Calculate the Molar Solubility of CuS at a Fixed pH
Use this premium chemistry calculator to estimate the molar solubility of copper(II) sulfide, CuS, when the pH is fixed and sulfide is distributed among H2S, HS–, and S2-. The tool applies acid-base speciation and the solubility product relationship in one step.
Results
Enter a pH and click the calculate button to see the molar solubility of CuS, the sulfide fraction present as S2-, and the supporting equilibrium values.
Expert Guide: How to Calculate the Molar Solubility of CuS at a Fixed pH
Calculating the molar solubility of copper(II) sulfide at a fixed pH is a classic equilibrium problem that combines two major ideas from general and analytical chemistry: solubility product equilibria and acid-base speciation. On paper, CuS looks simple because its dissolution expression is only one line long. In reality, the chemistry becomes more interesting the moment pH is specified, because sulfide does not remain entirely as S2- in water. Instead, dissolved sulfur is redistributed among H2S, HS–, and S2-, and that redistribution has a strong effect on how much CuS can dissolve.
The central dissolution equilibrium is:
CuS(s) ⇌ Cu2+ + S2-
The corresponding solubility product expression is:
Ksp = [Cu2+][S2-]
If sulfide behaved as a spectator ion, the problem would be straightforward. But sulfide is a basic anion, so in acidic or even mildly neutral solution it is largely protonated. That means the free concentration of S2- is only a fraction of the total dissolved sulfide. This is exactly why pH matters. When pH falls, the amount of free S2- becomes extremely small, which allows more solid CuS to dissolve before the Ksp condition is reached.
The Key Chemical Relationships
The dissolved sulfide system is governed by two acid dissociation reactions:
- H2S ⇌ H+ + HS–, with Ka1
- HS– ⇌ H+ + S2-, with Ka2
At a fixed pH, the fraction of total dissolved sulfide present as S2- is commonly written as α2:
α2 = (Ka1Ka2) / ([H+]2 + Ka1[H+] + Ka1Ka2)
If the molar solubility of CuS is s, then each mole of CuS that dissolves produces one mole of Cu2+ and one mole of total dissolved sulfide. Therefore:
- [Cu2+] = s
- Total dissolved sulfide = s
- [S2-] = α2s
Substitute those expressions into the Ksp relationship:
Ksp = s(α2s) = α2s2
So the molar solubility at fixed pH is:
s = √(Ksp / α2)
This formula is the heart of the calculator above. Once pH is known, [H+] can be found. Once [H+] is known, α2 can be computed. Once α2 is known, the molar solubility follows immediately.
Why CuS Solubility Depends So Strongly on pH
Copper(II) sulfide is famous for being extremely insoluble, with a very small Ksp. However, “extremely insoluble” does not mean “independent of conditions.” The free sulfide ion concentration is heavily suppressed in acidic media because S2- quickly protonates to HS– and H2S. From the perspective of the solubility product, this removal of free S2- lets more CuS dissolve. In basic solution, by contrast, a larger fraction of total sulfur remains as S2-, and the molar solubility drops.
This pattern is common for salts that contain a basic anion or an acidic cation. Any process that reduces the concentration of one dissolved product shifts dissolution forward. In the CuS system, lowering the pH effectively ties up S2- into protonated forms, making the solid appear “more soluble” on a total dissolved basis.
Worked Example at pH 7.00
Let us use typical instructional values:
- Ksp(CuS) = 8.0 × 10-37
- Ka1 = 9.1 × 10-8
- Ka2 = 1.2 × 10-13
- pH = 7.00, so [H+] = 1.0 × 10-7 M
First calculate α2:
α2 = (9.1 × 10-8 × 1.2 × 10-13) / ((1.0 × 10-7)2 + (9.1 × 10-8)(1.0 × 10-7) + (9.1 × 10-8)(1.2 × 10-13))
This gives α2 of roughly 5.7 × 10-7. That means only a tiny fraction of dissolved sulfide exists as free S2- at neutral pH. Then:
s = √((8.0 × 10-37) / (5.7 × 10-7))
The result is approximately 1.2 × 10-15 M. This is still very small, but it is many orders of magnitude larger than the value you would estimate if you incorrectly assumed all dissolved sulfide stayed as S2-.
Step-by-Step Method You Can Use on Exams
- Write the dissolution equilibrium for CuS and the Ksp expression.
- Recognize that sulfide is a diprotic base and must be treated with acid-base equilibria.
- Convert pH to [H+] using [H+] = 10-pH.
- Compute the sulfide fraction α2 using Ka1, Ka2, and [H+].
- Set [Cu2+] = s and [S2-] = α2s.
- Substitute into Ksp = [Cu2+][S2-] to get Ksp = α2s2.
- Solve for s = √(Ksp/α2).
- Check whether your answer is chemically reasonable. Solubility should usually increase as pH decreases.
Speciation Data Table for the Sulfide System
The table below uses Ka1 = 9.1 × 10-8 and Ka2 = 1.2 × 10-13. Values are approximate and show how the S2- fraction changes dramatically with pH.
| pH | [H+] (M) | Approx. α2 for S2- | Interpretation |
|---|---|---|---|
| 4 | 1.0 × 10-4 | 1.1 × 10-16 | Almost no dissolved sulfide exists as free S2-. |
| 7 | 1.0 × 10-7 | 5.7 × 10-7 | Neutral solution still strongly suppresses S2-. |
| 10 | 1.0 × 10-10 | 1.3 × 10-3 | Basic conditions increase the free sulfide fraction. |
| 12 | 1.0 × 10-12 | 5.4 × 10-1 | A substantial portion of dissolved sulfide is now S2-. |
Estimated CuS Molar Solubility Versus pH
The next table combines the same speciation assumptions with Ksp(CuS) = 8.0 × 10-37. It illustrates the trend expected from the equation s = √(Ksp/α2).
| pH | α2 for S2- | Estimated Molar Solubility, s (M) | Trend Summary |
|---|---|---|---|
| 4 | 1.1 × 10-16 | 2.7 × 10-10 | Acidic solution strongly increases total dissolved CuS. |
| 7 | 5.7 × 10-7 | 1.2 × 10-15 | Neutral pH gives an extremely small but measurable theoretical value. |
| 10 | 1.3 × 10-3 | 7.9 × 10-17 | More free sulfide means lower solubility. |
| 12 | 5.4 × 10-1 | 1.2 × 10-18 | In strongly basic conditions CuS remains exceptionally insoluble. |
Common Mistakes to Avoid
- Ignoring sulfide protonation. This is the most common error. If you set [S2-] = s at all pH values, your answer can be wrong by many orders of magnitude.
- Using pOH instead of pH. The α2 expression requires [H+], so make sure you convert correctly.
- Mixing up Ka and pKa. The formula uses Ka values, not pKa values, unless you convert first.
- Forgetting units. Solubility is reported in mol/L or M.
- Assuming constants are universal. Ksp and Ka values vary somewhat with temperature and source.
How to Interpret the Calculator Output
The calculator returns the molar solubility s, the hydrogen ion concentration, the S2- fraction α2, and the predicted free S2- concentration at equilibrium. It also plots estimated molar solubility across the full pH range from 0 to 14. This chart is helpful because it gives immediate visual intuition: the curve usually slopes downward with increasing pH, reflecting the reduced need for CuS to dissolve when sulfide remains in the free S2- form.
In laboratory and environmental chemistry, this kind of analysis matters because metal sulfides play roles in precipitation, wastewater treatment, qualitative inorganic analysis, geochemistry, and ore processing. The exact numeric value depends on which thermodynamic constants you adopt, but the direction of the pH effect is robust: lower pH typically increases the apparent molar solubility of CuS.
Practical Limitations and Advanced Notes
This simplified model is excellent for teaching and many problem-solving contexts, but advanced systems may require additional chemistry. Copper can form hydroxo complexes, sulfide can participate in metal complexation, ionic strength can change effective activities, and oxidation of sulfide can occur in aerated systems. In real environmental matrices, the “true” dissolved copper concentration may deviate from the idealized classroom value. Still, for a fixed-pH equilibrium calculation built around Ksp and sulfide acid-base chemistry, the method used here is the standard and correct starting point.
Authoritative Reference Links
- U.S. Environmental Protection Agency: pH Overview
- U.S. Geological Survey: pH and Water
- NIST Chemistry WebBook: Hydrogen Sulfide Data
Educational note: equilibrium constants in the literature can differ by source, ionic strength convention, and temperature. If your instructor or textbook provides specific values, use those values in the calculator inputs.