Practical examples of Ksp calculation for sparingly soluble salts

Real examples of Ksp calculation for sparingly soluble salts

Let’s skip the vague theory and start with actual numbers. The best examples of Ksp calculation for sparingly soluble salts almost always follow the same pattern:

• Write the dissolution equation.
• Express Ksp in terms of ion concentrations.
• Relate those concentrations to molar solubility.
• Solve the resulting algebra.

We’ll build from simple 1:1 salts up to pH‑dependent and common‑ion situations, using real Ksp values reported in standard references like the NIST Chemistry WebBook and general chemistry texts.

Example of Ksp calculation: AgCl in pure water

Silver chloride is a classic sparingly soluble salt. At 25 °C, its Ksp is about 1.8 × 10⁻¹⁰.

Dissolution equation
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

If the molar solubility of AgCl is \(s\) mol/L in pure water, then at equilibrium:

• [Ag⁺] = s
• [Cl⁻] = s

So the Ksp expression is:

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

Set this equal to the tabulated value:

\[ s^2 = 1.8 \times 10^{-10} \]
\[ s = \sqrt{1.8 \times 10^{-10}} \approx 1.34 \times 10^{-5} \, \text{mol/L} \]

That’s the molar solubility of AgCl in pure water at 25 °C. This is one of the simplest examples of Ksp calculation for sparingly soluble salts and a good baseline for everything that follows.

Examples of Ksp calculation for sparingly soluble salts with 1:2 stoichiometry (CaF₂)

When a salt produces more than one ion of a given type, the algebra gets a bit more interesting. Calcium fluoride is a great example of this.

At 25 °C, Ksp(CaF₂) ≈ 3.9 × 10⁻¹¹.

Dissolution equation
CaF₂(s) ⇌ Ca²⁺(aq) + 2 F⁻(aq)

Let molar solubility be \(s\) mol/L. Then:

• [Ca²⁺] = s
• [F⁻] = 2s

The Ksp expression becomes:

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

Set equal to Ksp:

\[ 4s^3 = 3.9 \times 10^{-11} \]
\[ s^3 = 9.75 \times 10^{-12} \]
\[ s = (9.75 \times 10^{-12})^{1/3} \approx 2.15 \times 10^{-4} \, \text{mol/L} \]

This example of Ksp calculation shows how stoichiometric coefficients turn into exponents and multipliers in the Ksp expression, a pattern you’ll see in many other sparingly soluble salts.

Best examples of Ksp calculation using given solubility (PbCl₂)

Sometimes you’re given the solubility in g/L or mol/L and asked to back‑calculate Ksp. Lead(II) chloride is a standard case.

Suppose the molar solubility of PbCl₂ at 25 °C is measured to be 1.6 × 10⁻² mol/L in pure water.

Dissolution equation
PbCl₂(s) ⇌ Pb²⁺(aq) + 2 Cl⁻(aq)

From the stoichiometry:

• [Pb²⁺] = 1.6 × 10⁻² M
• [Cl⁻] = 2 × 1.6 × 10⁻² = 3.2 × 10⁻² M

Ksp is then:

\[ K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 = (1.6 \times 10^{-2})(3.2 \times 10^{-2})^2 \]

\[ K_{sp} = (1.6 \times 10^{-2})(1.024 \times 10^{-3}) \approx 1.64 \times 10^{-5} \]

This is one of the best examples of Ksp calculation for sparingly soluble salts when lab data are given and you need to connect experimental solubility to an equilibrium constant.

Examples include common‑ion effect: AgCl in 0.10 M NaCl

Real solutions rarely contain just water and one salt. The common‑ion effect is where Ksp calculations start to feel like chemistry instead of pure algebra.

Let’s revisit AgCl, but now dissolve it in 0.10 M NaCl instead of pure water. The NaCl provides a common ion, Cl⁻.

Dissolution equation (same as before):
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

Let the solubility of AgCl in 0.10 M NaCl be \(s\) mol/L. Then:

• [Ag⁺] = s
• [Cl⁻] ≈ 0.10 + s ≈ 0.10 M (since s will be much smaller than 0.10)

Ksp expression:

\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = s(0.10) = 1.8 \times 10^{-10} \]

Solve for s:

\[ s = \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9} \, \text{mol/L} \]

Compare that to 1.34 × 10⁻⁵ M in pure water. The solubility dropped by about four orders of magnitude due to the common Cl⁻ ion. This is a textbook example of Ksp calculation for sparingly soluble salts in the presence of a common ion, and it’s directly relevant to controlling precipitation in analytical chemistry.

Example of Ksp calculation with pH dependence: CaF₂ in acidic solution

Some sparingly soluble salts contain basic anions that react with H⁺. Fluoride is one of them, which makes CaF₂ a good example of pH‑dependent solubility.

We’ll keep the same Ksp (3.9 × 10⁻¹¹ at 25 °C), but now consider a solution buffered at pH 2.00. At pH 2, [H⁺] = 1.0 × 10⁻² M.

Relevant equilibria:

• CaF₂(s) ⇌ Ca²⁺ + 2 F⁻
• F⁻ + H⁺ ⇌ HF (with Ka(HF) ≈ 6.8 × 10⁻⁴)

Because F⁻ is protonated to HF, the free [F⁻] is reduced, which drives more CaF₂ to dissolve.

A full treatment uses both Ksp and the acid–base equilibrium, but the key idea is this: as [H⁺] increases, [F⁻] from CaF₂ is partially “removed” via HF formation. In the Ksp expression

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

[F⁻] is no longer simply 2s. Instead, you solve a coupled system where total fluoride (from CaF₂) is split between F⁻ and HF. This is one of the more realistic examples of Ksp calculation for sparingly soluble salts, because many natural waters are not at neutral pH.

For a detailed thermodynamic background on such equilibria in natural waters, the U.S. Geological Survey provides accessible resources on aqueous geochemistry and speciation models (usgs.gov).

Best examples of Ksp calculation tied to real‑world chemistry

Ksp isn’t just an exam topic; it shows up everywhere from geology to medicine. A few real examples:

Barium sulfate (BaSO₄) and medical imaging

BaSO₄ is used as a contrast agent in gastrointestinal X‑ray imaging because it’s extremely insoluble yet chemically stable in the body.

At 25 °C, Ksp(BaSO₄) ≈ 1.1 × 10⁻¹⁰.

Dissolution equation
BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq)

Let molar solubility be \(s\):

• [Ba²⁺] = s
• [SO₄²⁻] = s

Ksp expression:

\[ K_{sp} = s^2 = 1.1 \times 10^{-10} \]
\[ s = \sqrt{1.1 \times 10^{-10}} \approx 1.05 \times 10^{-5} \, \text{mol/L} \]

That tiny solubility is exactly why BaSO₄ can be ingested safely in controlled medical settings, while soluble barium salts are toxic. Organizations like the U.S. National Library of Medicine and NIH discuss these toxicity differences in more detail (nih.gov). This is a great example of Ksp calculation helping explain why one form of a metal is medically useful and another is hazardous.

Calcium phosphate and kidney stones

Calcium phosphate minerals, such as hydroxyapatite, have low Ksp values and are involved in bone formation as well as kidney stone formation. While the full chemistry of kidney stones is complex, the basic idea is that when the ionic product [Ca²⁺]·[PO₄³⁻]ⁿ exceeds Ksp, precipitation becomes favorable.

Medical resources like the National Institute of Diabetes and Digestive and Kidney Diseases (NIDDK) at NIH discuss how urinary composition affects stone risk (niddk.nih.gov). Under the hood, these are real examples of Ksp calculation for sparingly soluble salts happening inside the human body.

Examples of Ksp calculation for sparingly soluble salts in mixed‑ion solutions

Let’s push a bit further with a mixed‑ion scenario involving Mg(OH)₂.

At 25 °C, Ksp(Mg(OH)₂) ≈ 1.8 × 10⁻¹¹.

Dissolution equation
Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2 OH⁻(aq)

Case 1: Pure water
Let solubility be \(s\):

• [Mg²⁺] = s
• [OH⁻] = 2s

Ksp:

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

\[ 4s^3 = 1.8 \times 10^{-11} \Rightarrow s^3 = 4.5 \times 10^{-12} \Rightarrow s \approx 1.65 \times 10^{-4} \, \text{mol/L} \]

Case 2: In 0.010 M NaOH
Now there’s a common OH⁻ ion.

Let additional solubility from Mg(OH)₂ be \(x\):

• [Mg²⁺] = x
• [OH⁻] ≈ 0.010 + 2x ≈ 0.010 M (since 2x ≪ 0.010)

Ksp:

\[ K_{sp} = x(0.010)^2 = 1.8 \times 10^{-11} \]

\[ x = \frac{1.8 \times 10^{-11}}{1.0 \times 10^{-4}} = 1.8 \times 10^{-7} \, \text{mol/L} \]

The solubility dropped from about 1.65 × 10⁻⁴ M to 1.8 × 10⁻⁷ M when NaOH was added. This is another sharp, quantitative example of Ksp calculation for sparingly soluble salts illustrating the common‑ion effect.

2024–2025 context: Why Ksp problems still matter

In current chemistry teaching (2024–2025), Ksp calculations are still a standard part of AP Chemistry, IB Chemistry, and first‑year college courses. But they’re also increasingly tied to modern topics:

• Water treatment and environmental chemistry: Predicting when heavy metal salts will precipitate in wastewater treatment plants or natural waters relies on the same Ksp math you’re practicing here. Agencies like the U.S. Environmental Protection Agency (EPA) and USGS routinely work with solubility equilibria when modeling contaminant behavior.
• Materials and energy: Battery chemistry, scale formation in geothermal plants, and corrosion control all involve sparingly soluble salts and their Ksp values.
• Biomedical applications: From contrast agents like BaSO₄ to biomineralization in bones and teeth, Ksp is baked into modern bioinorganic chemistry.

So when you work through these examples of Ksp calculation for sparingly soluble salts, you’re not just prepping for an exam; you’re using the same logic researchers and engineers apply in real systems.

FAQ: Short, exam‑style questions about Ksp examples

Q1: Can you give a quick example of using Ksp to predict precipitation?
Suppose you mix 0.010 M AgNO₃ and 0.010 M NaCl in equal volumes. After mixing, each ion is 0.0050 M. The ionic product is Q = [Ag⁺][Cl⁻] = (0.0050)(0.0050) = 2.5 × 10⁻⁵. Since Ksp(AgCl) = 1.8 × 10⁻¹⁰, Q ≫ Ksp, so AgCl will precipitate. This is one of the simplest examples of Ksp calculation for sparingly soluble salts used to decide if a precipitate forms.

Q2: How do examples of Ksp calculation change when the stoichiometry is 1:3, like Fe(OH)₃?
For Fe(OH)₃(s) ⇌ Fe³⁺ + 3 OH⁻, if the solubility is s, then [Fe³⁺] = s and [OH⁻] = 3s, so Ksp = s(3s)³ = 27s⁴. The algebra is a bit more involved (fourth root instead of square or cube root), but the logic is identical.

Q3: Are there real examples of Ksp calculation used in medicine or health?
Yes. Predicting when calcium phosphate or calcium oxalate will precipitate as kidney stones involves comparing ionic products to Ksp values, along with complex speciation. Medical organizations like the NIH and Mayo Clinic discuss risk factors such as urinary calcium and oxalate concentrations; underneath, the same Ksp framework is being applied.

Q4: How do I know when I can ignore activity coefficients in Ksp examples?
Most classroom examples of Ksp calculation for sparingly soluble salts assume ideal behavior and low ionic strength, so you use concentrations directly. In high‑ionic‑strength solutions (like seawater or brines), you should technically use activities instead of concentrations. Advanced texts and research papers, including those referenced by USGS and major universities, discuss how to correct for this.

Q5: What are some good practice problems beyond these examples?
Look for problems that mix Ksp with acid–base equilibria, buffer solutions, and complex ion formation (like AgCl in ammonia). These combined‑equilibrium problems give the best examples of how Ksp interacts with the rest of solution chemistry and reflect how real systems behave.

If you can work comfortably through these examples of examples of Ksp calculation for sparingly soluble salts—pure water, common‑ion, pH‑dependent, and mixed‑ion scenarios—you’re in solid shape for both exams and real‑world chemistry problems.