Calculate The Ph When 100Ml Of 0.1M Ca Oh 2

Use this premium calculator to determine the pH, pOH, hydroxide concentration, and hydroxide moles for calcium hydroxide solutions. The default example is the classic chemistry question: calculate the pH when 100 mL of 0.1 M Ca(OH)2 is given.

How to calculate the pH when 100 mL of 0.1 M Ca(OH)2 is given

If you need to calculate the pH when 100 mL of 0.1 M Ca(OH)2 is placed in solution, the chemistry is straightforward once you remember one essential idea: calcium hydroxide is a strong base that releases two hydroxide ions for every formula unit that dissociates. That one stoichiometric fact changes the answer significantly compared with a base like sodium hydroxide, which contributes only one hydroxide ion per molecule.

The problem usually appears in general chemistry, analytical chemistry, and titration practice because it checks whether you understand the relationship between molarity, volume, moles, ion dissociation, pOH, and pH. The default values in the calculator above are already set to the classic example, but it is helpful to understand the logic behind the answer so you can solve similar problems confidently by hand.

Final answer for the standard problem

For 100 mL of 0.1 M Ca(OH)2, the hydroxide concentration is 0.200 M OH-, the pOH = 0.699, and the pH = 13.301 at 25°C.

Step by step solution

1. Convert volume to liters.
   100 mL = 0.100 L
2. Calculate moles of Ca(OH)2.
   Moles = molarity × volume = 0.1 mol/L × 0.100 L = 0.0100 mol Ca(OH)2
3. Use dissociation stoichiometry.
   Ca(OH)2 → Ca2+ + 2OH-
   So each mole of Ca(OH)2 gives 2 moles of OH-.
   Moles of OH- = 2 × 0.0100 = 0.0200 mol OH-
4. Find hydroxide concentration.
   [OH-] = moles / volume = 0.0200 / 0.100 = 0.200 M
5. Calculate pOH.
   pOH = -log(0.200) = 0.699
6. Calculate pH.
   At 25°C, pH + pOH = 14.00
   pH = 14.00 – 0.699 = 13.301

Why calcium hydroxide gives two hydroxide ions

Many students miss the factor of 2. Calcium hydroxide is written as Ca(OH)2 because one calcium ion is paired with two hydroxide ions. When it dissociates completely in the standard strong-base approximation, each dissolved unit contributes one Ca2+ ion and two OH- ions. That means the hydroxide ion concentration is twice the molarity of the dissolved Ca(OH)2, assuming complete dissociation and no additional dilution or side reactions.

In this example, the Ca(OH)2 molarity is 0.1 M, so the hydroxide ion concentration becomes 0.2 M. That is the concentration you must use in the logarithm when calculating pOH. If you mistakenly use 0.1 M instead of 0.2 M, you would get pOH = 1 and pH = 13, which is close but not correct.

Quick formula you can reuse

For a calcium hydroxide solution at 25°C under the strong-base assumption:

• [OH-] = 2 × M(Ca(OH)2)
• pOH = -log[OH-]
• pH = 14 – pOH

Notice something useful: if you are already given the molarity of the final Ca(OH)2 solution, you often do not even need the volume to determine pH. Volume becomes crucial when the question asks for moles, when dilution occurs, or when a reaction with another acid or base is involved. In the exact problem here, the 100 mL volume helps you verify the mole calculation, but the pH can also be inferred directly from the 0.1 M concentration.

Worked interpretation of the result

A pH of 13.301 means the solution is strongly basic. On the conventional pH scale, values above 7 are basic, and values near 13 indicate a high concentration of hydroxide ions. This is exactly what we expect from a strong metal hydroxide. Calcium hydroxide, also called slaked lime in many practical settings, is used in water treatment, construction chemistry, food processing under controlled conditions, and environmental applications where alkalinity is beneficial.

The computed pH is based on the usual classroom assumptions: complete dissociation, ideal behavior, and a 25°C environment where pH + pOH = 14. In real laboratory work, concentrated ionic solutions can show small deviations from ideality because activity effects matter, but for general chemistry calculations, the textbook answer remains 13.301.

Comparison table: strong bases and hydroxide yield

Base	Formula	OH- ions released per formula unit	If base concentration = 0.10 M, ideal [OH-]	pOH at 25°C	pH at 25°C
Sodium hydroxide	NaOH	1	0.10 M	1.000	13.000
Potassium hydroxide	KOH	1	0.10 M	1.000	13.000
Calcium hydroxide	Ca(OH)2	2	0.20 M	0.699	13.301
Barium hydroxide	Ba(OH)2	2	0.20 M	0.699	13.301

Common mistakes when solving this question

• Forgetting to convert mL to L. You must use liters in the molarity formula unless the units are handled explicitly.
• Ignoring the factor of 2. Calcium hydroxide contributes two hydroxide ions, not one.
• Using pH = -log[OH-]. That equation gives pOH, not pH.
• Rounding too early. Keep extra digits through the logarithm step to avoid small answer drift.
• Confusing concentration with moles. 0.0200 mol OH- is not the same as 0.200 M OH-. One is amount, the other is amount per liter.

What if the volume changes?

If the solution concentration stays at 0.1 M and the solution is already prepared at that concentration, changing the total sample volume alone does not change the pH. For example, 50 mL of a 0.1 M Ca(OH)2 solution and 500 mL of a 0.1 M Ca(OH)2 solution have the same ideal hydroxide concentration, so they have the same pH. However, the total number of moles present changes proportionally with volume.

This distinction matters because many questions ask either for concentration-based properties like pH or for amount-based properties like moles available for neutralization. In this problem, 100 mL is enough to contain 0.0100 mol of Ca(OH)2, which corresponds to 0.0200 mol of OH-.

Data table: volume, moles, and pH for 0.1 M Ca(OH)2

Volume of 0.1 M Ca(OH)2	Moles of Ca(OH)2	Moles of OH- produced	Ideal [OH-]	pH at 25°C
50 mL	0.0050 mol	0.0100 mol	0.200 M	13.301
100 mL	0.0100 mol	0.0200 mol	0.200 M	13.301
250 mL	0.0250 mol	0.0500 mol	0.200 M	13.301
1000 mL	0.1000 mol	0.2000 mol	0.200 M	13.301

Why the pH scale matters in real applications

The pH value of a calcium hydroxide solution is not just a classroom number. It has practical meaning in environmental systems, cement and concrete chemistry, water treatment, and laboratory neutralization workflows. A pH near 13.3 indicates a highly alkaline solution that can neutralize acids effectively and alter the chemistry of dissolved metals and minerals. In engineering and environmental science, alkaline reagents are often selected based on how many hydroxide ions they can contribute per mole, which is another reason understanding Ca(OH)2 stoichiometry is useful.

For a student, the broader lesson is that pH problems are often stoichiometry problems first and logarithm problems second. Once you correctly count moles of OH-, the rest becomes systematic. This is especially true for polyhydroxide bases such as Ca(OH)2 and Ba(OH)2, where ion count directly affects the answer.

Authoritative chemistry references

If you want deeper background on acids, bases, pH, and water chemistry, these sources are excellent starting points:

• U.S. Environmental Protection Agency: What is pH?
• LibreTexts Chemistry hosted by academic institutions
• U.S. Geological Survey: pH and Water

Summary

To calculate the pH when 100 mL of 0.1 M Ca(OH)2 is given, first find the moles of calcium hydroxide, then double that amount to account for the two hydroxide ions released per formula unit. In 0.100 L of 0.1 M solution, there are 0.0100 moles of Ca(OH)2, which produce 0.0200 moles of OH-. Dividing by the solution volume gives [OH-] = 0.200 M. Then pOH = -log(0.200) = 0.699 and pH = 14.000 – 0.699 = 13.301 at 25°C.

The key takeaway is simple: calcium hydroxide is a strong base with two hydroxide ions per formula unit. If you remember that one fact, the entire calculation becomes fast, accurate, and easy to repeat for similar problems.