Calculate OH and pH for 0.035 M Na2S Solution
This premium calculator estimates hydroxide concentration, pOH, and pH for sodium sulfide solutions using sulfide hydrolysis chemistry at 25 degrees C. The default setup is already tuned for a 0.035 M Na2S solution, which is the target case in this guide.
Na2S pH Calculator
Expert Guide: How to Calculate OH and pH for 0.035 M Na2S Solution
If you need to calculate OH and pH for 0.035 M Na2S solution, the key idea is that sodium sulfide is not just a neutral salt dissolved in water. It produces the sulfide ion, S2-, and that ion is a substantial base. Once dissolved, S2- reacts with water to generate hydroxide, OH–, and that is why the solution becomes strongly basic. The most reliable classroom approach is to treat the first hydrolysis step as the dominant source of hydroxide and then check whether the second hydrolysis contributes meaningfully. For a 0.035 M solution, the first step dominates by a very wide margin.
Start by writing the dissociation and hydrolysis reactions clearly. Sodium sulfide dissociates completely in water:
Na2S → 2 Na+ + S2-
The relevant acid-base reaction is then:
S2- + H2O ⇌ HS– + OH–
This equilibrium is governed by the base constant of sulfide, not by the neutral salt itself. Because sulfide is the conjugate base of HS–, the needed constant comes from the second acid dissociation constant of hydrogen sulfide:
Kb1 = Kw / Ka2
Step 1: Identify the correct equilibrium constants
Hydrogen sulfide, H2S, is diprotic, so it has two acid dissociation constants. Typical values used in general chemistry are approximately Ka1 = 9.1 × 10-8 and Ka2 = 1.2 × 10-13. At 25 degrees C, Kw = 1.0 × 10-14. From this, the first hydrolysis constant for sulfide is:
Kb1 = 1.0 × 10-14 / 1.2 × 10-13 = 0.0833
That value is fairly large for a weak-base equilibrium. It tells you immediately that hydrolysis is substantial, which means the common weak-base shortcut x = √(KbC) is not appropriate here. Instead, a quadratic solution gives a better answer.
| Quantity | Typical value at 25 degrees C | Role in the calculation |
|---|---|---|
| Na2S concentration, C | 0.035 M | Initial concentration of sulfide from complete salt dissociation |
| Ka1 of H2S | 9.1 × 10-8 | Used to estimate the second hydrolysis of HS– |
| Ka2 of H2S | 1.2 × 10-13 | Used to calculate Kb1 for S2- |
| Kw | 1.0 × 10-14 | Connects acid and base constants through Kb = Kw / Ka |
| Kb1 of S2- | 0.0833 | Main driver of OH– production |
| Kb2 of HS– | 1.10 × 10-7 | Usually a tiny correction relative to the first step |
Step 2: Set up the ICE table for the dominant hydrolysis step
For the reaction S2- + H2O ⇌ HS– + OH–, let x be the amount of sulfide that reacts:
- Initial: [S2-] = 0.035, [HS–] = 0, [OH–] = 0
- Change: -x, +x, +x
- Equilibrium: [S2-] = 0.035 – x, [HS–] = x, [OH–] = x
Substitute into the equilibrium expression:
Kb1 = x2 / (0.035 – x) = 0.0833
Rearranging gives a quadratic equation:
x2 + 0.0833x – 0.0029155 = 0
Solving the quadratic produces x ≈ 0.0266 M. This means the first hydrolysis step alone generates about 0.0266 M hydroxide.
Step 3: Convert hydroxide concentration to pOH and pH
Once you know [OH–], the rest is straightforward:
- pOH = -log[OH–]
- pOH = -log(0.0266) ≈ 1.58
- pH = 14.00 – 1.58 = 12.42
So, the standard textbook answer for the pH of a 0.035 M Na2S solution is approximately 12.42. The hydroxide concentration is approximately 2.66 × 10-2 M.
Step 4: Decide whether the second hydrolysis matters
After the first step, you have a significant amount of HS– in solution. HS– can also act as a base:
HS– + H2O ⇌ H2S + OH–
The base constant for that step is:
Kb2 = Kw / Ka1 = 1.0 × 10-14 / 9.1 × 10-8 ≈ 1.10 × 10-7
Because this value is tiny and the solution already contains a large hydroxide concentration from the first hydrolysis step, the second reaction is strongly suppressed. In practice, the added OH– is negligible compared with 0.0266 M. This is why most chemistry courses stop after the first hydrolysis for this kind of problem.
Common mistake: treating Na2S like a simple strong base
A frequent error is to assume Na2S behaves exactly like two moles of NaOH per mole of salt. That is not correct. The sodium ions are spectators, but the sulfide ion is a weak base in equilibrium with water, not a source of automatic stoichiometric hydroxide. Another mistake is to ignore the large value of Kb1 and use the weak-base approximation x = √(KbC). If you do that here, you get a value too high and even physically impossible because x would exceed the initial concentration. The proper quadratic method avoids that problem.
What the species distribution looks like
At equilibrium in a 0.035 M Na2S solution, the sulfur-containing species are mostly split between S2- and HS–, with only a trace amount of H2S. Using the standard constants and the dominant-step solution, you get approximately:
- [S2-] ≈ 0.0084 M
- [HS–] ≈ 0.0266 M
- [H2S] extremely small
- [OH–] ≈ 0.0266 M
This distribution makes intuitive sense. The first hydrolysis is favorable enough to convert most, but not all, of the sulfide into bisulfide, while the second hydrolysis remains too weak to generate much H2S in such a strongly basic medium.
| Na2S concentration (M) | Calculated [OH–] from first hydrolysis (M) | Calculated pOH | Calculated pH |
|---|---|---|---|
| 0.005 | 0.00374 | 2.43 | 11.57 |
| 0.010 | 0.00710 | 2.15 | 11.85 |
| 0.035 | 0.02656 | 1.58 | 12.42 |
| 0.050 | 0.03656 | 1.44 | 12.56 |
| 0.100 | 0.06591 | 1.18 | 12.82 |
Why 0.035 M Na2S ends up near pH 12.4
The central reason is the relationship between Ka2 of H2S and Kb1 of S2-. Since Ka2 is very small, the conjugate base S2- is relatively strong for a weak-base species. However, the initial concentration is also moderate, not extremely high. The final pH therefore lands in the low 12s instead of pushing all the way to 13 or higher. This is exactly the kind of result expected for a basic salt with meaningful, but not complete, hydrolysis.
Step-by-step summary you can use on homework or exams
- Write the complete dissociation: Na2S → 2 Na+ + S2-.
- Use the hydrolysis reaction: S2- + H2O ⇌ HS– + OH–.
- Compute Kb1 = Kw / Ka2.
- Set up an ICE table with initial sulfide concentration 0.035 M.
- Solve x2 / (0.035 – x) = 0.0833 using the quadratic formula.
- Get x ≈ 0.0266 M, which is approximately [OH–].
- Find pOH = -log(0.0266) ≈ 1.58.
- Find pH = 14.00 – 1.58 ≈ 12.42.
When your answer may differ slightly
You may see small variations in reported answers, often ranging from about pH 12.40 to 12.44. These differences usually come from one of three sources: different reference values for Ka1 and Ka2, rounding during the quadratic solution, or whether the second hydrolysis step is included as a tiny refinement. In most academic settings, any carefully justified answer near 12.42 is considered correct for 0.035 M Na2S at 25 degrees C.
Practical interpretation
A pH above 12 means the solution is strongly caustic and should be handled with the same chemical respect given to other highly basic laboratory solutions. In applied settings, sulfide-containing waters also raise broader environmental and safety issues because sulfide chemistry can shift with pH, oxidation conditions, and dissolved metals. That is one reason pH control matters so much in water treatment, industrial processing, and analytical chemistry.
Authoritative references for deeper study
- USGS Water Science School: pH and Water
- U.S. EPA: Alkalinity and Acid Neutralizing Capacity
- Princeton University: pH overview
Final answer
For a 0.035 M Na2S solution at 25 degrees C, using typical textbook constants for hydrogen sulfide, the calculated hydroxide concentration is about 0.0266 M, the pOH is about 1.58, and the pH is about 12.42. That is the target result this calculator returns by default.