Best Examples of Predicting Solubility Using Ksp Values

Starting with simple examples of predicting solubility using Ksp values

Let’s start with the kind of example of predicting solubility using Ksp values that shows up in every general chemistry course: a sparingly soluble salt in pure water.

Take silver chloride, AgCl(s). Its solubility product at 25 °C is about:

\[ K_{sp}(\text{AgCl}) \approx 1.8 \times 10^{-10} \]

AgCl dissolves according to:

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

If the molar solubility of AgCl is \(s\), then at equilibrium:

• \([\text{Ag}^+] = s\)
• \([\text{Cl}^-] = s\)

and

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

So:

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

This is a clean example of predicting solubility using Ksp values: you’ve just shown that, at 25 °C, you can dissolve only about \(1.3 \times 10^{-5}\) moles of AgCl per liter of water before it stops dissolving.

Real examples: Will a precipitate form? Using Q vs Ksp

Most interesting examples of predicting solubility using Ksp values are not about pure water; they’re about mixing solutions and asking whether a precipitate appears. That’s where the ion product \(Q\) comes in.

Example 1 – Mixing silver nitrate and sodium chloride

Suppose you mix 50.0 mL of 0.010 M AgNO₃ with 50.0 mL of 0.010 M NaCl. Does AgCl precipitate?

After mixing, total volume is 0.100 L. Each ion is diluted by a factor of 2:

• \([\text{Ag}^+] = 0.010 / 2 = 0.0050\,\text{M}\)
• \([\text{Cl}^-] = 0.010 / 2 = 0.0050\,\text{M}\)

Compute the ion product \(Q\):

\[ Q = [\text{Ag}^+][\text{Cl}^-] = (0.0050)(0.0050) = 2.5 \times 10^{-5} \]

Compare with \(K_{sp} = 1.8 \times 10^{-10}\).

• \(Q \gg K_{sp}\) → the solution is supersaturated → AgCl precipitates.

This is one of the best examples to show the logic:

• If \(Q > K_{sp}\): precipitation occurs.
• If \(Q = K_{sp}\): the system is at equilibrium.
• If \(Q < K_{sp}\): no precipitate forms; more solid could dissolve.

Example 2 – Same ions, lower concentrations

Now mix 50.0 mL of 1.0 × 10⁻⁴ M AgNO₃ with 50.0 mL of 1.0 × 10⁻⁴ M NaCl.

After mixing:

• \([\text{Ag}^+] = 5.0 \times 10^{-5}\,\text{M}\)
• \([\text{Cl}^-] = 5.0 \times 10^{-5}\,\text{M}\)

\[ Q = (5.0 \times 10^{-5})(5.0 \times 10^{-5}) = 2.5 \times 10^{-9} \]

Now \(Q = 2.5 \times 10^{-9}\) is still greater than \(1.8 \times 10^{-10}\), so a small amount of AgCl will still precipitate.

If you push the concentrations even lower, you eventually get \(Q < K_{sp}\), and then no precipitate forms. These simple solution-mixing problems are classic examples of predicting solubility using Ksp values in both textbooks and lab work.

Examples of predicting solubility using Ksp values in hard water

Hard water is loaded with calcium and magnesium ions. When those ions meet carbonate or sulfate, you get scale: CaCO₃ or CaSO₄ deposits on pipes and heating elements.

Example 3 – Calcium carbonate in hard water

Consider a water sample with:

• \([\text{Ca}^{2+}] = 2.0 \times 10^{-3}\,\text{M}\)
• \([\text{CO}_3^{2-}] = 1.0 \times 10^{-4}\,\text{M}\)

The solubility product for CaCO₃ at 25 °C is approximately:

\[ K_{sp}(\text{CaCO}_3) \approx 4.8 \times 10^{-9} \]

Dissolution:

\[ \text{CaCO}_3(s) \rightleftharpoons \text{Ca}^{2+} + \text{CO}_3^{2-} \]

Compute \(Q\):

\[ Q = [\text{Ca}^{2+}][\text{CO}_3^{2-}] = (2.0 \times 10^{-3})(1.0 \times 10^{-4}) = 2.0 \times 10^{-7} \]

Since \(Q > K_{sp}\), CaCO₃ will precipitate. This is a very real example of predicting solubility using Ksp values in environmental and water-treatment chemistry.

In practice, utilities and engineers use Ksp-based models (often wrapped into software, not done by hand) to estimate when scaling will occur in municipal systems or industrial boilers. The underlying logic is still the same Q vs Ksp comparison you do in class.

For more background on water chemistry and hardness, the U.S. Geological Survey has accessible material on water properties and ions in natural systems: https://www.usgs.gov/special-topics/water-science-school.

Using Ksp to estimate molar solubility: more examples

Another category of examples of predicting solubility using Ksp values involves converting Ksp into a molar solubility in different conditions.

Example 4 – Lead(II) iodide solubility in pure water

Lead(II) iodide, PbI₂, dissolves as:

\[ \text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+} + 2\text{I}^- \]

\[ K_{sp}(\text{PbI}_2) \approx 7.1 \times 10^{-9} \]

Let \(s\) be the molar solubility of PbI₂ in pure water. At equilibrium:

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

Then:

\[ K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2 = s(2s)^2 = 4s^3 \]

So:

\[ s = \sqrt[3]{\dfrac{K_{sp}}{4}} = \sqrt[3]{\dfrac{7.1 \times 10^{-9}}{4}} \approx 1.2 \times 10^{-3} \; \text{M} \]

That means you can dissolve about \(1.2 \times 10^{-3}\) moles of PbI₂ per liter before reaching equilibrium. Converting to grams per liter is just stoichiometry from there.

This style of calculation is a standard example of predicting solubility using Ksp values in any Ksp problem set.

Example 5 – Common ion effect: PbI₂ in KI solution

Now, dissolve PbI₂ not in pure water but in 0.10 M KI. The iodide ion is already present at a high concentration, which suppresses solubility.

Assume the additional iodide from PbI₂ is small compared to 0.10 M. Then at equilibrium:

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

Use the same Ksp expression:

\[ K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2 \approx s(0.10)^2 = 0.010s \]

So:

\[ s \approx \dfrac{7.1 \times 10^{-9}}{0.010} = 7.1 \times 10^{-7}\,\text{M} \]

Compared to \(1.2 \times 10^{-3}\,\text{M}\) in pure water, the solubility has dropped by about three orders of magnitude. This is not just an academic trick; the common ion effect is used in qualitative analysis labs to control which ions precipitate and which stay in solution.

The logic here is another clear example of predicting solubility using Ksp values: you plug realistic ion concentrations into the Ksp expression and solve for the unknown solubility.

pH-dependent solubility: hydroxides and carbonates

Many real systems involve salts that contain basic anions like OH⁻, CO₃²⁻, or PO₄³⁻. Their solubility depends strongly on pH because those anions react with H⁺.

Example 6 – Aluminum hydroxide in acidic vs basic solutions

Aluminum hydroxide, Al(OH)₃, is used as an antacid and also appears in water treatment. Its dissolution is:

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

\[ K_{sp}(\text{Al(OH)}_3) \approx 3 \times 10^{-34} \]

In a strongly basic solution where \([\text{OH}^-]\) is large, the Ksp expression:

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

forces \([\text{Al}^{3+}]\) to be tiny. In acidic solution, though, H⁺ consumes OH⁻, lowering \([\text{OH}^-]\), and Le Châtelier’s principle pushes more Al(OH)₃ to dissolve.

A simplified way to view this as a Ksp problem is:

• In basic conditions, plug a large \([\text{OH}^-]\) into the Ksp expression and solve for \([\text{Al}^{3+}]\).
• In acidic conditions, \([\text{OH}^-]\) is related to pH via \(K_w\), and you again solve for \([\text{Al}^{3+}]\).

These are more advanced examples of predicting solubility using Ksp values, but they show how Ksp interacts with acid–base equilibria.

If you’re interested in how metal hydroxides and pH are handled in environmental and health contexts, the U.S. Environmental Protection Agency and the National Institutes of Health both host reference material on metal exposure and water chemistry, for instance: https://www.ncbi.nlm.nih.gov/books/.

Example 7 – Calcium carbonate and blood pH (conceptual)

Calcium carbonate also shows up in biology, from marine shells to kidney stones. While the full biological picture is complex, you can still frame a simple example of predicting solubility using Ksp values:

• CaCO₃(s) is in equilibrium with Ca²⁺ and CO₃²⁻.
• CO₃²⁻ is tied to the carbonic acid/bicarbonate system, which depends on pH and CO₂.

As pH drops (more acidic), some CO₃²⁻ is converted to HCO₃⁻ and H₂CO₃, lowering \([\text{CO}_3^{2-}]\). To maintain the Ksp expression \(K_{sp} = [\text{Ca}^{2+}][\text{CO}_3^{2-}]\), more CaCO₃ dissolves.

This same logic is used in ocean chemistry to discuss the impact of rising atmospheric CO₂ on carbonate shells and coral skeletons. While ocean models use far more sophisticated thermodynamics, Ksp is still at the core.

For an accessible scientific overview of carbonate chemistry and health-related mineral equilibria, you can explore educational materials from major universities such as MIT OpenCourseWare: https://ocw.mit.edu.

Industrial and analytical chemistry: more real examples

To make this feel less like a classroom exercise and more like actual chemistry, let’s look at a few real examples of predicting solubility using Ksp values in applied settings.

Example 8 – Removing heavy metals from wastewater

Imagine an industrial facility that needs to remove copper(II) from wastewater by precipitating it as copper(II) hydroxide, Cu(OH)₂.

Dissolution:

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

\[ K_{sp}(\text{Cu(OH)}_2) \approx 2.2 \times 10^{-20} \]

Suppose the target is to reduce \([\text{Cu}^{2+}]\) to \(1.0 \times 10^{-6}\,\text{M}\). What [OH⁻] is needed at equilibrium?

Use the Ksp expression:

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

Rearrange for \([\text{OH}^-]\):

\[ [\text{OH}^-]^2 = \dfrac{K_{sp}}{[\text{Cu}^{2+}]} = \dfrac{2.2 \times 10^{-20}}{1.0 \times 10^{-6}} = 2.2 \times 10^{-14} \]

\[ [\text{OH}^-] = \sqrt{2.2 \times 10^{-14}} \approx 1.5 \times 10^{-7}\,\text{M} \]

That corresponds to a pOH of about 6.82, or pH ≈ 7.18. In real plants, safety margins, competing equilibria, and complex formation are all considered, but this calculation is the backbone of the design.

Again, this is a practical example of predicting solubility using Ksp values: you set the desired ion concentration and back-calculate the conditions needed to reach it.

Example 9 – Qualitative analysis: separating cations

Classic qualitative analysis schemes (still taught in many undergraduate labs) rely heavily on Ksp. For instance, consider separating Ag⁺ from Ba²⁺ by adding Cl⁻.

• AgCl has \(K_{sp} \approx 1.8 \times 10^{-10}\).
• BaCl₂ is very soluble; Ba²⁺ does not form a sparingly soluble chloride under these conditions.

By carefully adding Cl⁻ to a solution that contains both ions, you can selectively precipitate AgCl while Ba²⁺ remains in solution. The decision about how much chloride to add, and at what concentrations the precipitation becomes significant, is based on comparing Q for AgCl to its Ksp while keeping an eye on competing equilibria.

These lab protocols are living examples of predicting solubility using Ksp values that students still perform today.

Why Ksp still matters in 2024–2025

If you look at current chemistry and environmental science research, you’ll see Ksp quietly doing work behind the scenes. From modeling battery materials to understanding mineral scaling in desalination plants, solubility products are built into the thermodynamic databases used by simulation software.

Recent trends include:

• Water reuse and desalination: Engineers use Ksp-based models to predict scale formation in reverse osmosis membranes and high-recovery systems.
• Green chemistry and remediation: Ksp helps guide the choice of precipitating agents to immobilize toxic metals in soils and groundwater.
• Biomineralization and medicine: Ksp is part of the thermodynamic background behind models of bone mineral, dental enamel, and kidney stone formation and dissolution. Health-focused sites such as Mayo Clinic and NIH-linked resources discuss these processes qualitatively, while the underlying math still involves solubility equilibria.

Even if you’re not running these big simulations yourself, the logic in all of the examples of predicting solubility using Ksp values above is exactly what those models are automating at scale.

FAQ: common questions about Ksp and solubility

How do I know if a precipitate will form when I mix two solutions?

Calculate the ion product \(Q\) using the actual concentrations after mixing, then compare it to Ksp. If \(Q > K_{sp}\), a precipitate forms; if \(Q < K_{sp}\), it doesn’t. Many of the best examples of predicting solubility using Ksp values are just careful Q vs Ksp comparisons.

Can you give a quick example of using Ksp to find molar solubility?

Yes. For a 1:1 salt like AgCl, \(K_{sp} = s^2\), where \(s\) is the molar solubility. So \(s = \sqrt{K_{sp}}\). That’s exactly what we did in the AgCl example earlier.

Why does adding a common ion reduce solubility?

Adding a common ion increases its concentration in solution, which pushes the equilibrium toward the solid side to keep the Ksp expression satisfied. In the PbI₂ and KI example, the large iodide concentration forces the dissolved lead concentration to drop dramatically.

Are Ksp values affected by temperature?

Yes. Ksp values are temperature-dependent because they’re equilibrium constants. Most tables list values at 25 °C. If you change temperature significantly, you should look up the appropriate Ksp or use data that includes temperature dependence.

Where can I find reliable Ksp data and more worked examples?

University general chemistry courses and open textbooks are good sources. Many chemistry departments, such as those at MIT, UC Berkeley, or other major universities, publish free tables and problem sets online. These often include multiple examples of predicting solubility using Ksp values with step-by-step solutions.

Bottom line: once you’re comfortable translating a formula into a Ksp expression and comparing Q to Ksp, you can tackle a wide range of real and theoretical problems. The examples of predicting solubility using Ksp values above are not just homework fodder; they’re snapshots of how chemists think about what dissolves and what doesn’t, from your tap water to industrial reactors.