Calculate the pH of a 0.24 M Ba(OH)2 Solution
Use this premium chemistry calculator to find hydroxide concentration, pOH, and pH for barium hydroxide solutions using the strong base dissociation model.
Ba(OH)2 pH Calculator
Click Calculate pH to solve for a 0.24 M Ba(OH)2 solution or enter your own concentration.
How to calculate the pH of a 0.24 M Ba(OH)2 solution
If you need to calculate the pH of a 0.24 M Ba(OH)2 solution, the key idea is that barium hydroxide is a strong base that dissociates essentially completely in water under ordinary general chemistry conditions. That means each formula unit of Ba(OH)2 releases one barium ion and two hydroxide ions. Because pH depends on either hydrogen ion concentration or, for bases, the hydroxide concentration through pOH, the stoichiometric factor of 2 is the most important step. Once you account for that, the rest of the calculation is straightforward.
For a concentration of 0.24 M Ba(OH)2, the hydroxide concentration is not 0.24 M. Instead, it is double that amount because each mole of Ba(OH)2 produces 2 moles of OH-. This gives:
Then calculate pOH from the hydroxide ion concentration:
At 25 C, the standard relationship between pH and pOH is:
So the final pH is:
Rounded appropriately, the pH of a 0.24 M Ba(OH)2 solution is 13.68.
Why Ba(OH)2 gives two hydroxide ions
Students often miss this because they focus only on the molarity of the original compound. Barium hydroxide is written as Ba(OH)2, and the subscript 2 on hydroxide means that one dissolved formula unit provides two OH- ions. The dissociation equation is:
This is why the hydroxide concentration is always twice the molarity of dissolved Ba(OH)2, assuming complete dissociation. In introductory chemistry, that assumption is standard for strong bases such as sodium hydroxide, potassium hydroxide, and barium hydroxide. The same logic also applies when comparing bases with one hydroxide versus two hydroxides. For example, a 0.24 M NaOH solution would give 0.24 M OH-, but a 0.24 M Ba(OH)2 solution gives 0.48 M OH- and therefore has a higher pH.
Step by step method
- Write the balanced dissociation equation: Ba(OH)2 → Ba2+ + 2OH-.
- Start with the molarity of barium hydroxide: 0.24 M.
- Multiply by 2 to get hydroxide concentration: [OH-] = 0.48 M.
- Calculate pOH using pOH = -log10[OH-].
- Use pH = 14.00 – pOH at 25 C.
- Round to a sensible number of decimal places, giving pH ≈ 13.68.
Worked example in detail
Let us solve it carefully, as you might in a homework set or exam response.
- Given concentration of Ba(OH)2 = 0.24 mol/L
- Each mole of Ba(OH)2 produces 2 moles of OH-
- Therefore [OH-] = 2 × 0.24 = 0.48 mol/L
- pOH = -log10(0.48) = 0.318758763
- pH = 14.000000000 – 0.318758763 = 13.681241237
Final answer: pH = 13.68.
Comparison table: Ba(OH)2 versus other common strong bases
The table below shows why the hydroxide stoichiometry matters. The pH values assume ideal complete dissociation at 25 C.
| Base | Initial concentration (M) | OH- ions released per formula unit | Resulting [OH-] (M) | pOH | pH |
|---|---|---|---|---|---|
| NaOH | 0.24 | 1 | 0.24 | 0.6198 | 13.3802 |
| KOH | 0.24 | 1 | 0.24 | 0.6198 | 13.3802 |
| Ca(OH)2 | 0.24 | 2 | 0.48 | 0.3188 | 13.6812 |
| Ba(OH)2 | 0.24 | 2 | 0.48 | 0.3188 | 13.6812 |
This comparison makes the main concept very clear. Bases that release two hydroxide ions per formula unit produce a higher hydroxide concentration at the same starting molarity than bases that release only one. Since pOH decreases as hydroxide concentration increases, pH rises accordingly.
What assumptions are used in this calculation?
Most classroom pH calculations use a short list of assumptions so the problem can be solved cleanly:
- Complete dissociation: Ba(OH)2 is treated as a strong base in water.
- Ideal behavior: activities are approximated by concentrations.
- 25 C reference: pH + pOH = 14.00 is used.
- No competing equilibria: precipitation, complexation, or high ionic strength corrections are ignored.
These are appropriate assumptions for typical general chemistry exercises. In advanced analytical chemistry or physical chemistry, activity coefficients and ionic strength effects can matter, especially at higher concentrations. However, for the expected level of this problem, using concentration directly is the correct method.
Common mistakes students make
- Forgetting the factor of 2. The most common mistake is using [OH-] = 0.24 M instead of 0.48 M.
- Calculating pH directly from the base molarity. You must find hydroxide concentration first.
- Mixing up pH and pOH. Strong bases are often easiest to solve by finding pOH first, then converting to pH.
- Using the wrong logarithm sign. Remember pOH = -log10[OH-].
- Ignoring temperature assumptions. The relation pH + pOH = 14.00 is the usual 25 C classroom approximation.
Reference table: Ba(OH)2 concentration and expected pH
The following data points show how pH changes as the concentration of barium hydroxide changes. These values are computed with complete dissociation at 25 C and are useful for checking homework answers.
| Ba(OH)2 concentration (M) | Calculated [OH-] (M) | Calculated pOH | Calculated pH |
|---|---|---|---|
| 0.001 | 0.002 | 2.6990 | 11.3010 |
| 0.010 | 0.020 | 1.6990 | 12.3010 |
| 0.050 | 0.100 | 1.0000 | 13.0000 |
| 0.100 | 0.200 | 0.6990 | 13.3010 |
| 0.240 | 0.480 | 0.3188 | 13.6812 |
| 0.500 | 1.000 | 0.0000 | 14.0000 |
Why this problem matters in chemistry
Questions like this test more than just calculator skills. They check whether you understand ionic dissociation, stoichiometry in aqueous solution, logarithmic scales, and the relationship between pH and pOH. Being able to move from a chemical formula to ion concentrations is essential in acid-base chemistry, equilibrium, titrations, buffer analysis, and solubility problems.
For example, if you later work on neutralization reactions, you will need to know that one mole of Ba(OH)2 can neutralize two moles of monoprotic acid because it supplies two moles of hydroxide. The pH calculation here is a direct extension of the same stoichiometric thinking. Once you master this style of reasoning, you will find it much easier to solve titration curves, precipitation problems, and laboratory calculations.
Authority sources for acid-base and water chemistry
For trustworthy chemistry background, you can review educational and government resources such as:
- LibreTexts Chemistry for detailed explanations of pH, pOH, and strong base calculations.
- U.S. Environmental Protection Agency for practical information on pH and water chemistry.
- U.S. Geological Survey for foundational pH concepts and water science.
- University of California, Berkeley Chemistry for university level chemistry context.
Quick answer summary
To calculate the pH of a 0.24 M Ba(OH)2 solution, first recognize that barium hydroxide is a strong base that releases two hydroxide ions per formula unit. Multiply 0.24 M by 2 to get [OH-] = 0.48 M. Then compute pOH = -log10(0.48) ≈ 0.3188. Finally, use pH = 14.00 – 0.3188 = 13.6812. The final answer is 13.68.
When to be cautious with a simple pH calculation
Although the standard classroom answer is 13.68, highly concentrated real solutions may deviate somewhat from ideal behavior. In more advanced settings, chemists use activity rather than raw concentration, because ions interact with one another in solution. Ionic strength, temperature changes, and nonideal effects can make the measured pH differ slightly from the simple textbook value. That said, none of those refinements change the proper method expected for this problem. If your assignment or exam asks for the pH of 0.24 M Ba(OH)2, the intended approach is almost always complete dissociation and the 25 C pH plus pOH equals 14 relationship.
Final takeaway
The whole problem can be reduced to one powerful idea: count the number of hydroxide ions released per formula unit. Since Ba(OH)2 releases two OH- ions, the hydroxide concentration is double the stated molarity of the base. From there, the pOH and pH follow immediately. For 0.24 M Ba(OH)2, the result is a strongly basic solution with pH approximately 13.68.