Calculate the pH of 0.010 M Ba(OH)2
Use this premium chemistry calculator to determine hydroxide concentration, pOH, and pH for aqueous barium hydroxide. The default example is 0.010 M Ba(OH)2, a classic strong-base dissociation problem.
How to calculate the pH of 0.010 M Ba(OH)2
To calculate the pH of 0.010 M Ba(OH)2, you use the fact that barium hydroxide is treated as a strong base in introductory and general chemistry. That means it dissociates essentially completely in water:
Ba(OH)2 -> Ba2+ + 2OH–
The key detail is the coefficient on hydroxide. Every one mole of dissolved barium hydroxide yields two moles of OH–. So for a 0.010 M solution:
[OH–] = 2 x 0.010 = 0.020 M
Next, calculate pOH:
pOH = -log(0.020) = 1.699
At 25 degrees C, pH + pOH = 14. Therefore:
pH = 14.000 – 1.699 = 12.301
So the answer is:
The pH of 0.010 M Ba(OH)2 is approximately 12.30.
Why Ba(OH)2 gives two hydroxide ions
Barium hydroxide contains one barium ion with a +2 charge and two hydroxide ions, each carrying a -1 charge. The formula must remain electrically neutral, so two hydroxides are required for every barium ion. In practice, that means the hydroxide concentration is double the formal molarity of dissolved Ba(OH)2, assuming complete dissociation.
This is the single most common place students make mistakes. They often plug 0.010 directly into the pOH formula and get a pOH of 2.00, then a pH of 12.00. That would be the result for a strong monohydroxide base at 0.010 M, such as NaOH, not for Ba(OH)2. Because barium hydroxide contributes twice as much OH–, its pH is slightly higher.
Step by step method
- Write the balanced dissociation equation: Ba(OH)2 -> Ba2+ + 2OH–.
- Multiply the base molarity by 2 to find hydroxide concentration.
- Use pOH = -log[OH–].
- At 25 degrees C, compute pH = 14.00 – pOH.
- Round according to the precision of the given concentration.
Worked example for 0.010 M Ba(OH)2
Let us walk through the entire calculation in a compact but rigorous way.
- Given concentration of Ba(OH)2: 0.010 M
- Stoichiometric hydroxide factor: 2
- Hydroxide concentration: 0.020 M
- pOH = -log(0.020) = 1.699
- pH = 14.000 – 1.699 = 12.301
Rounded to two decimal places, the pH is 12.30. Rounded to three decimal places, the pH is 12.301.
| Quantity | Expression | Value for 0.010 M Ba(OH)2 |
|---|---|---|
| Base concentration | CBa(OH)2 | 0.010 M |
| Hydroxide stoichiometric factor | 2 OH– per formula unit | 2 |
| Hydroxide concentration | [OH–] = 2C | 0.020 M |
| pOH | -log(0.020) | 1.699 |
| pH at 25 degrees C | 14.000 – 1.699 | 12.301 |
Comparing Ba(OH)2 with other strong bases
It helps to compare barium hydroxide with more familiar strong bases. Sodium hydroxide and potassium hydroxide each provide one hydroxide ion per formula unit. Calcium hydroxide and barium hydroxide provide two. For equal formal molarities, the dihydroxide bases generate higher hydroxide concentration and therefore slightly lower pOH and higher pH.
| Base | Formal concentration | OH– released per mole | [OH–] | pOH | pH at 25 degrees C |
|---|---|---|---|---|---|
| NaOH | 0.010 M | 1 | 0.010 M | 2.000 | 12.000 |
| KOH | 0.010 M | 1 | 0.010 M | 2.000 | 12.000 |
| Ca(OH)2 | 0.010 M | 2 | 0.020 M | 1.699 | 12.301 |
| Ba(OH)2 | 0.010 M | 2 | 0.020 M | 1.699 | 12.301 |
Chemistry behind the calculation
The pH scale is logarithmic, which means even a small numerical change in pH reflects a noticeable change in hydrogen ion concentration. In a basic solution, chemists often find pOH first because the dissolved species directly contributes hydroxide ions. Once pOH is known, pH is obtained from the relationship between hydrogen ions and hydroxide ions in water.
At 25 degrees C, the ionic product of water is Kw = 1.0 x 10-14. This leads to the familiar relation:
pH + pOH = 14.00
Because Ba(OH)2 is a strong electrolyte under standard classroom assumptions, its dissociation is not handled with an equilibrium expression in these basic examples. Instead, the hydroxide concentration comes directly from stoichiometry. More advanced contexts may consider activity effects, ionic strength, and temperature-specific values of Kw, but for standard educational problems the straightforward approach is correct.
Why the answer is not exactly 12.00
If you mistakenly assume [OH–] equals 0.010 M, then pOH = 2.00 and pH = 12.00. But that ignores the second hydroxide group. Since [OH–] is actually 0.020 M, pOH becomes 1.699, which pushes pH to 12.301. The logarithmic scale matters here. Doubling hydroxide concentration does not increase pH by a full unit, but it does increase it by about 0.301.
Common mistakes students make
- Forgetting the coefficient 2: The most frequent error is using 0.010 M as the hydroxide concentration instead of 0.020 M.
- Using pH = -log[OH–]: That expression gives pOH, not pH.
- Skipping the temperature assumption: The relation pH + pOH = 14.00 is exact only at 25 degrees C in typical coursework.
- Rounding too early: Keep more digits in intermediate calculations and round only at the end.
- Confusing molarity of the compound with molarity of ions: Ionic concentrations must reflect dissociation stoichiometry.
What the answer means in practical terms
A pH of about 12.30 indicates a strongly basic solution. Such a solution has a low hydrogen ion concentration and relatively high hydroxide concentration. In practical laboratory settings, a solution in this pH range can be corrosive to skin and eyes and should be handled with proper personal protective equipment. It is also chemically reactive toward acids and can influence precipitation reactions involving metal ions.
For example, barium hydroxide is often discussed in the context of acid-base neutralization and ionic reaction stoichiometry. Because it fully dissociates into Ba2+ and OH–, it is a useful teaching example for connecting molarity, dissociation, ion concentration, and pH.
Extended formula approach
If the concentration is represented by C and the number of hydroxide ions released per formula unit is n, then:
[OH–] = nC
pOH = -log(nC)
pH = pKw – pOH
At 25 degrees C, pKw is 14.00. For Ba(OH)2, n = 2, so:
pH = 14.00 + log(2C)
Substituting C = 0.010:
pH = 14.00 + log(0.020) with the sign handled through pOH calculation, giving 12.301
Real data points and educational context
Students often learn pH through benchmark values. Pure water at 25 degrees C is pH 7.0. A 0.010 M strong acid like HCl has pH about 2.0, while a 0.010 M strong monohydroxide base like NaOH has pH about 12.0. A 0.010 M dihydroxide base such as Ba(OH)2 goes a little higher, reaching about 12.30 because the hydroxide concentration is doubled. That 0.30 increase is a useful mental anchor for remembering that doubling concentration shifts the logarithm by log(2), approximately 0.301.
Reference values for selected hydroxide concentrations
- [OH–] = 1.0 x 10-2 M gives pOH 2.000 and pH 12.000
- [OH–] = 2.0 x 10-2 M gives pOH 1.699 and pH 12.301
- [OH–] = 1.0 x 10-1 M gives pOH 1.000 and pH 13.000
Authority sources for pH, water chemistry, and educational reference
For trustworthy background on pH, acid-base chemistry, and water equilibrium, consult authoritative educational and government resources such as:
- USGS: pH and Water
- Chemistry LibreTexts educational reference
- U.S. EPA: Water Quality Criteria and chemistry context
These sources provide broader conceptual grounding, including what pH measures, why water chemistry matters, and how acid-base values are interpreted in environmental and laboratory settings.
Final answer summary
To calculate the pH of 0.010 M Ba(OH)2, first recognize that Ba(OH)2 dissociates completely to produce two hydroxide ions per mole. That gives [OH–] = 0.020 M. Then calculate pOH = -log(0.020) = 1.699. Finally, at 25 degrees C, subtract from 14.00 to get pH = 12.301. Therefore, the pH of 0.010 M Ba(OH)2 is 12.30 to two decimal places.
If you use the calculator above, you can also test how pH changes with concentration and visualize the difference between the formal Ba(OH)2 concentration and the actual hydroxide ion concentration. That makes this example a powerful study tool for stoichiometry, logarithms, and acid-base chemistry all at once.