Calculate the pH in a 0.00150 M Ba(OH)2 Solution
Use this interactive chemistry calculator to find hydroxide concentration, pOH, and pH for a barium hydroxide solution. The default setup solves the exact problem: calculate the pH in a 0.00150 M Ba(OH)2 solution, assuming complete dissociation at 25 degrees Celsius.
How to calculate the pH in a 0.00150 M Ba(OH)2 solution
To calculate the pH in a 0.00150 M Ba(OH)2 solution, you begin with the fact that barium hydroxide is a strong base. In most introductory and intermediate chemistry settings, strong bases are assumed to dissociate completely in water. That means each formula unit of Ba(OH)2 separates into one Ba2+ ion and two OH– ions:
Ba(OH)2 → Ba2+ + 2OH–
The key idea is that the hydroxide concentration is not the same as the original Ba(OH)2 concentration. Because there are two hydroxide ions released for every one unit of Ba(OH)2, you must multiply the molarity by 2:
[OH–] = 2 × 0.00150 = 0.00300 M
Next, calculate pOH:
pOH = -log(0.00300) = 2.523
At 25 degrees Celsius, the relationship between pH and pOH is:
pH + pOH = 14.00
So:
pH = 14.00 – 2.523 = 11.477
Why Ba(OH)2 gives twice as much hydroxide
This is one of the most common points of confusion in acid-base calculations. Many students see the molarity of the dissolved compound and immediately plug that concentration into the pOH formula. That would work for a base such as NaOH, where one mole of NaOH gives one mole of OH–. It does not work directly for Ba(OH)2, because the chemical formula contains two hydroxide groups.
In other words, stoichiometry controls the ion concentration. If you dissolve 0.00150 moles of Ba(OH)2 per liter, complete dissociation produces:
- 0.00150 M Ba2+
- 0.00300 M OH–
That doubling step is the reason the pH is significantly higher than it would be if you incorrectly treated the hydroxide concentration as 0.00150 M. Acid-base problems involving Ca(OH)2, Sr(OH)2, and Ba(OH)2 often test this exact concept.
Step-by-step method you can use on homework and exams
- Write the balanced dissociation equation for Ba(OH)2.
- Use stoichiometry to determine hydroxide ion concentration.
- Calculate pOH using pOH = -log[OH–].
- Use pH = 14.00 – pOH at 25 degrees Celsius.
- Round according to the requested precision.
Worked example with the exact numbers
Suppose the solution concentration is 0.00150 M Ba(OH)2. Since each mole of Ba(OH)2 yields 2 moles of OH–, the hydroxide concentration becomes 0.00300 M. The negative base-10 logarithm of 0.00300 is 2.523. Subtracting from 14.00 gives a pH of 11.477. In a standard chemistry classroom, this would be reported as 11.48 or 11.477 depending on rounding rules.
Common mistakes when solving this problem
- Forgetting the coefficient 2: Ba(OH)2 releases two hydroxide ions, not one.
- Calculating pH directly from Ba(OH)2 concentration: you must first convert to [OH–].
- Mixing up pH and pOH: strong bases are often easiest to handle through pOH first.
- Using the wrong pH plus pOH relationship: 14.00 is standard at 25 degrees Celsius, but pKw changes slightly with temperature.
- Rounding too early: keep extra digits until the final line to avoid small but noticeable errors.
Comparison table: strong bases and hydroxide yield
| Base | Dissociation pattern | OH– released per mole of base | If base concentration = 0.00150 M, then [OH–] |
|---|---|---|---|
| NaOH | NaOH → Na+ + OH– | 1 | 0.00150 M |
| KOH | KOH → K+ + OH– | 1 | 0.00150 M |
| Ca(OH)2 | Ca(OH)2 → Ca2+ + 2OH– | 2 | 0.00300 M |
| Ba(OH)2 | Ba(OH)2 → Ba2+ + 2OH– | 2 | 0.00300 M |
Interpreting the result: is pH 11.477 strongly basic?
Yes. A pH of 11.477 is definitely basic and far above neutral pH 7. In practical terms, this means the solution has a high hydroxide ion concentration compared with pure water. While this is not as extreme as concentrated laboratory hydroxide solutions, it is still chemically basic enough to affect indicators, react with acids, and require safe handling procedures in a real laboratory.
The reason pH values above 11 matter is that the pH scale is logarithmic. A shift of one pH unit corresponds to a tenfold change in hydrogen ion activity. So even a moderate-looking numerical increase reflects a large chemical difference in acidity or basicity.
Comparison data table: pH values for selected Ba(OH)2 concentrations at 25 degrees Celsius
| Ba(OH)2 concentration (M) | [OH–] (M) | pOH | pH |
|---|---|---|---|
| 0.000100 | 0.000200 | 3.699 | 10.301 |
| 0.000500 | 0.001000 | 3.000 | 11.000 |
| 0.00150 | 0.00300 | 2.523 | 11.477 |
| 0.00500 | 0.01000 | 2.000 | 12.000 |
| 0.0100 | 0.0200 | 1.699 | 12.301 |
What assumptions are built into the standard answer?
The textbook answer of about 11.48 depends on a few standard assumptions. First, it assumes barium hydroxide behaves as a strong electrolyte, meaning dissociation is effectively complete. Second, it assumes the solution is dilute enough that basic stoichiometric treatment is valid for classroom calculations. Third, it assumes the temperature is 25 degrees Celsius, where pKw is taken as 14.00. Finally, it ignores small activity effects that become more important in advanced physical chemistry.
In general chemistry, those assumptions are exactly what instructors expect unless a problem explicitly asks for a more rigorous treatment. If you are studying for AP Chemistry, college general chemistry, nursing chemistry, or allied health chemistry, this strong-base method is the correct one for a problem written this way.
How this problem connects to broader acid-base chemistry
This calculation sits at the intersection of three foundational chemistry skills: dissociation, stoichiometry, and logarithms. Once you understand those three elements, many acid-base questions become much easier. For example, the same logic can be used to compare strong monoprotic bases and strong dibasic bases, estimate neutralization outcomes, and predict indicator color changes.
It also teaches an important habit: always inspect the chemical formula before starting. If the species can generate more than one H+ or OH–, that stoichiometric multiplier can change the final answer significantly. Students who slow down for this initial inspection usually avoid the most common error.
Quick reference formula set
- Ba(OH)2 → Ba2+ + 2OH–
- [OH–] = 2 × [Ba(OH)2]
- pOH = -log[OH–]
- pH = 14.00 – pOH at 25 degrees Celsius
Authoritative chemistry references
If you want to verify acid-base principles, solution chemistry, or pH definitions from trustworthy educational and government-backed sources, these references are excellent starting points:
- Chemistry LibreTexts for detailed university-level explanations of strong bases, pH, and logarithmic calculations.
- U.S. Environmental Protection Agency for practical pH background and water chemistry context.
- NIST Chemistry WebBook for authoritative chemical data resources maintained by the U.S. government.
Final takeaway
When asked to calculate the pH in a 0.00150 M Ba(OH)2 solution, the most important step is recognizing that barium hydroxide contributes two hydroxide ions per formula unit. That doubles the hydroxide concentration to 0.00300 M. From there, the pOH is 2.523 and the pH at 25 degrees Celsius is 11.477. If you remember that one stoichiometric detail, the rest of the problem becomes a clean and fast logarithm exercise.