M $\text{K}_{2}\text{CrO}_{4}$ .

💡 Show Solution

Dissolution: $\text{Ag}_{2}\text{CrO}_{4}(s) \rightleftharpoons 2\,\text{Ag}^{+} + \text{CrO}_{4}^{2-}$ .

(a) Pure water. Let $s$ = molar solubility. Then $[\text{Ag}^{+}] = 2s$ , $[\text{CrO}_{4}^{2-}] = s$ .

$K_{sp} = (2s)^{2}(s) = 4s^{3} = 1.1 \times 10^{-12}$ → $s^{3} = 2.75 \times 10^{-13}$ → $\boxed{s = 6.5 \times 10^{-5}\,\text{M}}$ .

(b) 0.010 M $\text{K}_{2}\text{CrO}_{4}$ (common ion = $\text{CrO}_{4}^{2-}$ ). Now $[\text{CrO}_{4}^{2-}] \approx 0.010$ M (the trace from $\text{Ag}_{2}\text{CrO}_{4}$ is negligible). Let $s' =$ new molar solubility, so $[\text{Ag}^{+}] = 2s'$ .

$K_{sp} = (2s')^{2}(0.010) = 4s'^{2}(0.010) = 1.1 \times 10^{-12}$

$s'^{2} = 2.75 \times 10^{-11}$ → $\boxed{s' = 5.2 \times 10^{-6}\,\text{M}}$ .

The common ion drops solubility by ~12×.

10Problem 10hard

❓ Question:

$\text{CaF}_{2}$ has $K_{sp} = 3.9 \times 10^{-11}$ . (a) Calculate its molar solubility in pure water. (b) Qualitatively explain why $\text{CaF}_{2}$ is more soluble in 0.10 M HCl than in pure water, but $\text{CaCl}_{2}$ does not show such a pH dependence.

💡 Show Solution

(a) $\text{CaF}_{2}(s) \rightleftharpoons \text{Ca}^{2+} + 2\,\text{F}^{-}$ . Let $s$ = molar solubility. Then $[\text{Ca}^{2+}] = s$ , $[\text{F}^{-}] = 2s$ .

$K_{sp} = s(2s)^{2} = 4s^{3} = 3.9 \times 10^{-11}$ → $s^{3} = 9.75 \times 10^{-12}$ → $\boxed{s = 2.1 \times 10^{-4}\,\text{M}}$ .

(b) $\text{F}^{-}$ is the conjugate base of the weak acid HF (pKa ≈ 3.2). In 0.10 M HCl, the high $[\text{H}^{+}]$ converts much of the dissolved $\text{F}^{-}$ to undissociated HF, removing $\text{F}^{-}$ from the equilibrium. By Le Châtelier, dissolution shifts right and solubility increases substantially.

For $\text{CaCl}_{2}$ , the anion is $\text{Cl}^{-}$ , the conjugate base of the strong acid HCl — it has essentially no tendency to combine with $\text{H}^{+}$ . So acid does not consume $\text{Cl}^{-}$ and the dissolution equilibrium is unaffected by pH. ( $\text{CaCl}_{2}$ is also extremely soluble in water for this kind of analysis to apply, but the key point is the lack of pH coupling.)

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⚠️ Common Mistakes: Solubility Equilibrium: Ksp, Common-Ion Effect & Precipitation

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📝 Worked Example: Stoichiometry — Limiting Reagent

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$2$ mol of $H_2$ reacts with $1$ mol of $O_2$ . How many grams of water are produced? Which is the limiting reagent? ( $2H_2 + O_2 \to 2H_2O$ )

2Determine the limiting reagent

3Calculate moles of product

4Convert moles to grams

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Introduction to Chemical Equilibrium Reaction Quotient and Le Chatelier's Principle ICE Tables and Equilibrium Calculations Solubility Equilibria and K_sp

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Solubility Equilibrium: Ksp, Common-Ion Effect & Precipitation

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Solubility Equilibrium: Ksp, Common-Ion Effect & Precipitation

1. KspK_{sp}Ksp​ Expressions — the foundation

2. Molar Solubility from KspK_{sp}Ksp​

3. The Common-Ion Effect

4. QQQ vs. KspK_{sp}Ksp​ — Will a Precipitate Form?

5. pH Effects on Solubility

6. Selective Precipitation

7. Synthesis & AP Review

Common mistakes

Quick reference card

📚 Practice Problems

1Problem 1easy

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2Problem 2easy

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3Problem 3easy

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4Problem 4medium

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5Problem 5medium

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6Problem 6medium

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7Problem 7medium

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8Problem 8hard

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9Problem 9hard

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10Problem 10hard

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❓ Frequently Asked Questions

1. $K_{sp}$ Expressions — the foundation

2. Molar Solubility from $K_{sp}$

4. $Q$ vs. $K_{sp}$ — Will a Precipitate Form?