Best examples of calculating molar solubility from Ksp (step‑by‑step)

Why examples of calculating molar solubility from Ksp matter

Molar solubility is the number chemists really care about in practice: how many moles of a solid will dissolve per liter of solution. Ksp is the equilibrium constant that describes that dissolution. Turning one into the other is where students either lock in their understanding…or get lost.

Working through several examples of calculating molar solubility from Ksp forces you to:

• Translate a dissolution equation into an expression for Ksp.
• Connect stoichiometric coefficients to powers in that expression.
• Solve for the molar solubility, usually labeled s.
• Adjust for common ions, pH, and different ionic ratios.

Below, I’ll walk through a series of real examples, gradually increasing the difficulty so you can see the patterns.

Example of a simple 1:1 salt: AgCl in pure water

Let’s start with one of the classic examples of calculating molar solubility from Ksp: silver chloride, AgCl.

Dissolution equilibrium:

[
\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)
]

Ksp expression:

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

Suppose you look up the Ksp of AgCl at 25 °C in a data table (for instance, in a general chemistry textbook or a reference like the NIST database) and find:

[
K_{sp} = 1.8 \times 10^{-10}
]

Let s be the molar solubility of AgCl in pure water:

• When s moles per liter of AgCl dissolve, you get s M Ag⁺ and s M Cl⁻.

So:

[
[\text{Ag}^+] = s, \quad [\text{Cl}^-] = s
]

Plug into Ksp:

[
K_{sp} = s \cdot s = s^2
]

Solve for s:

[
s = \sqrt{1.8 \times 10^{-10}} \approx 1.3 \times 10^{-5} \text{ M}
]

So the molar solubility of AgCl in pure water at 25 °C is about \(1.3 \times 10^{-5}\) mol/L. This is a simple but powerful example of how Ksp translates directly into molar solubility when the ion ratio is 1:1.

Examples of calculating molar solubility from Ksp for 1:2 salts: CaF₂

Things get more interesting when the stoichiometry isn’t 1:1. One of the best examples is calcium fluoride, CaF₂.

Dissolution equilibrium:

[
\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\,\text{F}^-(aq)
]

Ksp expression:

[
K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2
]

Take a typical Ksp value at 25 °C:

[
K_{sp}(\text{CaF}_2) \approx 3.9 \times 10^{-11}
]

Let s be the molar solubility of CaF₂ in pure water.

• For every 1 mole of CaF₂ that dissolves, you get 1 mole of Ca²⁺ and 2 moles of F⁻.

So at equilibrium:

[
[\text{Ca}^{2+}] = s, \quad [\text{F}^-] = 2s
]

Plug into Ksp:

[
K_{sp} = s(2s)^2 = s \cdot 4s^2 = 4s^3
]

Solve for s:

[
4s^3 = 3.9 \times 10^{-11} \Rightarrow s^3 = 9.75 \times 10^{-12}
]

[
s = (9.75 \times 10^{-12})^{1/3} \approx 2.1 \times 10^{-4} \text{ M}
]

This example of a 1:2 salt shows why you must pay attention to coefficients: they become exponents in the Ksp expression and multipliers in the concentration terms.

Examples include 1:3 salts: Fe(OH)₃ in water

Now let’s push to a 1:3 ratio, another favorite in examples of calculating molar solubility from Ksp problems. Consider iron(III) hydroxide, Fe(OH)₃.

Dissolution equilibrium:

[
\text{Fe(OH)}_3(s) \rightleftharpoons \text{Fe}^{3+}(aq) + 3\,\text{OH}^-(aq)
]

Ksp expression:

[
K_{sp} = [\text{Fe}^{3+}][\text{OH}^-]^3
]

Use a typical Ksp value at 25 °C:

[
K_{sp}(\text{Fe(OH)}_3) \approx 2.8 \times 10^{-39}
]

Let s be the molar solubility in pure water.

• At equilibrium: \([\text{Fe}^{3+}] = s\) and \([\text{OH}^-] = 3s\).

Plug in:

[
K_{sp} = s(3s)^3 = s \cdot 27s^3 = 27s^4
]

[
27s^4 = 2.8 \times 10^{-39} \Rightarrow s^4 = 1.037 \times 10^{-40}
]

[
s = (1.037 \times 10^{-40})^{1/4} \approx 1.0 \times 10^{-10} \text{ M}
]

This is one of the best examples to illustrate how extremely small Ksp values translate into vanishingly small molar solubilities.

Common‑ion effect: example of CaF₂ in 0.10 M NaF

Real solutions rarely contain “just water.” Ions already present in solution suppress solubility via the common‑ion effect. Here’s an example of CaF₂ again, but this time in a solution that already has fluoride.

We’ll reuse:

[
\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\,\text{F}^-(aq),\quad K_{sp} = 3.9 \times 10^{-11}
]

Suppose the solution initially contains 0.10 M NaF, so \([\text{F}^-]_{initial} = 0.10\,\text{M}\).

Let s be the additional molar solubility of CaF₂ in this solution.

At equilibrium:

• \([\text{Ca}^{2+}] = s\)
• \([\text{F}^-] = 0.10 + 2s\)

Ksp expression:

[
3.9 \times 10^{-11} = s(0.10 + 2s)^2
]

Because \(s\) will be tiny compared with 0.10, we can approximate \(0.10 + 2s \approx 0.10\):

[
3.9 \times 10^{-11} \approx s(0.10)^2 = s(0.010)
]

[
s \approx \frac{3.9 \times 10^{-11}}{0.010} = 3.9 \times 10^{-9} \text{ M}
]

Compare that to \(2.1 \times 10^{-4}\) M in pure water. The molar solubility drops by about five orders of magnitude. This is one of the clearest examples of calculating molar solubility from Ksp where the common‑ion effect absolutely dominates.

pH‑dependent solubility: example of Mg(OH)₂ in basic solution

Hydroxide salts are favorites in problem sets because pH directly affects [OH⁻]. Here’s a classic example of Mg(OH)₂.

Dissolution equilibrium:

[
\text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\,\text{OH}^-(aq)
]

Ksp expression:

[
K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2
]

Use a typical Ksp:

[
K_{sp}(\text{Mg(OH)}_2) \approx 1.2 \times 10^{-11}
]

Case: Mg(OH)₂ in a solution buffered at pH 12.00

At pH 12.00:

[
\text{pOH} = 14.00 - 12.00 = 2.00
]

[
[\text{OH}^-] = 10^{-2.00} = 1.0 \times 10^{-2} \text{ M}
]

Let s be the molar solubility of Mg(OH)₂ in this solution.

At equilibrium:

• \([\text{Mg}^{2+}] = s\)
• \([\text{OH}^-] \approx 1.0 \times 10^{-2} \text{ M}\) (because the buffer dominates)

Plug into Ksp:

[
1.2 \times 10^{-11} = s(1.0 \times 10^{-2})^2 = s(1.0 \times 10^{-4})
]

[
s = \frac{1.2 \times 10^{-11}}{1.0 \times 10^{-4}} = 1.2 \times 10^{-7} \text{ M}
]

This example of pH‑dependent solubility shows how making the solution more basic (higher [OH⁻]) suppresses dissolution of a metal hydroxide.

For a contrast example, if you instead acidify the solution so OH⁻ is consumed, the solubility of Mg(OH)₂ would increase; that’s a common lab trick for dissolving hydroxide precipitates.

Mixed‑ion example: PbCl₂ in 0.10 M Pb(NO₃)₂

Let’s look at one more common‑ion case, this time with the cation as the common ion. Lead(II) chloride is a standard in examples of calculating molar solubility from Ksp.

Dissolution equilibrium:

[
\text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\,\text{Cl}^-(aq)
]

Ksp expression:

[
K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2
]

Use a typical Ksp value:

[
K_{sp}(\text{PbCl}_2) \approx 1.7 \times 10^{-5}
]

Now imagine a solution that already contains 0.10 M Pb(NO₃)₂, so \([\text{Pb}^{2+}]_{initial} = 0.10\,\text{M}\).

Let s be the additional molar solubility of PbCl₂ in this solution.

At equilibrium:

• \([\text{Pb}^{2+}] = 0.10 + s \approx 0.10\)
• \([\text{Cl}^-] = 2s\)

Plug into Ksp, using the approximation \(0.10 + s \approx 0.10\):

[
1.7 \times 10^{-5} \approx (0.10)(2s)^2 = 0.10 \cdot 4s^2 = 0.40s^2
]

[
s^2 \approx \frac{1.7 \times 10^{-5}}{0.40} = 4.25 \times 10^{-5}
]

[
s \approx 6.5 \times 10^{-3} \text{ M}
]

Again, the common‑ion effect significantly lowers the molar solubility compared with pure water.

Real‑world context: where these examples show up in 2024–2025

If you’re wondering whether all these examples of calculating molar solubility from Ksp are just textbook exercises, the short answer is no. In 2024–2025, this math shows up in:

• Water treatment and environmental chemistry – Predicting whether heavy metal ions like Pb²⁺ or Cd²⁺ will precipitate as sulfides or carbonates at a given pH and ion concentration. Agencies like the U.S. Environmental Protection Agency and research groups rely on solubility data to model contaminant behavior.
• Pharmaceutical formulation – Ksp and molar solubility help chemists design drugs that either stay dissolved (for absorption) or precipitate in a controlled way. Tutorials from universities, such as those hosted by MIT OpenCourseWare or Khan Academy, often use drug‑like molecules as examples.
• Materials and corrosion science – Predicting when metal hydroxides or carbonates form protective layers versus when they keep dissolving, which affects the lifetime of infrastructure.

If you want to cross‑check Ksp values or see how they’re used in more advanced equilibrium modeling, good starting points include general chemistry resources from UC Davis ChemWiki / LibreTexts and data compilations referenced in university courses.

Putting it together: pattern spotting across all these examples

Across all these examples of calculating molar solubility from Ksp, the workflow is nearly the same:

• Write the balanced dissolution equation.
• Express Ksp in terms of ion concentrations.
• Represent each ion concentration in terms of a single variable s (the molar solubility), taking stoichiometric ratios into account.
• Modify those expressions if there are common ions or fixed pH.
• Solve the resulting algebraic equation for s.

The examples include:

• Simple 1:1 salts like AgCl, where \(K_{sp} = s^2\).
• 1:2 salts like CaF₂, where \(K_{sp} = 4s^3\).
• 1:3 salts like Fe(OH)₃, where \(K_{sp} = 27s^4\).
• Common‑ion cases (CaF₂ in NaF, PbCl₂ in Pb(NO₃)₂).
• pH‑dependent solubility (Mg(OH)₂ in basic solution).

Once you recognize these patterns, most new problems just feel like variations on the same theme.

FAQ: common questions about Ksp and molar solubility

What is an example of using Ksp to compare solubilities?

A straightforward example of using Ksp to compare solubilities is to look at two salts with the same ion ratio, like AgCl and AgBr (both 1:1). If \(K_{sp}(\text{AgCl}) = 1.8 \times 10^{-10}\) and \(K_{sp}(\text{AgBr}) = 5.0 \times 10^{-13}\), then in pure water AgCl has the higher molar solubility because its Ksp is larger. When the stoichiometry is the same, larger Ksp generally means more soluble.

Are there examples of Ksp problems where approximations fail?

Yes. Approximations like “0.10 + s ≈ 0.10” or “0.10 + 2s ≈ 0.10” can fail when s is not tiny compared with the initial concentration. In those cases, you have to solve the full quadratic or higher‑order equation without dropping terms. Some instructors intentionally choose Ksp and initial concentrations so that s is a noticeable fraction of the starting value and the approximation gives the wrong answer.

Where can I find reliable Ksp values for more real examples?

Good sources for data include:

• General chemistry tables from university sites such as Harvard University’s chemistry resources
• Open educational resources like LibreTexts Chemistry
• Standard reference data sets used in college courses, which often trace back to NIST or IUPAC evaluations

These let you build your own examples of calculating molar solubility from Ksp using up‑to‑date values.

Do temperature changes affect these examples of molar solubility calculations?

Yes. Ksp is temperature‑dependent, so all examples of molar solubility calculations are tied to a specific temperature, usually 25 °C in textbooks. If the temperature changes, Ksp changes, and so does the calculated molar solubility. For some salts, solubility increases with temperature; for others, it can decrease.

How do these examples connect to real lab work?

In the lab, you might measure how much of a solid dissolves, then back‑calculate an experimental Ksp and compare it to literature values. Or you might predict whether a precipitate will form when you mix two solutions. Every one of the examples of calculating molar solubility from Ksp above maps directly to those decisions: Will a solid form? How much? At what pH does it start to dissolve again?

If you’re studying for an exam or just trying to feel more confident with equilibrium, revisiting these best examples of calculating molar solubility from Ksp and then tweaking the numbers (different Ksp, different common‑ion concentrations, different pH) is one of the fastest ways to make the math feel natural.