Calculate the pH of 0.0000000001 Ca(OH)2
Use this interactive calculator to find the pH of an extremely dilute calcium hydroxide solution and understand why water autoionization matters at very low concentrations.
pH Calculator Inputs
Enter the molar concentration of Ca(OH)2 before dissociation.
For very dilute bases such as 1.0 × 10-10 M Ca(OH)2, the exact method is the correct approach.
Calculated Results
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How to calculate the pH of 0.0000000001 Ca(OH)2
If you need to calculate the pH of 0.0000000001 Ca(OH)2, you are working with an extremely dilute solution of calcium hydroxide. Written in scientific notation, this concentration is 1.0 × 10-10 M Ca(OH)2. At first glance, the problem seems simple because calcium hydroxide is a strong base. A student might immediately say that each formula unit of Ca(OH)2 produces two hydroxide ions, so the hydroxide concentration should be 2.0 × 10-10 M, and then pOH = -log(2.0 × 10-10) = 9.699, giving pH = 4.301. But that answer is chemically impossible for a base. The reason is that the simple shortcut breaks down when the dissolved base is much more dilute than the hydroxide already associated with pure water equilibrium.
Pure water at 25 °C is not chemically empty. It self-ionizes slightly according to the equilibrium H2O ⇌ H+ + OH–. In neutral pure water, both [H+] and [OH–] are approximately 1.0 × 10-7 M, and the ion product is Kw = 1.0 × 10-14. Because 1.0 × 10-7 M is much larger than 2.0 × 10-10 M, the solution’s chemistry is dominated by water’s own ionization, not solely by the added base. That is why the correct pH comes out only slightly above 7, not strongly acidic and not strongly basic.
Step 1: Write the dissociation of calcium hydroxide
Calcium hydroxide is treated as a strong base in dilute aqueous solution:
Ca(OH)2 → Ca2+ + 2OH–
If the formal concentration of Ca(OH)2 is C = 1.0 × 10-10 M, then the hydroxide contributed by the solute is:
2C = 2.0 × 10-10 M
Step 2: Recognize that the usual shortcut fails
The usual strong-base shortcut assumes that the hydroxide generated by the base is so large that water’s contribution can be ignored. That is valid for many introductory problems, such as 0.010 M NaOH or 0.0010 M Ba(OH)2. It is not valid here. Since pure water already has about 1.0 × 10-7 M OH–, adding only 2.0 × 10-10 M from Ca(OH)2 perturbs neutrality only slightly.
Step 3: Use charge balance plus Kw
Let b represent the hydroxide concentration added directly by the dissolved calcium hydroxide:
b = 2C = 2.0 × 10-10 M
Because water still autoionizes, the total hydroxide concentration is not just b. Instead, we use:
- [OH–] = [H+] + b
- [H+][OH–] = Kw = 1.0 × 10-14
Substitute [OH–] = [H+] + b into the Kw expression:
[H+]([H+] + b) = 1.0 × 10-14
Let h = [H+]. Then:
h2 + bh – 1.0 × 10-14 = 0
Insert b = 2.0 × 10-10:
h2 + (2.0 × 10-10)h – 1.0 × 10-14 = 0
Now solve the quadratic:
h = [-b + √(b2 + 4Kw)] / 2
h = [-(2.0 × 10-10) + √((2.0 × 10-10)2 + 4.0 × 10-14)] / 2
This gives:
[H+] ≈ 9.99 × 10-8 M
Step 4: Compute pH
Now calculate pH from the hydrogen ion concentration:
pH = -log[H+]
pH = -log(9.99 × 10-8) ≈ 7.0004
That is the correct answer at 25 °C for a 1.0 × 10-10 M Ca(OH)2 solution under the standard idealized assumptions used in general chemistry.
Final answer
The pH of 0.0000000001 M Ca(OH)2 is approximately 7.0004. It is only slightly basic because the base is so dilute that the autoionization of water cannot be ignored.
Why the exact answer is only slightly above 7
This result surprises many learners because they expect any strong base to produce a pH much greater than 7. The missing insight is scale. A concentration of 1.0 × 10-10 M is one thousand times smaller than 1.0 × 10-7 M. Since neutral water already contains around 1.0 × 10-7 M H+ and OH– at 25 °C, adding only 2.0 × 10-10 M hydroxide causes a very small shift in the balance. The solution becomes basic, but only very slightly basic.
Comparison between the wrong shortcut and the correct method
| Method | Assumption | Calculated value | Why it succeeds or fails |
|---|---|---|---|
| Simple strong-base shortcut | [OH–] = 2C = 2.0 × 10-10 M | pOH = 9.699, pH = 4.301 | Fails because it ignores water autoionization and predicts an acidic pH for a basic solute. |
| Exact method with Kw | [OH–] = [H+] + 2C and [H+][OH–] = 1.0 × 10-14 | pH ≈ 7.0004 | Correct for extremely dilute strong bases because it includes water equilibrium. |
Useful concentration benchmarks
It helps to compare the added hydroxide from Ca(OH)2 with the natural ion concentrations in water. The table below shows why 1.0 × 10-10 M is in the “water-dominated” regime.
| Quantity | Value at 25 °C | Interpretation |
|---|---|---|
| Kw | 1.0 × 10-14 | The ion-product constant for water. |
| [H+] in pure water | 1.0 × 10-7 M | Defines pH 7.00 in ideal pure water at 25 °C. |
| [OH–] in pure water | 1.0 × 10-7 M | Equal to [H+] in neutral water. |
| Added OH– from 1.0 × 10-10 M Ca(OH)2 | 2.0 × 10-10 M | About 500 times smaller than 1.0 × 10-7 M. |
| Correct pH of 1.0 × 10-10 M Ca(OH)2 | 7.0004 | Only slightly basic because the perturbation is tiny. |
Step-by-step shortcut you can remember
- Convert the stated concentration into scientific notation if needed. Here, 0.0000000001 M = 1.0 × 10-10 M.
- Multiply by 2 because Ca(OH)2 releases two hydroxide ions per formula unit.
- Check whether 2C is much larger than 1.0 × 10-7 M. If not, water autoionization matters.
- Use the exact equation h(h + 2C) = Kw.
- Solve for h = [H+].
- Take pH = -log h.
Common mistakes students make
- Ignoring water autoionization: This is the biggest error in ultra-dilute acid-base problems.
- Forgetting the factor of 2: Ca(OH)2 produces two hydroxide ions, not one.
- Assuming every strong base creates a large pH: Strength and concentration are different ideas. A strong base can still be very dilute.
- Using pH + pOH = 14 without checking conditions: This relation is tied to Kw at 25 °C and idealized conditions.
- Reporting too many digits: For most educational settings, pH ≈ 7.0004 is appropriate.
When can you ignore water autoionization?
A good rule of thumb is that if the acid or base concentration is much larger than 1.0 × 10-6 M, the simple approximation is often acceptable in introductory chemistry. Once you move into the 10-7 M to 10-9 M range, water’s own equilibrium starts to matter significantly. For 1.0 × 10-10 M Ca(OH)2, it matters decisively.
Practical interpretation
In real laboratory systems, measured pH can be influenced by dissolved carbon dioxide from the air, activity effects, impurities, and instrumental limitations. A theoretical value such as 7.0004 assumes an idealized aqueous solution at 25 °C with no other acid-base species present. In practice, very dilute basic solutions can be difficult to prepare and maintain because even small contamination can shift the measured pH more than the calculated change from the solute itself.
Authoritative chemistry references
For readers who want to verify the underlying equilibrium concepts, these references are useful and authoritative:
- National Institute of Standards and Technology (NIST)
- University-level explanation of water autoionization from LibreTexts
- U.S. Geological Survey explanation of pH and water chemistry
Bottom line
To calculate the pH of 0.0000000001 Ca(OH)2, do not use the ordinary strong-base shortcut alone. Because the solution is extraordinarily dilute, you must account for the autoionization of water. Calcium hydroxide contributes 2.0 × 10-10 M hydroxide, but neutral water already contains about 1.0 × 10-7 M hydroxide at 25 °C. Solving the exact equilibrium relation gives a hydrogen ion concentration of about 9.99 × 10-8 M and a final pH of about 7.0004. That makes the solution slightly basic, which is exactly what chemistry predicts for a very weak perturbation of pure water by a strong but ultradilute base.