Calculate the pH of a 0.0845 M Solution of H2S
Use this premium weak-acid calculator to determine the pH of hydrogen sulfide solution from first principles. The default setup is preloaded for a 0.0845 M H2S solution at 25°C with accepted weak-acid dissociation values, and the visualization updates instantly when you calculate.
H2S pH Calculator
Hydrogen sulfide is a weak diprotic acid. For most introductory and many practical calculations, the first dissociation dominates the pH.
Results
Click Calculate pH to solve the default 0.0845 M H2S problem.
Visual Breakdown
This chart compares the initial concentration with the equilibrium concentrations predicted from the weak-acid calculation.
For a dilute weak acid like H2S, the equilibrium concentrations of H+ and HS– are very small compared with the initial H2S concentration, which is why the pH remains well above that of a strong acid at the same formal concentration.
Expert Guide: How to Calculate the pH of a 0.0845 M Solution of H2S
To calculate the pH of a 0.0845 M solution of H2S, you need to recognize the chemistry first: hydrogen sulfide is a weak diprotic acid, not a strong acid. That single observation determines the entire approach. If H2S were a strong acid, you could simply set the hydrogen ion concentration equal to the acid concentration. But H2S does not fully ionize in water. Instead, it establishes an equilibrium, and only a small fraction of the dissolved H2S molecules donate a proton to water. That means the pH must be found from an equilibrium expression rather than by direct stoichiometry.
The first dissociation step is the important one for this problem:
H2S ⇌ H+ + HS–
The second dissociation step exists too:
HS– ⇌ H+ + S2-
However, for a solution as acidic as this one, the second ionization is so small that it contributes negligibly to the overall hydrogen ion concentration. In most general chemistry and analytical chemistry settings, the pH of a moderately concentrated H2S solution is therefore determined almost entirely by the first acid dissociation constant, Ka1.
Step 1: Write the Acid Dissociation Expression
At 25°C, a commonly used value for the first dissociation constant of hydrogen sulfide is approximately Ka1 = 9.1 × 10-8. For the first dissociation, the equilibrium expression is:
Ka1 = [H+][HS–] / [H2S]
Let the initial concentration of H2S be 0.0845 M and let x be the amount that dissociates. Then at equilibrium:
- [H2S] = 0.0845 – x
- [H+] = x
- [HS–] = x
Substituting into the equilibrium expression gives:
9.1 × 10-8 = x² / (0.0845 – x)
Step 2: Decide Whether the Weak-Acid Approximation Is Valid
Because Ka1 is extremely small compared with the formal concentration, you would expect x to be much smaller than 0.0845. That allows the approximation:
0.0845 – x ≈ 0.0845
Then the equation simplifies to:
x² = (9.1 × 10-8)(0.0845)
So:
x = √(7.6895 × 10-9) ≈ 8.77 × 10-5 M
Since x is the equilibrium hydrogen ion concentration from the first dissociation, we have:
[H+] ≈ 8.77 × 10-5 M
Now calculate pH:
pH = -log(8.77 × 10-5) ≈ 4.06
Final answer: the pH of a 0.0845 M solution of H2S is approximately 4.06 when Ka1 is taken as 9.1 × 10-8 at 25°C.
Step 3: Confirm with the Quadratic Formula
For more rigorous work, especially if your instructor asks you not to assume x is negligible, solve the exact equilibrium expression:
9.1 × 10-8 = x² / (0.0845 – x)
Rearrange:
x² + (9.1 × 10-8)x – (7.6895 × 10-9) = 0
Apply the quadratic formula:
x = [-Ka + √(Ka² + 4KaC)] / 2
Substituting Ka = 9.1 × 10-8 and C = 0.0845 gives virtually the same result:
x ≈ 8.77 × 10-5 M
That confirms the approximation was excellent. In fact, the percent ionization is only around one tenth of one percent, so the change in [H2S] is tiny relative to the starting concentration.
Why the Second Dissociation Is Usually Ignored
Hydrogen sulfide is diprotic, but the second proton is much less acidic. A representative value for Ka2 is around 1.2 × 10-12. That means once HS– forms, it only very weakly donates the second proton under these conditions. The pH generated by the first dissociation already creates enough hydrogen ion concentration to suppress the second step even further through Le Châtelier’s principle. As a result, the concentration of S2- is extremely small and does not materially alter the pH in this problem.
Common Student Mistakes in H2S pH Problems
- Treating H2S like a strong acid. If you assume [H+] = 0.0845 M, you would get a pH near 1.07, which is wildly incorrect.
- Using both protons as if they fully dissociate. That would make the estimate even more wrong.
- Forgetting that H2S is weak and requires an equilibrium setup.
- Using pKa instead of Ka without converting properly. If you are given pKa1, use Ka1 = 10-pKa1.
- Ignoring units. The equilibrium constants are defined in the context of molar concentrations, so concentration must be entered consistently.
Numerical Comparison: Weak Acid Versus Strong Acid Assumptions
The table below shows why proper acid classification matters. A 0.0845 M strong monoprotic acid would produce a dramatically lower pH than H2S because it ionizes nearly completely.
| Case | Formal concentration | Assumption for [H+] | Calculated pH | Interpretation |
|---|---|---|---|---|
| 0.0845 M H2S | 0.0845 M | 8.77 × 10-5 M | 4.06 | Weak acid, partial ionization only |
| 0.0845 M HCl | 0.0845 M | 0.0845 M | 1.07 | Strong acid, near-complete ionization |
| 0.0845 M acetic acid | 0.0845 M | 1.23 × 10-3 M | 2.91 | Weak acid, but stronger than H2S |
This comparison is particularly useful because it shows that weak acids can have the same analytical concentration as strong acids but still generate far lower hydrogen ion concentrations. H2S, with Ka1 on the order of 10-7, is much weaker than acetic acid, whose Ka is about 1.8 × 10-5. That is why a 0.0845 M H2S solution still has a pH above 4.
Percent Ionization of 0.0845 M H2S
Percent ionization tells you what fraction of acid molecules actually dissociate:
Percent ionization = ([H+] / initial concentration) × 100
Using the values above:
Percent ionization = (8.77 × 10-5 / 0.0845) × 100 ≈ 0.104%
That tiny value is strong evidence that the weak-acid approximation is justified. It also offers chemical intuition: in a beaker of dissolved H2S, the overwhelming majority of species remain as molecular H2S, not as free ions.
| Quantity | Symbol | Value for 0.0845 M H2S | Notes |
|---|---|---|---|
| Initial acid concentration | C | 0.0845 M | Given in the problem |
| First dissociation constant | Ka1 | 9.1 × 10-8 | Representative 25°C value |
| Equilibrium [H+] | x | 8.77 × 10-5 M | From quadratic or √(KaC) |
| pH | -log[H+] | 4.06 | Main answer |
| Percent ionization | (x/C) × 100 | 0.104% | Shows weak dissociation |
How This Problem Is Solved in General Chemistry Courses
Most textbooks expect the following workflow for a weak-acid pH problem like this:
- Classify the acid as strong or weak.
- Write the relevant dissociation reaction.
- Set up an ICE table.
- Write the Ka expression.
- Solve for x using either the approximation or the quadratic formula.
- Convert x to pH using pH = -log[H+].
- Check whether the approximation was valid.
This process is not only correct for H2S, but also broadly useful for many weak acids such as HF, HCN, hypochlorous acid, and carboxylic acids. The only major differences are the Ka value and whether the acid has one acidic proton or more than one.
Real Chemical Context for Hydrogen Sulfide
Hydrogen sulfide is important beyond classroom chemistry. It appears in petroleum processing, geothermal systems, wastewater treatment, anaerobic biological environments, and occupational safety studies because it is both chemically relevant and hazardous. In aqueous systems, pH controls how sulfur species distribute between H2S and HS–. That matters in corrosion, odor control, gas release, and environmental transport. While your current problem is a straightforward pH calculation, it connects directly to real engineering and environmental chemistry decisions.
Authoritative References for Acid-Base and Sulfide Chemistry
If you want to cross-check equilibrium constants or review acid-base fundamentals from authoritative educational and government resources, these sources are useful:
- Chemistry LibreTexts educational reference
- U.S. Environmental Protection Agency materials on sulfide and water chemistry
- NIST Chemistry WebBook for chemical property reference data
Final Takeaway
When asked to calculate the pH of a 0.0845 M solution of H2S, the key is to treat H2S as a weak acid and use its first dissociation constant. Using Ka1 = 9.1 × 10-8, the equilibrium hydrogen ion concentration is approximately 8.77 × 10-5 M, which gives a pH of about 4.06. The second dissociation is negligible under these conditions, and the weak-acid approximation is highly valid because the percent ionization is only about 0.104%.
So if you need the concise answer for homework, test review, or lab preparation, this is the result you should remember: the pH of a 0.0845 M H2S solution is approximately 4.06, assuming standard weak-acid constants at room temperature.