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Solubility Equilibria

How much of an 'insoluble' salt actually dissolves, and when does a precipitate form?

IntermediateEquilibriumLast reviewed 6 October 2026
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What is it?

“Insoluble” salts are really very slightly soluble. When solid silver chloride is placed in water, a tiny amount dissolves until the solution is saturated. Then a dynamic equilibrium is set up between the solid and its ions:

AgCl(s)⇌AgX+(aq)+ClX−(aq)\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}

The equilibrium constant is the solubility product, KspK_\text{sp}:

Ksp=[AgX+][ClX−]=1.8×10−10K_\text{sp} = [\ce{Ag+}][\ce{Cl-}] = 1.8 \times 10^{-10}

The solid does not appear in the expression (its concentration is constant). Each ion concentration is raised to the power of its coefficient: for CaFX2(s)⇌CaX2+(aq)+2 FX−(aq)\ce{CaF2(s) <=> Ca^2+(aq) + 2F-(aq)}, Ksp=[CaX2+][FX−]2K_\text{sp} = [\ce{Ca^2+}][\ce{F-}]^2.

Key idea

KspK_\text{sp} is the ceiling for the product of the ion concentrations. If the ion product QQ is below KspK_\text{sp}, more solid can dissolve. If QQ is above KspK_\text{sp}, the solution is supersaturated and a precipitate forms until Q=KspQ = K_\text{sp}.

Why does it matter?

  • Health. Barium sulfate is swallowed for X-ray imaging of the gut: barium ions are toxic, but BaSOX4\ce{BaSO4} is so insoluble (Ksp≈10−10K_\text{sp} \approx 10^{-10}) that almost none dissolves. Kidney stones form when calcium oxalate exceeds its KspK_\text{sp}.
  • Teeth. Fluoride converts tooth enamel into a less soluble mineral, so it dissolves less in acid.
  • Analysis and clean-up. Chemists precipitate one ion while leaving others in solution, and remove heavy metals from waste water.

How does it work?

1. Molar solubility

The molar solubility, ss, is the number of moles that dissolve per litre of saturated solution. Write the ion concentrations in terms of ss:

SaltIonsKspK_\text{sp} in terms of ss
AgCl, BaSO₄ (1 : 1)ss and ssKsp=s2K_\text{sp} = s^2
CaF₂, PbI₂, Mg(OH)₂ (1 : 2)ss and 2s2sKsp=s(2s)2=4s3K_\text{sp} = s(2s)^2 = 4s^3
Ag₂CrO₄ (2 : 1)2s2s and ssKsp=4s3K_\text{sp} = 4s^3

Only compare KspK_\text{sp} values directly for salts with the same ion ratio; otherwise, compare their molar solubilities.

2. The common-ion effect

If the solution already contains one of the ions, the equilibrium shifts to the left (Le Chatelier’s principle), and less solid dissolves. Silver chloride is far less soluble in salt water than in pure water.

3. Will a precipitate form?

  1. Find each ion concentration after mixing (allow for the total volume).
  2. Calculate the ion product QQ in the same form as KspK_\text{sp}.
  3. Compare: Q>KspQ > K_\text{sp}: precipitate forms. Q<KspQ < K_\text{sp}: no precipitate. Q=KspQ = K_\text{sp}: saturated.

Think of it like this

KspK_\text{sp} is like the capacity of a crowded room. While there is space (Q below Ksp), more people can come in. Once it is full (Q = Ksp), anyone else who arrives must leave again. If you already filled the room with one group (a common ion), there is less space for anyone else.

More precisely

KspK_\text{sp} depends on temperature. Simple KspK_\text{sp} calculations ignore effects such as ion pairing and reactions of the ions with water: for example, the solubility of salts of weak acids (carbonates, phosphates, fluorides) increases in acid, because HX+\ce{H+} removes the anion. Strictly, equilibrium constants use activities rather than concentrations, which matters in concentrated solutions.

Visualise it

A scale of the ion product Q. Below Ksp the solution is unsaturated and more solid can dissolve. At Ksp the solution is saturated, in equilibrium with the solid. Above Ksp the solution is supersaturated and a precipitate forms until Q falls to Ksp.
Comparing the ion product Q with Ksp predicts what happens.

Worked example

Worked example: Molar solubility of AgCl

Question: Calculate the molar solubility of AgCl (Ksp=1.8×10−10K_\text{sp} = 1.8 \times 10^{-10}) in mol/L and in g/L.

  1. [AgX+]=[ClX−]=s[\ce{Ag+}] = [\ce{Cl-}] = s, so Ksp=s2K_\text{sp} = s^2.

  2. Molar solubility:

    s=1.8×10−10=1.3×10−5 mol/L\begin{aligned} &s = \sqrt{1.8 \times 10^{-10}} \\[4pt] &= 1.3 \times 10^{-5}\ \text{mol/L} \end{aligned}

  3. In g/L: 1.34×10−5 mol/L×143.32 g/mol=1.9×10−3 g/L1.34 \times 10^{-5}\ \text{mol/L} \times 143.32\ \text{g/mol} = 1.9 \times 10^{-3}\ \text{g/L}, about 2 mg per litre.

Worked example: A 1 : 2 salt

Question: Calculate the molar solubility of CaFX2\ce{CaF2} (Ksp=3.9×10−11K_\text{sp} = 3.9 \times 10^{-11}).

  1. [CaX2+]=s[\ce{Ca^2+}] = s and [FX−]=2s[\ce{F-}] = 2s, so Ksp=s(2s)2=4s3K_\text{sp} = s(2s)^2 = 4s^3.

  2. Solve:

    s=3.9×10−1143=2.1×10−4 mol/L\begin{aligned} &s = \sqrt[3]{\frac{3.9 \times 10^{-11}}{4}} \\[4pt] &= 2.1 \times 10^{-4}\ \text{mol/L} \end{aligned}

Worked example: Ksp from solubility

Question: The solubility of PbIX2\ce{PbI2} is 0.62 g/L. Calculate KspK_\text{sp}.

  1. Molar solubility: s=0.62 g/L461.0 g/mol=1.34×10−3 mol/Ls = \dfrac{0.62\ \text{g/L}}{461.0\ \text{g/mol}} = 1.34 \times 10^{-3}\ \text{mol/L}

  2. [PbX2+]=s[\ce{Pb^2+}] = s, [IX−]=2s[\ce{I-}] = 2s:

    Ksp=4s3=4×(1.345×10−3)3=9.7×10−9\small\begin{aligned} &K_\text{sp} = 4s^3 = 4 \times (1.345 \\[4pt] &\quad \times 10^{-3})^3 \\[4pt] &= 9.7 \times 10^{-9} \end{aligned}

Worked example: The common-ion effect

Question: Calculate the molar solubility of AgCl in 0.10 mol/L NaCl.

  1. [ClX−]≈0.10[\ce{Cl-}] \approx 0.10 mol/L (from the NaCl); [AgX+]=s[\ce{Ag+}] = s.

  2. Ksp=s×0.10K_\text{sp} = s \times 0.10, so

    s=1.8×10−100.10=1.8×10−9 mol/L\begin{aligned} &s = \frac{1.8 \times 10^{-10}}{0.10} \\[4pt] &= 1.8 \times 10^{-9}\ \text{mol/L} \end{aligned}

  3. That is about 7500 times less than in pure water (1.3×10−51.3 \times 10^{-5} mol/L).

Worked example: Will a precipitate form?

Question: Equal volumes of 0.0010 mol/L AgNOX3\ce{AgNO3} and 0.0010 mol/L NaCl are mixed. Does AgCl precipitate?

  1. Mixing equal volumes halves each concentration: [AgX+]=[ClX−]=5.0×10−4[\ce{Ag+}] = [\ce{Cl-}] = 5.0 \times 10^{-4} mol/L.

  2. Ion product:

    Q=(5.0×10−4)(5.0×10−4)=2.5×10−7\small\begin{aligned} &Q = (5.0 \times 10^{-4})(5.0 \\[4pt] &\quad \times 10^{-4}) \\[4pt] &= 2.5 \times 10^{-7} \end{aligned}

  3. Q=2.5×10−7Q = 2.5 \times 10^{-7} is greater than Ksp=1.8×10−10K_\text{sp} = 1.8 \times 10^{-10}: a precipitate forms.

Common mistake

Common mistake: Forgetting to double (and square) the second ion

For CaFX2\ce{CaF2}, [FX−]=2s[\ce{F-}] = 2s, and it is squared in KspK_\text{sp}: Ksp=s(2s)2=4s3K_\text{sp} = s(2s)^2 = 4s^3, not s2s^2 or 2s22s^2.

Common mistake: Comparing Ksp values of different types of salt

A smaller KspK_\text{sp} does not always mean a less soluble salt. Ag₂CrO₄ (Ksp=1.1×10−12K_\text{sp} = 1.1 \times 10^{-12}) is actually more soluble than AgCl (1.8×10−101.8 \times 10^{-10}): 6.5×10−56.5 \times 10^{-5} versus 1.3×10−51.3 \times 10^{-5} mol/L. Compare molar solubilities.

Common mistake: Ignoring dilution when solutions are mixed

Mixing two solutions increases the volume, so every concentration falls. Recalculate the concentrations before working out QQ.

Notation note

  • Solids are left out of KspK_\text{sp} expressions; ion concentrations are in mol/L at equilibrium.
  • KspK_\text{sp} values are usually quoted without units.

Remember this

Remember this

  • KspK_\text{sp} = product of the ion concentrations in a saturated solution, each raised to its coefficient.
  • 1 : 1 salts: Ksp=s2K_\text{sp} = s^2. 1 : 2 or 2 : 1 salts: Ksp=4s3K_\text{sp} = 4s^3.
  • A common ion lowers the solubility.
  • Q>KspQ > K_\text{sp}: precipitate. Q<KspQ < K_\text{sp}: no precipitate. Remember to allow for dilution on mixing.

Test yourself

Check your understanding before moving on.

Flashcards

10 cards

  1. Question
    What is Ksp?
    Answer

    The solubility product: the equilibrium constant for a slightly soluble salt dissolving, the product of its ion concentrations in a saturated solution.

  2. Question
    Write the Ksp expression for CaF₂.
    Answer

    Ksp=[CaX2+][FX−]2K_\text{sp} = [\ce{Ca^2+}][\ce{F-}]^2

  3. Question
    Why is the solid left out of the Ksp expression?
    Answer

    The concentration of a pure solid is constant.

  4. Question
    How is Ksp related to the molar solubility s for a 1 : 1 salt such as AgCl?
    Answer

    Ksp=s2K_\text{sp} = s^2

  5. Question
    How is Ksp related to s for a 1 : 2 salt such as CaF₂?
    Answer

    Ksp=s(2s)2=4s3K_\text{sp} = s(2s)^2 = 4s^3

  6. Question
    What is the common-ion effect?
    Answer

    A salt is less soluble in a solution that already contains one of its ions, because the equilibrium shifts to the left.

  7. Question
    When does a precipitate form?
    Answer

    When the ion product Q is greater than Ksp.

  8. Question
    Calculate the molar solubility of AgCl (Ksp = 1.8 × 10⁻¹⁰).
    Answer

    s=1.8×10−10=1.3×10−5s = \sqrt{1.8 \times 10^{-10}} = 1.3 \times 10^{-5} mol/L

  9. Question
    Why can't you always compare Ksp values directly to rank solubility?
    Answer

    Salts with different ion ratios have different Ksp formulas (s² vs 4s³); compare molar solubilities instead.

  10. Question
    What must you do to concentrations when two solutions are mixed before calculating Q?
    Answer

    Recalculate them for the new total volume (dilution).

Quiz

7 questions

  1. Question 1EasyWhat is the Ksp expression for Mg(OH)₂?
    Show answer

    Answer: Ksp = [Mg²⁺][OH⁻]²

    Mg(OH)₂(s) ⇌ Mg²⁺ + 2OH⁻. Each ion is raised to its coefficient, and the solid is left out.

  2. Question 2EasyWhat is the molar solubility of BaSO₄ (Ksp = 1.1 × 10⁻¹⁰)?
    Show answer

    Answer: 1.0 × 10⁻⁵ mol/L

    For a 1 : 1 salt, Ksp = s², so s = √(1.1 × 10⁻¹⁰) = 1.0 × 10⁻⁵ mol/L.

  3. Question 3MediumWhat is the molar solubility of CaF₂ (Ksp = 3.9 × 10⁻¹¹)?
    Show answer

    Answer: 2.1 × 10⁻⁴ mol/L

    Ksp = s(2s)² = 4s³, so s = ∛(3.9 × 10⁻¹¹ ÷ 4) = 2.1 × 10⁻⁴ mol/L. 6.2 × 10⁻⁶ wrongly uses s².

  4. Question 4MediumHow does adding NaCl affect the solubility of AgCl?
    Show answer

    Answer: It decreases it

    Cl⁻ is a common ion; the equilibrium AgCl(s) ⇌ Ag⁺ + Cl⁻ shifts left, so less AgCl dissolves. Ksp itself does not change.

  5. Question 5MediumIn a solution, [Ag⁺] = [Cl⁻] = 5.0 × 10⁻⁴ mol/L. Ksp(AgCl) = 1.8 × 10⁻¹⁰. What happens?
    Show answer

    Answer: A precipitate forms: Q is greater than Ksp

    Q = (5.0 × 10⁻⁴)² = 2.5 × 10⁻⁷, which is greater than 1.8 × 10⁻¹⁰, so AgCl precipitates.

  6. Question 6HardWhich salt is MORE soluble in water: AgCl (Ksp 1.8 × 10⁻¹⁰) or Ag₂CrO₄ (Ksp 1.1 × 10⁻¹²)?
    Show answer

    Answer: Ag₂CrO₄, because its molar solubility is larger

    AgCl: s = √Ksp = 1.3 × 10⁻⁵ mol/L. Ag₂CrO₄: s = ∛(Ksp ÷ 4) = 6.5 × 10⁻⁵ mol/L. Different ion ratios mean Ksp values cannot be compared directly.

  7. Question 7HardWhat is the molar solubility of AgCl in 0.10 mol/L NaCl? (Ksp = 1.8 × 10⁻¹⁰)
    Show answer

    Answer: 1.8 × 10⁻⁹ mol/L

    [Cl⁻] ≈ 0.10 mol/L, so s = Ksp ÷ 0.10 = 1.8 × 10⁻⁹ mol/L, thousands of times less than in pure water.

Notes and downloads

  • Worksheet

    Solubility Equilibria Worksheet

    8 questions on Ksp expressions, molar solubility, Ksp from solubility, the common-ion effect and predicting precipitates. Answer key included.

    IntermediateFree

References

  1. Brown, T. L.; LeMay, H. E., Jr.; Bursten, B. E.; Murphy, C. J.; Woodward, P. M.; Stoltzfus, M. W. Chemistry: The Central Science, 15th ed.; Pearson, 2022.

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