Molar Solubility Given pH and Ksp Calculator
Calculate the molar solubility of a metal hydroxide in a solution with known pH. This calculator uses the fixed-pH approximation for salts of the form M(OH)n, where pH controls the hydroxide concentration and Ksp controls equilibrium.
Enter the solubility product constant.
Valid range: 0 to 14 at 25°C.
The equilibrium is M(OH)n(s) ⇌ Mn+ + nOH–.
This calculator assumes pKw = 14.00 at 25°C.
Used for display only. It does not affect the math.
Results
Enter Ksp, pH, and the hydroxide stoichiometric coefficient to calculate the molar solubility.
How to Calculate Molar Solubility Given pH and Ksp
Calculating molar solubility from pH and Ksp is a classic equilibrium problem in general chemistry, analytical chemistry, environmental chemistry, and water treatment. The key idea is that solubility depends on the concentration of ions already present in solution. If pH is known, then the concentration of H+ or OH– is known, and that can shift a solubility equilibrium dramatically. For salts that contain hydroxide, pH is especially important because OH– is directly part of the dissolution expression. In a more basic solution, the common-ion effect suppresses dissolution. In a more acidic solution, H+ removes OH–, so the compound often becomes more soluble.
This page focuses on a practical and common case: a metal hydroxide with formula M(OH)n. The equilibrium is:
M(OH)n(s) ⇌ Mn+(aq) + nOH–(aq)
The corresponding solubility product expression is:
Ksp = [Mn+][OH–]n
If the pH of the solution is fixed by a buffer or otherwise specified, then [OH–] can be determined from pH. Once [OH–] is known, solving for the metal-ion concentration is straightforward. Since the molar solubility equals the concentration of dissolved formula units, the result is:
s = Ksp / [OH–]n
Step-by-Step Method
- Write the balanced dissolution equation for the solid.
- Write the Ksp expression based on the ion stoichiometry.
- Use the given pH to calculate pOH, typically with pOH = 14.00 – pH at 25°C.
- Convert pOH to hydroxide concentration using [OH–] = 10-pOH.
- Substitute [OH–] into the Ksp expression.
- Solve for the dissolved metal concentration, which equals molar solubility for M(OH)n.
Core Equations You Need
1. pH and pOH relationship
At 25°C, water has pKw = 14.00, so:
- pH + pOH = 14.00
- [H+] = 10-pH
- [OH–] = 10-pOH
2. Solubility product for hydroxides
For the generic solid M(OH)n:
- M(OH)n(s) ⇌ Mn+ + nOH–
- Ksp = [Mn+][OH–]n
- s = [Mn+] = Ksp / [OH–]n when pH fixes [OH–]
3. Why this shortcut works
In many textbook and lab problems, pH is imposed externally, usually by a buffer or a large reservoir. That means the solution’s OH– concentration is treated as already known and not significantly changed by the small amount of dissolving solid. Under that condition, the Ksp equation becomes a direct algebra problem instead of a polynomial equilibrium problem.
Worked Example
Suppose you want the molar solubility of Mg(OH)2 in a solution with pH 10.50, and the Ksp is 5.6 × 10-12.
- Write the equilibrium: Mg(OH)2(s) ⇌ Mg2+ + 2OH–
- Write the expression: Ksp = [Mg2+][OH–]2
- Find pOH: 14.00 – 10.50 = 3.50
- Find hydroxide concentration: [OH–] = 10-3.50 = 3.16 × 10-4 M
- Solve for solubility: s = 5.6 × 10-12 / (3.16 × 10-4)2
- Result: s = 5.6 × 10-5 M
This result shows why pH matters. If the solution were even more basic, [OH–] would rise and the solubility would drop further. If the solution were acidic, hydroxide would be neutralized and the solubility would rise.
Comparison Table: How pH Changes Hydroxide Concentration
| pH | pOH | [OH–] in mol/L | Change in [OH–] vs Previous pH Unit | Implication for M(OH)2 Solubility |
|---|---|---|---|---|
| 8.0 | 6.0 | 1.0 × 10-6 | Base reference point | Relatively higher than in strongly basic media |
| 9.0 | 5.0 | 1.0 × 10-5 | 10 times larger | For n = 2, solubility becomes 100 times smaller |
| 10.0 | 4.0 | 1.0 × 10-4 | 10 times larger | For n = 2, solubility becomes another 100 times smaller |
| 11.0 | 3.0 | 1.0 × 10-3 | 10 times larger | For n = 2, solubility is 10,000 times lower than at pH 9 |
| 12.0 | 2.0 | 1.0 × 10-2 | 10 times larger | Very strongly suppressed by the common-ion effect |
The statistical pattern in the table is important: every 1-unit increase in pH changes [OH–] by a factor of 10. For compounds where n = 2, this means solubility changes by a factor of 102 = 100 for each pH unit. For n = 3, the effect is even stronger, with a factor of 1000 per pH unit. That is why metal hydroxides can precipitate sharply over narrow pH ranges in environmental and industrial systems.
Comparison Table: Stoichiometry and pH Sensitivity
| Hydroxide Type | General Ksp Form | Solubility Formula at Fixed pH | Effect of 1 pH Unit Increase | Typical Interpretation |
|---|---|---|---|---|
| M(OH) | Ksp = [M+][OH–] | s = Ksp / [OH–] | Solubility drops 10 times | Moderate pH sensitivity |
| M(OH)2 | Ksp = [M2+][OH–]2 | s = Ksp / [OH–]2 | Solubility drops 100 times | Strong pH sensitivity |
| M(OH)3 | Ksp = [M3+][OH–]3 | s = Ksp / [OH–]3 | Solubility drops 1000 times | Extremely strong pH dependence |
| M(OH)4 | Ksp = [M4+][OH–]4 | s = Ksp / [OH–]4 | Solubility drops 10,000 times | Very steep precipitation behavior |
Common Mistakes Students Make
- Using pH directly as [OH–]. pH is logarithmic, not a concentration. You must convert properly.
- Forgetting to calculate pOH. At 25°C, pOH = 14 – pH.
- Ignoring stoichiometry. If the compound is M(OH)3, the hydroxide term is cubed.
- Using Ksp as if it were solubility. Ksp is an equilibrium constant, not the molar solubility itself.
- Confusing fixed-pH and pure-water cases. In pure water, the dissolving salt itself contributes OH–. In fixed-pH problems, [OH–] is usually imposed externally.
When This Calculator Is Most Accurate
This tool is designed for conditions where the pH of the solution is already established and remains effectively constant during dissolution. Typical examples include buffered lab systems, environmental waters where pH is measured independently, and instructional chemistry problems that explicitly state the pH. The model is especially useful when:
- The compound is a metal hydroxide or another equilibrium directly controlled by OH–.
- The solution has enough buffering capacity that dissolution does not significantly alter pH.
- The ionic strength is low enough that activities can be approximated by concentrations.
- The temperature is near 25°C, so pKw ≈ 14.00 is appropriate.
Advanced Perspective: Why pH Controls Solubility
From Le Châtelier’s principle, adding a product ion shifts equilibrium toward the solid. In hydroxide systems, OH– is a product of dissolution. When pH rises, the solution already contains more OH–. That suppresses additional dissolution. In acidic systems, H+ reacts with OH– to form water, effectively removing a product and pulling the equilibrium toward more dissolved ions. This is why many metal hydroxides are more soluble in acidic water and far less soluble in alkaline water.
The effect is not linear. Because pH uses a base-10 logarithmic scale, small numerical changes in pH create large concentration changes. Then Ksp amplifies those changes again through the stoichiometric exponent n. A trivalent hydroxide, for example, can show thousand-fold solubility shifts for each pH unit. That kind of sensitivity matters in wastewater precipitation, corrosion control, geochemical transport, and pharmaceutical formulation.
Practical Applications
Environmental Chemistry
In natural waters and treatment systems, metal mobility depends heavily on pH. Hydroxide precipitation is routinely used to remove dissolved metals from wastewater. Knowing Ksp and pH helps predict whether a metal remains dissolved or precipitates as a hydroxide solid.
Analytical Chemistry
Selective precipitation relies on controlling pH so that one ion precipitates while another stays in solution. Since each metal hydroxide has a different Ksp, pH can become a separation tool.
Materials and Process Chemistry
Chemists use pH-sensitive precipitation to prepare solids with specific particle sizes and purities. Solubility calculations are central when designing wash steps, filtration conditions, and crystal growth procedures.
Authoritative References
- U.S. Geological Survey (USGS): pH and Water
- U.S. Environmental Protection Agency (EPA): pH Overview
- University of Washington Chemistry Department
Final Takeaway
To calculate molar solubility given pH and Ksp, the main job is to convert pH into the relevant ion concentration and then substitute that value into the equilibrium expression. For metal hydroxides, the process is especially direct: determine [OH–] from pH, apply the Ksp expression, and solve for the dissolved metal concentration. Once you understand that pH changes ion concentrations logarithmically and that stoichiometric exponents magnify the effect, these problems become much easier to solve consistently and correctly.
If you are working with a hydroxide such as M(OH)2 or M(OH)3, this calculator gives you a fast, visual way to see how solubility responds across the pH scale. It is a practical bridge between equilibrium theory and real chemistry decisions.