Calculate the pH of a 0.05 M Calcium Hydroxide Solution
Use this interactive calculator to find hydroxide concentration, pOH, and pH for aqueous calcium hydroxide, then explore the chemistry behind the answer in the expert guide below.
How to calculate the pH of a 0.05 M calcium hydroxide solution
Calcium hydroxide, written as Ca(OH)2, is a strong base that dissociates in water to produce calcium ions and hydroxide ions. If you are asked to calculate the pH of a 0.05 M calcium hydroxide solution, the key idea is that each formula unit of Ca(OH)2 releases two hydroxide ions. That stoichiometric relationship is what makes this question different from the pH of a 0.05 M solution of sodium hydroxide, which releases only one hydroxide ion per formula unit.
The balanced dissociation equation is:
Ca(OH)2 → Ca2+ + 2OH-
Because the hydroxide coefficient is 2, a 0.05 M calcium hydroxide solution produces:
[OH-] = 2 × 0.05 = 0.10 M
Once you know the hydroxide concentration, calculating pOH is straightforward:
pOH = -log(0.10) = 1.00
At 25 C, water obeys the familiar relationship:
pH + pOH = 14.00
So the final answer is:
pH = 14.00 – 1.00 = 13.00
Step by step solution with chemical reasoning
1. Identify whether the compound is an acid or a base
Calcium hydroxide is a base because it contains hydroxide ions. In aqueous solution it contributes OH–, which lowers hydrogen ion concentration and raises pH. It is commonly called slaked lime and is used in water treatment, construction materials, and some industrial neutralization processes.
2. Determine the dissociation stoichiometry
The formula Ca(OH)2 contains two hydroxide groups. That means one dissolved unit produces two hydroxide ions. This is the single most important step in the problem. Students often see 0.05 M and immediately calculate pOH from 0.05, which would be incorrect for calcium hydroxide. The correct hydroxide concentration is double the compound molarity.
3. Convert base concentration to hydroxide concentration
The formal concentration of the base is 0.05 M. Multiply by the number of hydroxide ions released per formula unit:
- Ca(OH)2 concentration = 0.05 mol/L
- OH– per formula unit = 2
- Hydroxide concentration = 0.10 mol/L
4. Calculate pOH
Now apply the definition of pOH:
- Take the base-10 logarithm of the hydroxide concentration.
- Change the sign to negative.
- Since log(0.10) = -1, pOH = 1.
5. Convert pOH to pH
At 25 C, pH and pOH are linked by the ionic product of water. The practical classroom relationship is pH + pOH = 14. Therefore:
- pOH = 1.00
- pH = 14.00 – 1.00 = 13.00
Why the answer is so basic
A pH of 13 indicates a strongly basic solution. This makes sense because 0.10 M hydroxide ion concentration is very high compared with neutral water, which has hydroxide concentration near 1.0 × 10-7 M at 25 C. In other words, this solution contains roughly a million times more hydroxide than neutral water. That dramatic increase drives pOH downward and pH upward.
Comparison table: calcium hydroxide vs other common bases
| Base | Formal concentration | OH- released per unit | Resulting [OH-] | pOH at 25 C | pH at 25 C |
|---|---|---|---|---|---|
| NaOH | 0.05 M | 1 | 0.05 M | 1.30 | 12.70 |
| KOH | 0.05 M | 1 | 0.05 M | 1.30 | 12.70 |
| Ca(OH)2 | 0.05 M | 2 | 0.10 M | 1.00 | 13.00 |
| Ba(OH)2 | 0.05 M | 2 | 0.10 M | 1.00 | 13.00 |
This comparison shows why stoichiometry matters. Calcium hydroxide and barium hydroxide both deliver twice as much hydroxide as sodium hydroxide or potassium hydroxide at the same formal molarity. As a result, their pH is higher.
Important caveat: solubility and ideal assumptions
In introductory chemistry, problems like this usually assume complete dissociation and treat the stated concentration as the effective dissolved concentration. That is the correct approach for most textbook and exam settings. However, in more advanced solution chemistry, calcium hydroxide raises a practical question: is a 0.05 M solution realistic under the specific conditions? Since calcium hydroxide has finite solubility in water, the actual dissolved concentration in a saturated sample may differ depending on temperature and experimental setup.
Still, if the problem explicitly states 0.05 M calcium hydroxide solution, then the intended calculation is almost always:
- Assume 0.05 mol/L is dissolved.
- Assume full dissociation into Ca2+ and 2OH–.
- Use pOH and pH definitions at 25 C.
Data table: pH values for several calcium hydroxide concentrations
| Ca(OH)2 concentration | [OH-] after dissociation | pOH | pH |
|---|---|---|---|
| 0.001 M | 0.002 M | 2.70 | 11.30 |
| 0.005 M | 0.010 M | 2.00 | 12.00 |
| 0.010 M | 0.020 M | 1.70 | 12.30 |
| 0.050 M | 0.100 M | 1.00 | 13.00 |
| 0.100 M | 0.200 M | 0.70 | 13.30 |
These values help you see the logarithmic nature of the pH scale. Increasing concentration does not raise pH in a linear way. Instead, pH shifts according to the negative logarithm of hydroxide concentration. Doubling concentration changes pOH by a smaller amount than many learners first expect.
Common mistakes when solving this problem
Forgetting the coefficient of 2
This is the most frequent error. If you use 0.05 M directly as [OH–], you get pOH = 1.30 and pH = 12.70, which would be the result for a monohydroxide base like NaOH at the same concentration, not for Ca(OH)2.
Using pH = -log[OH-]
That formula is not correct. The negative logarithm of hydroxide concentration gives pOH, not pH. You must subtract pOH from 14 at 25 C.
Ignoring temperature assumptions
In most general chemistry work, 25 C is assumed unless another temperature is given. At other temperatures, the relation pH + pOH = 14 is not exact because the ionic product of water changes slightly. For classroom calculations, 25 C remains the standard default.
Confusing M with moles
Molarity means moles per liter. The problem states concentration, not the total number of moles in a specific vessel. Unless volume is part of the question, you should work directly with molarity.
Short derivation you can memorize
- Write dissociation: Ca(OH)2 → Ca2+ + 2OH–
- Find hydroxide concentration: [OH–] = 2 × 0.05 = 0.10 M
- Find pOH: pOH = -log(0.10) = 1
- Find pH: pH = 14 – 1 = 13
Real-world context for calcium hydroxide pH
Calcium hydroxide is important in environmental engineering, wastewater treatment, agriculture, and materials science. In water treatment, basic compounds are often used to adjust alkalinity and pH. In construction, lime-based materials rely on strongly basic chemistry during curing and interaction with carbon dioxide. In laboratory work, calcium hydroxide solutions and suspensions illustrate strong-base dissociation, common-ion effects, and solubility equilibrium.
Because pH control matters in public health and environmental systems, authoritative organizations publish extensive guidance on acids, bases, water quality, and solution chemistry. If you want a deeper technical foundation, the following sources are reliable starting points:
- U.S. Environmental Protection Agency: Water Quality Criteria
- U.S. Geological Survey: pH and Water
- LibreTexts Chemistry hosted by educational institutions
Exam strategy for similar questions
If you encounter a pH calculation involving a metal hydroxide, start by checking how many hydroxide ions appear in the formula. Magnesium hydroxide, calcium hydroxide, strontium hydroxide, and barium hydroxide all have the general form M(OH)2. That means each dissolved formula unit contributes two hydroxide ions. If the formula were Al(OH)3, the stoichiometric factor would be three, though in real chemistry there may be additional complications involving solubility and amphoterism. For most first-pass calculations, count hydroxide groups first, then proceed to pOH and pH.
Final answer summary
To calculate the pH of a 0.05 M calcium hydroxide solution, you multiply the base molarity by 2 because each unit of Ca(OH)2 produces two hydroxide ions. That gives [OH–] = 0.10 M. The pOH is then 1.00, and the pH at 25 C is 13.00. This result reflects the strong basicity of the solution and the importance of dissociation stoichiometry in acid-base calculations.