Calculate Molar Solubility of Ca(OH)2 at pH 4
Use this premium calculator to estimate the molar solubility of calcium hydroxide in a fixed-pH environment, compare it with pure-water solubility, and visualize how pH changes the equilibrium dramatically.
Calcium Hydroxide Solubility Calculator
Solubility vs pH Chart
Results
Core Equilibrium
Dissolution: Ca(OH)2(s) ⇌ Ca2+ + 2OH–
Ksp expression: Ksp = [Ca2+][OH–]2
Buffered pH model: [OH–] = 10-(14 – pH), so s = Ksp / [OH-]^2
Pure water model: Ksp = s(2s)2 = 4s3, so s = (Ksp / 4)1/3
How to Calculate the Molar Solubility of Ca(OH)2 at pH 4
Calculating the molar solubility of calcium hydroxide, Ca(OH)2, at pH 4 is a classic equilibrium problem that combines solubility product chemistry with acid-base relationships. The result is striking: because calcium hydroxide produces hydroxide ions, placing it in a strongly acidic environment can increase its theoretical solubility by an enormous amount. That happens because the acid removes OH– from solution, shifting the dissolution equilibrium to the right according to Le Chatelier’s principle.
If you are trying to calculate molar solubility of ca oh 2 ph 4, the single most important question is this: is the solution buffered so the pH remains fixed at 4.00? If the answer is yes, then the calculation is straightforward. If the answer is no, then the pH will rise as Ca(OH)2 dissolves, and the final equilibrium must be treated differently. For textbook and exam-style questions, the phrase “at pH 4” usually means the pH is fixed externally.
Step 1: Write the Dissolution Reaction
Calcium hydroxide is a sparingly soluble ionic solid. Its dissolution equilibrium is:
Ca(OH)2(s) ⇌ Ca2+ + 2OH-
The solubility product expression is:
Ksp = [Ca2+][OH-]^2
At 25 C, a commonly used value is Ksp = 5.5 × 10^-6. Some books and data tables list slightly different values, but this is a standard instructional estimate.
Step 2: Convert pH 4 to Hydroxide Concentration
At 25 C, the relationship between pH and pOH is:
pH + pOH = 14
So if the pH is 4.00:
- pOH = 14.00 – 4.00 = 10.00
- [OH-] = 10^-10 M
This hydroxide concentration is extremely low, which is why the calculated solubility becomes so large under the fixed-pH assumption.
Step 3: Substitute into the Ksp Expression
If the pH is held constant by a strong acid or a buffer, then the equilibrium hydroxide concentration is fixed at 1.0 × 10-10 M. Let the molar solubility be s, so:
- [Ca2+] = s
- [OH-] = 1.0 × 10^-10 M
Now insert these into the Ksp expression:
5.5 × 10^-6 = s(1.0 × 10^-10)^2
5.5 × 10^-6 = s(1.0 × 10^-20)
s = 5.5 × 10^14 M
This is the formal textbook answer for a strictly fixed pH of 4. However, it is physically unrealistic in ordinary water because such a concentration cannot actually be achieved. Instead, it tells you that under highly acidic conditions, the dissolution is no longer limited by hydroxide accumulation. In real systems, other constraints appear first, such as finite acid supply, solution nonideality, ionic strength effects, concentration limits, and practical mass balance restrictions.
Why the Answer Becomes So Large
Calcium hydroxide releases hydroxide ions, making the solution more basic. In an acidic environment, H+ reacts with OH– to form water:
H+ + OH- → H2O
That consumption of OH– keeps the hydroxide concentration low, which keeps the ion product smaller than it would be in neutral water. As a result, more Ca(OH)2 can dissolve before the solubility product limit is reached. This is the same general reason weak bases and basic salts often become more soluble in acidic media.
Pure Water Comparison
To understand how dramatic the pH effect is, compare the pH 4 case with dissolution in pure water. In pure water, if the molar solubility is s, then:
- [Ca2+] = s
- [OH-] = 2s
So:
Ksp = s(2s)^2 = 4s^3
s = (Ksp / 4)^(1/3)
Using Ksp = 5.5 × 10^-6:
s ≈ 0.0111 M
That means the fixed-pH 4 estimate is many orders of magnitude larger than the pure-water solubility. This is exactly what the calculator and chart above are designed to show.
| Condition | Key Assumption | [OH–] Used | Calculated Molar Solubility of Ca(OH)2 |
|---|---|---|---|
| Pure water, 25 C | [OH–] comes only from dissolved Ca(OH)2 | 2s | 0.0111 M |
| Buffered solution at pH 4.00 | pH is held constant externally | 1.0 × 10-10 M | 5.5 × 1014 M |
| Buffered solution at pH 7.00 | Neutral pH maintained by external control | 1.0 × 10-7 M | 5.5 × 108 M |
| Buffered solution at pH 10.00 | Mildly basic fixed pH | 1.0 × 10-4 M | 550 M |
Interpreting the Result Correctly
Students sometimes see the huge number at pH 4 and think they made an algebra mistake. Usually they did not. The giant value is the mathematical consequence of forcing [OH–] to remain at 10-10 M while also using the Ksp expression. In real laboratory chemistry, you would never obtain 5.5 × 1014 moles of dissolved Ca(OH)2 per liter. The number exceeds any realistic concentration limit and therefore serves as a signal that the fixed-pH assumption dominates the equilibrium model.
In other words, the calculation is still chemically meaningful. It shows that under sufficiently acidic and externally controlled conditions, Ksp is not the practical limiting factor. The true limit would instead come from the amount of available acid, the volume of solution, precipitation of other phases, ionic activity corrections, and nonideal behavior at high concentration.
When You Should Use This Formula
- Use the buffered pH model when a problem explicitly states the pH is fixed or the solution is buffered.
- Use the pure-water model when Ca(OH)2 dissolves in plain water with no other significant acid-base source.
- Use a full equilibrium and mass-balance approach when the problem gives a limited amount of acid rather than a fixed pH.
- Consider activity coefficients and ionic strength corrections for advanced analytical or geochemical work.
Real Data That Helps Explain the Trend
The pH scale is logarithmic, so every 1-unit drop in pH means a tenfold increase in hydrogen ion concentration and, at 25 C, a tenfold decrease in hydroxide ion concentration. Because the Ksp expression contains [OH–]2, each 1-unit decrease in pH increases the calculated molar solubility by a factor of 100 under the fixed-pH model. This is why the curve is so steep.
| pH | pOH | [OH–] (M) | Predicted Solubility s = Ksp / [OH–]2 | Relative Increase vs pH 12 |
|---|---|---|---|---|
| 12 | 2 | 1.0 × 10-2 | 0.055 M | 1× |
| 10 | 4 | 1.0 × 10-4 | 550 M | 10,000× |
| 8 | 6 | 1.0 × 10-6 | 5.5 × 106 M | 100,000,000× |
| 6 | 8 | 1.0 × 10-8 | 5.5 × 1010 M | 1.0 × 1012× |
| 4 | 10 | 1.0 × 10-10 | 5.5 × 1014 M | 1.0 × 1016× |
Common Mistakes in Ca(OH)2 Solubility Problems
- Forgetting the coefficient 2 on OH–. Calcium hydroxide produces two hydroxide ions for every formula unit that dissolves.
- Mixing up pH and pOH. At 25 C, pOH = 14 – pH.
- Using [OH–] = 2s when pH is fixed. In a buffered solution, [OH–] is determined by pH, not by 2s.
- Ignoring the physical meaning of the answer. A very large computed value can still be the correct equilibrium result under the assumptions given.
- Confusing Ksp with actual concentration limits. Ksp predicts equilibrium ion products, not whether a concentration is practical in real laboratory conditions.
Why This Matters in Chemistry and Environmental Science
Calcium hydroxide appears in water treatment, cement chemistry, environmental remediation, and analytical chemistry. Understanding how pH changes solubility helps explain why alkaline solids dissolve differently in acidic groundwater, why lime reacts effectively in acidic waste streams, and why buffering conditions matter when predicting precipitation or dissolution. Reliable pH interpretation is also essential in water analysis and geochemical modeling.
For deeper reading on pH fundamentals and water chemistry, see the USGS page on pH and water. For thermochemical and substance reference data related to calcium hydroxide, the NIST Chemistry WebBook entry for calcium hydroxide is useful. For an academic overview of water quality chemistry and acid-base behavior, the U.S. EPA water quality criteria resources provide additional context.
Final Answer for the Standard Textbook Setup
If the question is simply, “calculate the molar solubility of Ca(OH)2 at pH 4,” and you are expected to assume a fixed pH with Ksp = 5.5 × 10^-6, then the answer is:
Molar solubility = 5.5 × 10^14 M
That is the equilibrium answer under the fixed-pH assumption. If you are working on a real system rather than a textbook idealization, mention that the value is theoretical and not physically attainable in ordinary aqueous solution because other constraints would dominate first.
Quick Summary
- Write the dissolution equilibrium for Ca(OH)2.
- Convert pH 4 to pOH 10, then find [OH–] = 10-10 M.
- Use Ksp = [Ca2+][OH-]^2.
- Set [Ca2+] = s and solve s = Ksp / [OH-]^2.
- Using 5.5 × 10-6 gives s = 5.5 × 10^14 M.