Calculate the pH of 1.0 × 10-12 M Ca(OH)2
This premium calculator uses the exact equilibrium approach at 25°C, including water autoionization. That matters because 1.0 × 10-12 M calcium hydroxide is so dilute that the added hydroxide is much smaller than the 1.0 × 10-7 M H+ and OH– already present in pure water.
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How to Calculate the pH of 1.0 × 10-12 M Ca(OH)2 Correctly
If you are trying to calculate the pH of 1.0 × 10-12 M calcium hydroxide, the most important idea is that this is an extremely dilute basic solution. At concentrations like this, you cannot use the shortcut that works for ordinary strong bases. The correct answer depends on the autoionization of water, and that changes the result dramatically.
Calcium hydroxide, Ca(OH)2, is a strong base in the sense that the dissolved portion dissociates essentially completely into calcium ions and hydroxide ions:
Ca(OH)2 → Ca2+ + 2OH–
Many students see the coefficient 2 in front of OH– and immediately write:
[OH–] = 2.0 × 10-12 M
That stoichiometric statement describes only the hydroxide added by the base. It does not describe the final hydroxide concentration in water. Pure water already contains H+ and OH– due to the equilibrium:
H2O ⇌ H+ + OH–
At 25°C, the ion-product constant is:
Kw = [H+][OH–] = 1.0 × 10-14}
In pure water, that means:
[H+] = [OH–] = 1.0 × 10-7 M
Notice what this means numerically: the hydroxide contributed by 1.0 × 10-12 M Ca(OH)2 is only 2.0 × 10-12 M, which is tiny compared with 1.0 × 10-7 M. Since the added base is five orders of magnitude smaller than the hydroxide associated with neutral water, ignoring water autoionization gives a misleading answer.
Bottom line: the pH is not strongly basic and it is definitely not acidic. The correct pH is just slightly above 7 because the solution contains a tiny excess of hydroxide over pure water.
Step-by-Step Derivation
1. Define the formal concentration
Let the formal concentration of calcium hydroxide be:
C = 1.0 × 10-12 M
Assuming complete dissociation of the dissolved Ca(OH)2:
- [Ca2+] = C = 1.0 × 10-12 M
- The base contributes 2C = 2.0 × 10-12 M hydroxide equivalents
2. Write the charge balance
The total positive charge must equal the total negative charge:
2[Ca2+] + [H+] = [OH–]
Since [Ca2+] = C:
[OH–] = [H+] + 2C
3. Combine with the water equilibrium
Use:
[H+][OH–] = Kw
Substitute the charge-balance expression for [OH–]:
[H+]([H+] + 2C) = Kw
That expands to a quadratic:
[H+]2 + 2C[H+] – Kw = 0
4. Insert the numbers
With C = 1.0 × 10-12 and Kw = 1.0 × 10-14:
[H+]2 + 2.0 × 10-12[H+] – 1.0 × 10-14 = 0
5. Solve for [H+]
Using the quadratic formula:
[H+] = (-2C + √((2C)2 + 4Kw)) / 2
Numerically:
[H+] = (-2.0 × 10-12 + √(4.0 × 10-24 + 4.0 × 10-14)) / 2
This gives approximately:
[H+] ≈ 9.9999 × 10-8 M
6. Convert to pH
pH = -log[H+]
pH ≈ 7.000004
So the correct answer is:
pH of 1.0 × 10-12 M Ca(OH)2 at 25°C ≈ 7.000004
Why the Shortcut Fails
The common shortcut for strong bases is:
- Find [OH–] from stoichiometry
- Compute pOH = -log[OH–]
- Use pH = 14 – pOH
If you apply that blindly here:
- [OH–] = 2.0 × 10-12 M
- pOH = 11.6990
- pH = 2.3010
That result says a strong base creates an acidic solution, which is physically unreasonable. The mistake happens because the shortcut assumes the hydroxide coming from the base dominates the hydroxide in water. That assumption is false at ultra-low concentrations.
| Method | Assumed [OH–] | Calculated pH | Interpretation |
|---|---|---|---|
| Simple stoichiometric shortcut | 2.0 × 10-12 M | 2.3010 | Incorrect because it ignores water autoionization |
| Exact equilibrium method | ≈ 1.00001 × 10-7 M | ≈ 7.000004 | Correct, slightly basic as expected |
Comparison Data Across Different Ca(OH)2 Concentrations
The reason this topic matters is that the error becomes large only when the formal base concentration approaches or drops below 10-7 M. At ordinary concentrations, the shortcut is often fine. At ultra-dilute concentrations, it becomes unreliable.
| Formal Ca(OH)2 Concentration | Added OH– from Base | Approximate pH Using Shortcut | Exact pH at 25°C | Practical Takeaway |
|---|---|---|---|---|
| 1.0 × 10-1 M | 2.0 × 10-1 M | 13.3010 | 13.3010 | Water contribution is negligible |
| 1.0 × 10-6 M | 2.0 × 10-6 M | 8.3010 | 8.3010 | Shortcut still works well |
| 1.0 × 10-12 M | 2.0 × 10-12 M | 2.3010 | 7.000004 | Must use exact equilibrium |
Real Reference Values That Help Put This Result in Context
It helps to compare the final pH to real-world benchmarks. According to the U.S. Environmental Protection Agency, the recommended secondary drinking water pH range is 6.5 to 8.5. According to the USGS, a pH of 7 is neutral at 25°C. That means a solution with pH 7.000004 is still essentially neutral for most practical purposes, even though it is mathematically slightly basic.
| Reference Condition | pH | [H+] at 25°C | [OH–] at 25°C | Context |
|---|---|---|---|---|
| Lower end of EPA secondary drinking water range | 6.5 | 3.16 × 10-7 M | 3.16 × 10-8 M | Still acceptable aesthetically in many systems |
| Neutral pure water | 7.0 | 1.00 × 10-7 M | 1.00 × 10-7 M | USGS neutral reference point |
| 1.0 × 10-12 M Ca(OH)2 | 7.000004 | ≈ 9.9999 × 10-8 M | ≈ 1.00001 × 10-7 M | Slightly basic, but nearly neutral |
| Upper end of EPA secondary drinking water range | 8.5 | 3.16 × 10-9 M | 3.16 × 10-6 M | Noticeably more basic than this dilute Ca(OH)2 solution |
Common Mistakes Students Make
Ignoring the concentration scale
Before doing any pH shortcut, compare the acid or base concentration to 1.0 × 10-7 M. If it is near that magnitude or lower, water autoionization may matter.
Using pH + pOH = 14 mechanically
The identity itself is valid at 25°C, but the concentration plugged into pOH must be the actual equilibrium hydroxide concentration, not merely the amount contributed by the solute.
Forgetting the stoichiometric factor of 2
Ca(OH)2 gives two hydroxides per formula unit. That matters in moderate or high concentrations. In this ultra-dilute case, even doubling the base concentration still leaves it far below the hydroxide level associated with neutral water.
Confusing dissolved amount with saturated solubility
This problem usually assumes a stated dissolved concentration. In a separate solubility problem, you would also have to consider dissolution equilibrium. Here, the focus is the pH resulting from a very low given concentration.
Practical Rule for Ultra-Dilute Strong Acids and Bases
A useful rule of thumb is this:
- If the formal acid or base concentration is much larger than 10-7 M, the shortcut usually works.
- If it is close to 10-7 M or smaller, use the exact equilibrium method.
- If the shortcut produces a physically absurd answer, that is a sign you ignored water autoionization.
Final Answer
For a solution that is 1.0 × 10-12 M in Ca(OH)2 at 25°C, assuming complete dissociation of the dissolved calcium hydroxide and accounting for water autoionization:
[H+] ≈ 9.9999 × 10-8 M
[OH–] ≈ 1.00001 × 10-7 M
pH ≈ 7.000004
This means the solution is slightly basic, but so close to neutral that the difference is negligible in many practical settings. The key lesson is not just the number, but the method: at very low concentrations, exact equilibrium beats the usual shortcut.