Calculate the pH of 0.0067 M Ba(OH)2 Carefully
Use this interactive calculator to find hydroxide concentration, pOH, and final pH for barium hydroxide solutions. The default example is 0.0067 M Ba(OH)2, treated carefully as a strong base that releases two OH– ions per formula unit.
Visual Result Snapshot
This chart compares the input Ba(OH)2 molarity, the resulting hydroxide concentration, pOH, and pH on a single visual panel for quick interpretation.
How to calculate the pH of 0.0067 M Ba(OH)2 carefully
If you want to calculate the pH of 0.0067 M Ba(OH)2 carefully, the most important idea is that barium hydroxide is a strong base and dissociates essentially completely in dilute aqueous solution. That means you should not treat its hydroxide concentration as equal to the molarity of the compound. Instead, each formula unit of Ba(OH)2 releases two hydroxide ions. This is the detail students often miss, and it is exactly why the word “careful” matters in this calculation.
The dissociation equation is:
Ba(OH)2(aq) → Ba2+(aq) + 2OH–(aq)
Because the coefficient in front of OH– is 2, a 0.0067 M solution of barium hydroxide produces:
[OH–] = 2 × 0.0067 = 0.0134 M
Once you know hydroxide concentration, compute pOH:
pOH = -log(0.0134) ≈ 1.87
Then use the standard 25°C relationship:
pH + pOH = 14
Therefore:
pH = 14 – 1.87 = 12.13
Why Ba(OH)2 requires extra attention
Many pH problems involving strong bases are simple, but they become slightly more subtle when the base contributes more than one hydroxide ion. Sodium hydroxide, NaOH, contributes one OH– per formula unit, so a 0.0067 M NaOH solution would have [OH–] = 0.0067 M. Barium hydroxide is different because the formula contains two hydroxide groups. As a result, the hydroxide concentration doubles relative to the formal molarity of the dissolved base.
This distinction matters because pH is logarithmic. Even though doubling concentration may look small in ordinary arithmetic, it creates a meaningful shift in pOH and therefore pH. If someone forgets the factor of 2 and directly uses 0.0067 M as [OH–], they will get:
- Incorrect pOH = -log(0.0067) ≈ 2.17
- Incorrect pH = 14 – 2.17 ≈ 11.83
That is off by about 0.30 pH units, which is a significant error in chemistry. Since pH is logarithmic, a 0.30 unit difference corresponds to about a factor of two in hydrogen ion concentration. In lab work, titration calculations, equilibrium setup, and analytical chemistry, that difference is too large to ignore.
Step-by-step method
- Identify the solute as barium hydroxide, Ba(OH)2.
- Recognize that Ba(OH)2 is a strong base in introductory chemistry problems and dissociates completely.
- Write the dissociation equation: Ba(OH)2 → Ba2+ + 2OH–.
- Multiply the base molarity by 2 to get hydroxide concentration.
- Use pOH = -log[OH–].
- Use pH = 14 – pOH at 25°C.
- Round carefully, usually to two decimal places unless your instructor specifies otherwise.
Worked example with 0.0067 M
Start with the concentration:
C = 0.0067 M
Because each unit of Ba(OH)2 produces 2 OH–:
[OH–] = 2C = 2(0.0067) = 0.0134 M
Next:
pOH = -log(0.0134) = 1.8729…
Finally:
pH = 14.0000 – 1.8729 = 12.1271…
Rounded appropriately:
pH = 12.13
Comparison table: correct vs common mistakes
| Method | [OH–] used | pOH | pH | Comment |
|---|---|---|---|---|
| Correct Ba(OH)2 method | 0.0134 M | 1.87 | 12.13 | Uses 2 OH– per formula unit |
| Common mistake | 0.0067 M | 2.17 | 11.83 | Forgets to double hydroxide concentration |
| Difference | Factor of 2 | 0.30 unit | 0.30 unit | Large enough to matter in chemistry calculations |
Important chemistry context
In most general chemistry settings, strong bases such as LiOH, NaOH, KOH, Sr(OH)2, and Ba(OH)2 are treated as fully dissociated. However, the stoichiometric coefficient of hydroxide must still be respected. This is why strong-base pH problems often reduce to a simple stoichiometry step followed by a logarithm. The challenge is not advanced equilibrium math. The challenge is correctly identifying how many moles of OH– are produced per mole of base.
For Ba(OH)2, one mole of solute generates two moles of hydroxide ions. That ratio drives the rest of the calculation. In symbolic form, for a strong metal hydroxide M(OH)n, the hydroxide concentration is:
[OH–] = n × C
Here, n = 2 and C = 0.0067 M, so [OH–] = 0.0134 M.
Comparison table: strong bases and hydroxide yield
| Base | Formula | OH– released per mole | If formal concentration is 0.0067 M, then [OH–] | Approximate pH at 25°C |
|---|---|---|---|---|
| Sodium hydroxide | NaOH | 1 | 0.0067 M | 11.83 |
| Potassium hydroxide | KOH | 1 | 0.0067 M | 11.83 |
| Calcium hydroxide | Ca(OH)2 | 2 | 0.0134 M | 12.13 |
| Barium hydroxide | Ba(OH)2 | 2 | 0.0134 M | 12.13 |
Frequent student errors when solving this problem
1. Treating molarity of Ba(OH)2 as hydroxide molarity
This is by far the most common mistake. Always inspect the formula. If there are two hydroxide groups, then one mole of dissolved compound contributes two moles of hydroxide ions.
2. Using pH = -log[OH–]
That formula gives pOH, not pH. For a basic solution at 25°C, you first compute pOH, then subtract from 14 to get pH.
3. Forgetting the temperature assumption
The relationship pH + pOH = 14 is strictly based on the ionic product of water at approximately 25°C in standard coursework. In more advanced chemistry, the value changes slightly with temperature. Since this problem is a standard general chemistry question, using 14 is appropriate unless a different temperature is specified.
4. Rounding too early
If you round [OH–] or pOH too aggressively before the final subtraction, you can introduce small but avoidable error. Carry a few extra digits through the intermediate steps and round only at the end.
Real-world interpretation of the answer
A pH of 12.13 means the solution is strongly basic. This is far above neutral pH 7. On the pH scale, each whole unit corresponds to a tenfold change in hydrogen ion concentration, so values above 12 indicate a substantial excess of hydroxide ions. In practical terms, such a solution is chemically caustic and must be handled carefully in a laboratory setting with proper eye and skin protection. Strong basic solutions can damage tissue, alter indicator colors dramatically, and react readily with acids.
Although the molarity 0.0067 M may appear small compared with concentrated stock solutions, once dissociation is considered, the hydroxide concentration is still high enough to produce a strongly alkaline pH. This is a useful reminder that pH behavior depends not just on the stated molarity of the compound, but also on the number of acidic or basic particles released in solution.
Authority sources for related chemistry concepts
- LibreTexts Chemistry for foundational pH, pOH, and strong electrolyte explanations.
- U.S. Environmental Protection Agency for water chemistry and pH background in environmental systems.
- NIST Chemistry WebBook for authoritative chemical data resources.
- USGS Water Science School for pH concepts and interpretation.
Quick summary formula set
- Ba(OH)2 → Ba2+ + 2OH–
- [OH–] = 2 × 0.0067 = 0.0134 M
- pOH = -log(0.0134) ≈ 1.87
- pH = 14 – 1.87 ≈ 12.13
Final takeaway
To calculate the pH of 0.0067 M Ba(OH)2 carefully, you must account for complete dissociation and the fact that barium hydroxide provides two hydroxide ions per dissolved formula unit. Once that stoichiometric detail is included, the rest of the problem is straightforward. The correct hydroxide concentration is 0.0134 M, the pOH is about 1.87, and the final pH is about 12.13. If you remember to multiply by the hydroxide coefficient before taking the logarithm, you will avoid the most common error and solve this type of problem reliably.