Calculate the pH and S2- in a 0.10 M H2S Solution
Use this premium weak-acid equilibrium calculator to estimate pH, hydrosulfide concentration, sulfide ion concentration, and species distribution for aqueous hydrogen sulfide. The default setup is 0.10 M H2S with accepted dissociation constants, but you can modify the inputs for classwork, lab prep, or process calculations.
H2S Equilibrium Calculator
Enter the concentration and dissociation constants, then calculate pH and sulfide species.
Equilibrium Snapshot
The calculator models these stepwise acid equilibria in water.
The chart compares equilibrium concentrations of H2S, HS-, and S2-. In a 0.10 M H2S solution, S2- is extremely small because the second dissociation is much weaker than the first.
How to calculate the pH and S2- in a 0.10 M H2S solution
Hydrogen sulfide, H2S, is a classic example of a diprotic weak acid. That means it can donate two protons in stepwise fashion, but neither ionization is complete in water. If you are trying to calculate the pH and the sulfide ion concentration, S2-, in a 0.10 M H2S solution, the most important idea is that the first dissociation controls the pH while the second dissociation is so weak that only a tiny amount of S2- forms under acidic conditions.
At 25 C, a common set of values used in general chemistry and analytical chemistry is Ka1 about 9.1 × 10-8 and Ka2 about 1.2 × 10-13. Those numbers immediately tell you the chemical story. The first ionization of H2S is weak, but still strong enough to create measurable H+ and HS-. The second ionization of HS- is much weaker, so the equilibrium lies overwhelmingly toward HS- rather than S2- in an acidic solution.
Step 1: Write the two acid dissociation reactions
Because H2S is diprotic, the chemistry happens in two stages:
- H2S ⇌ H+ + HS-
- HS- ⇌ H+ + S2-
The equilibrium expressions are:
- Ka1 = [H+][HS-] / [H2S]
- Ka2 = [H+][S2-] / [HS-]
For a 0.10 M starting solution of H2S, the first equilibrium controls the hydrogen ion concentration. Since Ka1 is small, the acid only dissociates slightly, so the standard weak acid setup works very well.
Step 2: Solve for the hydrogen ion concentration from the first dissociation
Let x be the amount of H2S that dissociates in the first step. Then at equilibrium:
- [H+] = x
- [HS-] = x
- [H2S] = 0.10 – x
Substitute into the Ka1 expression:
Ka1 = x2 / (0.10 – x)
With Ka1 = 9.1 × 10-8:
9.1 × 10-8 = x2 / (0.10 – x)
Because x is much smaller than 0.10, many textbooks use the weak-acid approximation:
x ≈ √(Ka1C) = √((9.1 × 10-8)(0.10)) = √(9.1 × 10-9) ≈ 9.54 × 10-5
Then the pH is:
pH = -log(9.54 × 10-5) ≈ 4.02
If you prefer an exact solution, solve the quadratic equation:
x = [-Ka1 + √(Ka12 + 4Ka1C)] / 2
For this system, the quadratic result is essentially the same to practical significant figures. That is why the square-root method is commonly accepted in introductory work.
Step 3: Calculate the sulfide ion concentration S2-
Now use the second dissociation:
Ka2 = [H+][S2-] / [HS-]
Rearrange to solve for sulfide:
[S2-] = Ka2[HS-] / [H+]
From the first equilibrium, [HS-] and [H+] are both approximately x, so their ratio is close to 1. Therefore:
[S2-] ≈ Ka2 = 1.2 × 10-13 M
This result surprises many students at first. Even though the solution started with 0.10 M H2S, the amount of free S2- is not remotely close to 0.10 M. It is tiny because the solution is acidic and because Ka2 is extremely small. Most sulfur remains as molecular H2S, with a small amount as HS-, and almost none as S2-.
Why S2- is so small in acidic solution
The second ionization of a diprotic acid becomes less favorable when the solution already contains substantial H+. In this case the first dissociation generates enough acidity to suppress the second one. Since pH is only around 4, the environment is far from basic, and sulfide ion is strongly disfavored. This is why S2- becomes important only at much higher pH values, where deprotonation of HS- is more favorable.
| Parameter | Typical value at 25 C | Meaning |
|---|---|---|
| Initial [H2S] | 0.10 M | Starting concentration of dissolved hydrogen sulfide |
| Ka1 | 9.1 × 10-8 | First dissociation constant for H2S |
| pKa1 | 7.04 | -log(Ka1) |
| Ka2 | 1.2 × 10-13 | Second dissociation constant for HS- |
| pKa2 | 12.92 | -log(Ka2) |
| Calculated pH | About 4.02 | Acidity of the 0.10 M H2S solution |
| Calculated [S2-] | About 1.2 × 10-13 M | Equilibrium sulfide ion concentration |
Species distribution in a 0.10 M H2S solution
After calculating x from the first dissociation, you can estimate the equilibrium concentrations of all major sulfur-containing species:
- [H2S] ≈ 0.10 – 9.54 × 10-5 = 0.09990 M
- [HS-] ≈ 9.54 × 10-5 M
- [S2-] ≈ 1.2 × 10-13 M
These values show clearly that H2S remains the dominant species by many orders of magnitude. The concentration of HS- is small but measurable, while S2- is essentially negligible under these conditions.
| Species | Approximate concentration | Approximate fraction of total sulfur |
|---|---|---|
| H2S | 0.09990 M | 99.90% |
| HS- | 9.54 × 10-5 M | 0.095% |
| S2- | 1.2 × 10-13 M | About 1.2 × 10-10% |
Common mistakes when solving this problem
- Assuming H2S is a strong acid. It is not. You cannot set [H+] = 0.10 M.
- Ignoring the stepwise nature. H2S ionizes in two stages, and Ka1 and Ka2 are very different.
- Overestimating S2-. In acidic solution, S2- is extremely small because Ka2 is tiny and the existing H+ suppresses formation of S2-.
- Forgetting the square-root approximation limits. It works well here because dissociation is much less than 5% of the initial concentration.
- Mixing up HS- and S2-. Most ionized sulfur in this solution is HS-, not S2-.
Approximation check
It is always good practice to verify that the weak-acid approximation is legitimate. If x ≈ 9.54 × 10-5 M, then:
(x / 0.10) × 100% ≈ 0.095%
That is far below 5%, so replacing 0.10 – x with 0.10 is justified. This confirms that the simple approach is appropriate and accurate for this concentration range.
How pH changes if the H2S concentration changes
For weak acids, pH does not decrease linearly with concentration. Instead, [H+] tends to scale with the square root of concentration when the approximation holds. This is why increasing H2S from 0.010 M to 0.10 M does not make the pH ten times lower. It changes by about half a log unit.
Using the same Ka1 value, the following comparison illustrates the trend:
- 0.001 M H2S gives a pH around 4.52
- 0.010 M H2S gives a pH around 4.27
- 0.10 M H2S gives a pH around 4.02
- 1.0 M H2S gives a pH around 3.77
This pattern is exactly what you expect from weak-acid behavior. The solution becomes more acidic as concentration rises, but not nearly as dramatically as a strong acid would.
Why this problem matters in chemistry and environmental science
Calculations involving H2S are relevant in general chemistry, geochemistry, wastewater treatment, corrosion science, and environmental monitoring. Hydrogen sulfide occurs in anaerobic waters, petroleum processing, natural gas systems, and sulfur-rich groundwater. Understanding whether sulfur is present as H2S, HS-, or S2- matters because each species behaves differently in volatility, toxicity, metal binding, and precipitation reactions.
For example, metal sulfide precipitation depends strongly on sulfide speciation. If the pH is low, the free sulfide ion concentration is extremely small, which changes how quickly solids may form and how much sulfide is available to react with dissolved metals. That is one reason pH control is central in sulfide chemistry.
Authoritative references for sulfide chemistry
For deeper reading, consult these reliable sources:
U.S. Environmental Protection Agency
U.S. Geological Survey
LibreTexts Chemistry educational resource
Final takeaway
To calculate the pH and S2- in a 0.10 M H2S solution, start with the first dissociation because it determines [H+]. Using Ka1 = 9.1 × 10-8, you find [H+] ≈ 9.54 × 10-5 M and pH ≈ 4.02. Then apply Ka2 = 1.2 × 10-13 to the second dissociation. Since [HS-] and [H+] are nearly equal from the first step, the sulfide ion concentration is approximately 1.2 × 10-13 M. In other words, a 0.10 M H2S solution is mildly acidic, dominated by undissociated H2S, with only a trace amount of HS- and an almost negligible amount of S2-.