Calculate The Ph Of A 0.0679M Solution Of Hydrogen Sulfide

Use this premium calculator to estimate the pH of an aqueous hydrogen sulfide solution using either the exact weak-acid equilibrium method or the common square-root approximation. The default settings are preloaded for a 0.0679 M H₂S solution and a first dissociation constant consistent with standard general chemistry references.

Hydrogen Sulfide pH Calculator

For a weak diprotic acid like H₂S, the first dissociation controls most of the pH in moderately concentrated solutions. The second dissociation is so small that its effect on pH is negligible for this case, but it is shown for completeness.

Calculated Results

Expert Guide: How to Calculate the pH of a 0.0679 M Solution of Hydrogen Sulfide

Hydrogen sulfide, written as H₂S, is a weak diprotic acid. That means it can donate two protons in water, but it does not dissociate completely the way a strong acid such as hydrochloric acid does. If you are asked to calculate the pH of a 0.0679 M solution of hydrogen sulfide, the key idea is that the first dissociation step dominates the hydrogen ion concentration, while the second dissociation contributes so little that it can usually be ignored for introductory and most intermediate chemistry problems.

For the first equilibrium, hydrogen sulfide behaves as:

H₂S ⇌ H⁺ + HS⁻

The acid dissociation constant for this first step at room temperature is typically taken as about Ka₁ = 9.1 × 10^-8. Because this value is small, only a tiny fraction of the dissolved H₂S molecules ionize. That immediately tells us two things. First, the solution is acidic, but not nearly as acidic as a strong acid of the same formal concentration. Second, weak-acid equilibrium methods, not strong-acid shortcuts, must be used.

Step 1: Set up the weak-acid equilibrium expression

Let the initial concentration of hydrogen sulfide be 0.0679 M. If an amount x dissociates, the equilibrium concentrations become:

• [H₂S] = 0.0679 – x
• [H⁺] = x
• [HS⁻] = x

Insert these into the equilibrium expression:

Ka₁ = [H⁺][HS⁻] / [H₂S] = x² / (0.0679 – x)

Using Ka₁ = 9.1 × 10^-8 gives:

9.1 × 10^-8 = x² / (0.0679 – x)

Step 2: Decide whether the approximation is valid

Because Ka is much smaller than the initial concentration, many chemistry courses use the approximation that x is tiny compared with 0.0679. In that case:

0.0679 – x ≈ 0.0679

So the equation simplifies to:

x² = (9.1 × 10^-8)(0.0679)

x = √(6.1789 × 10^-9) ≈ 7.86 × 10^-5 M

Since x equals the equilibrium hydrogen ion concentration for the first dissociation, we now calculate pH:

pH = -log[H⁺] = -log(7.86 × 10^-5) ≈ 4.10

That is the standard textbook answer: the pH of a 0.0679 M hydrogen sulfide solution is about 4.10.

Step 3: Verify with the exact quadratic solution

If you want a more rigorous calculation, solve the equilibrium equation without approximation:

x² + Ka₁x – Ka₁C = 0

where C = 0.0679 M and Ka₁ = 9.1 × 10^-8.

Using the quadratic formula:

x = [-Ka₁ + √(Ka₁² + 4Ka₁C)] / 2

This gives essentially the same answer, with [H⁺] very close to 7.85 × 10^-5 M and pH still about 4.10. The reason both answers agree is that the approximation is excellent in this case. The percent ionization is only a small fraction of 1%, so neglecting x in the denominator is justified.

Why the second dissociation can usually be ignored

Hydrogen sulfide is diprotic, which means there is also a second equilibrium:

HS⁻ ⇌ H⁺ + S²⁻

The second dissociation constant is much smaller, commonly around Ka₂ = 1.2 × 10^-13. By the time this second step could occur, the solution already contains hydrogen ions from the first dissociation. That suppresses the second ionization strongly. In practical terms, the second step contributes almost nothing to the total [H⁺] in a 0.0679 M H₂S solution. Therefore, for pH calculations at this concentration, using only Ka₁ is the accepted and chemically correct approach.

Exact answer and practical answer

• Formal concentration: 0.0679 M H₂S
• Ka₁ used: 9.1 × 10^-8
• Calculated [H⁺]: about 7.85 to 7.86 × 10^-5 M
• Calculated pH: about 4.10
• Percent ionization: about 0.116%

Comparison table: exact vs approximate method

Method	Equation Used	[H⁺] Result	pH Result	Practical Use
Approximation	x = √(KaC)	7.86 × 10^-5 M	4.10	Fast hand calculation in general chemistry
Exact quadratic	x = [-Ka + √(Ka² + 4KaC)] / 2	7.85 × 10^-5 M	4.10	Best for calculator or software validation

How weak is hydrogen sulfide compared with common acids?

A useful way to understand the result is to compare H₂S with other familiar acids. A strong acid at 0.0679 M would produce [H⁺] very close to 0.0679 M, corresponding to a pH near 1.17. Hydrogen sulfide, in contrast, gives [H⁺] around 7.85 × 10^-5 M, which is nearly 860 times lower in hydrogen ion concentration than a solution with pH 1.17? Actually, because the difference is nearly 2.94 pH units, the hydrogen ion concentration is lower by about 10^2.94 ≈ 870 times. This highlights why H₂S must be treated as a weak acid.

Acid System	Representative Acid Constant or Behavior	Approximate [H⁺] at 0.0679 M	Approximate pH
Strong monoprotic acid	Essentially complete dissociation	6.79 × 10^-2 M	1.17
Hydrogen sulfide, H₂S	Ka₁ = 9.1 × 10^-8	7.85 × 10^-5 M	4.10
Pure water at 25°C	Kw = 1.0 × 10^-14	1.0 × 10^-7 M	7.00

Common mistakes students make

1. Treating H₂S as a strong acid. This leads to a wildly incorrect pH close to 1 instead of about 4.10.
2. Using both protons as if they dissociate completely. Hydrogen sulfide is diprotic, but both protons are weakly acidic, especially the second one.
3. Ignoring Ka and using concentration directly. For weak acids, concentration alone does not determine [H⁺]; equilibrium controls the extent of ionization.
4. Forgetting the 5% rule check. The approximation is acceptable only when x is sufficiently small relative to the initial concentration.
5. Using inconsistent Ka values. Different references may list slightly different constants because of temperature and ionic strength effects. The final pH usually changes only slightly.

5% rule check for the approximation

One standard validation step is to compute the percent ionization:

percent ionization = (x / 0.0679) × 100%

Using x ≈ 7.85 × 10^-5 M:

percent ionization ≈ (7.85 × 10^-5 / 0.0679) × 100% ≈ 0.116%

Because 0.116% is far below 5%, the approximation is clearly valid. This is why the shortcut and the exact solution agree so closely.

What does the pH actually mean here?

A pH of about 4.10 means the solution is acidic enough to increase hydrogen ion concentration significantly relative to neutral water, but still far less acidic than a strong acid solution of the same concentration. In environmental chemistry and aqueous equilibrium problems, this distinction is very important because pH controls speciation, corrosion behavior, gas absorption, sulfide distribution, and biological compatibility. Hydrogen sulfide also has practical significance in wastewater treatment, petroleum systems, geothermal fluids, and natural waters.

Molality vs molarity note

The problem statement uses “0.0679m,” which can sometimes mean molality. In many classroom pH problems, however, that notation is used loosely when the intended meaning is concentration in solution. For dilute aqueous systems, treating 0.0679 m as approximately 0.0679 M usually produces a very similar pH result. If the problem comes from a rigorous physical chemistry context, then activity effects, density, and ionic strength may matter. In a standard chemistry setting, using 0.0679 M with Ka₁ at 25°C is the expected interpretation.

Authoritative references for pH and hydrogen sulfide chemistry

• USGS: pH and Water
• U.S. EPA: Hydrogen Sulfide Technical Overview
• NIH NCBI Bookshelf: Hydrogen Sulfide Profile

Worked summary

Here is the shortest complete route to the answer:

1. Write the first dissociation: H₂S ⇌ H⁺ + HS⁻
2. Use Ka₁ = x² / (0.0679 – x)
3. Approximate 0.0679 – x ≈ 0.0679 because Ka is small
4. Solve x = √[(9.1 × 10^-8)(0.0679)]
5. Get x ≈ 7.86 × 10^-5 M
6. Compute pH = -log(7.86 × 10^-5) ≈ 4.10

If you are answering a quiz, homework, or exam question asking to calculate the pH of a 0.0679 M solution of hydrogen sulfide, the correct reported result is usually pH = 4.10, with a note that the first dissociation dominates and the second dissociation is negligible. If your instructor expects the quadratic method, it will still round to the same pH value in most cases.

Final answer: The pH of a 0.0679 M hydrogen sulfide solution is approximately 4.10 when calculated using the standard first dissociation constant of H₂S at about 25°C.