Molar Solubility from Ksp and pH Calculator
Estimate molar solubility from a solubility product constant and solution pH. This premium calculator supports three practical chemistry cases: pH-independent salts, metal hydroxides in buffered media, and salts whose anions are protonated in acidic solution.
How to calculate molar solubility from Ksp and pH
Molar solubility is the number of moles of a slightly soluble compound that dissolve per liter of solution at equilibrium. In many chemistry problems, the starting point is the solubility product constant, Ksp. If the compound dissolves into ions, Ksp links the equilibrium concentrations of those ions. Once you know the dissolution stoichiometry, you can convert Ksp into a molar solubility value, commonly represented by s. When pH matters, however, the calculation changes because hydrogen ions or hydroxide ions can remove one of the dissolved ions from equilibrium. That shifts the dissolution balance and usually changes the amount of solid that can dissolve.
This is why students often see two different styles of problems. In the first, the salt does not react with acid or base, so pH has little or no effect and the classic Ksp-only setup works. In the second, one ion is tied to acid-base chemistry. Hydroxides become less soluble as pH rises because hydroxide is a common ion. Salts with basic anions, such as fluoride or acetate, often become more soluble as pH drops because the anion is protonated, reducing its free concentration in solution. The calculator above is designed to help with exactly these practical cases.
Core idea: write the dissolution equation first
Every reliable molar solubility calculation begins with the balanced dissolution equation. For a generic salt:
The solubility product expression is:
If the molar solubility is s, then:
- [Mn+] = a s
- [Xm-] = b s
Substitute into the Ksp expression:
Now solve for s. This is the pH-independent case used for salts whose ions are not strongly affected by acid-base equilibria in the problem.
When pH changes the answer
pH enters the calculation when one of the dissolved ions is either a common ion already present or a species that reacts with H+ or OH–. There are two high-value scenarios to understand.
- Metal hydroxides in basic solution: compounds like Mg(OH)2 or Fe(OH)3 become less soluble as pH increases because OH– is part of the Ksp expression.
- Salts with basic anions in acidic solution: compounds containing F–, CN–, S2-, or acetate can dissolve more as pH decreases because the anion is protonated, lowering the concentration of the free ion that appears in Ksp.
The chemistry behind both cases is Le Chatelier’s principle. If one product ion is removed from the equilibrium expression, more solid can dissolve. If one product ion is added, less solid can dissolve.
Case 1: generic pH-independent molar solubility
Suppose a salt MaXb does not participate in any significant acid-base side equilibrium under the problem conditions. Then the derivation is straightforward. Example: if a 1:1 salt MX has Ksp = 1.0 × 10-10, then:
So s = √Ksp = 1.0 × 10-5 M. For a 1:2 salt like M(OH)2 or MX2, the algebra changes because the ion concentrations are not equal. If s is the molar solubility, then [M2+] = s and [X–] = 2s. That gives:
Therefore, s = (Ksp/4)1/3. This is why stoichiometry matters so much in Ksp calculations.
Case 2: metal hydroxides and fixed pH
For hydroxides, pH is often the deciding factor. At 25 C, pH and pOH are related by:
If the surrounding solution fixes pH, then [OH–] can be treated as known. For a hydroxide Ma(OH)b:
Since [M] = a s, the molar solubility under buffered conditions is:
This shows why high pH suppresses hydroxide solubility so dramatically. Because [OH–] appears in the denominator, even a one-unit pH increase can change the answer by an order of magnitude or more, depending on stoichiometry.
| pH | pOH | [OH–] (M) | Calculated s for Mg(OH)2, Ksp = 5.61 × 10-12 (M) |
|---|---|---|---|
| 9 | 5 | 1.0 × 10-5 | 5.61 × 10-2 |
| 10 | 4 | 1.0 × 10-4 | 5.61 × 10-4 |
| 11 | 3 | 1.0 × 10-3 | 5.61 × 10-6 |
| 12 | 2 | 1.0 × 10-2 | 5.61 × 10-8 |
These values use the buffered approximation for Mg(OH)2 where Ksp = [Mg2+][OH–]2 and [OH–] is set by pH. The trend is the point: a two-unit increase in pH can lower the calculated molar solubility by roughly four orders of magnitude.
Case 3: salts with anions that are protonated by acid
Now consider a salt containing a basic anion A–. In acidic solution, some of A– is converted into HA:
If Ka is the acid dissociation constant of HA, then the fraction of total dissolved anion present as free A– is:
For a salt MaAb, the free anion concentration is approximately b α s, while the cation concentration is a s. The Ksp expression becomes:
Solving for s gives:
As pH decreases, [H+] increases, α becomes smaller, and the apparent molar solubility rises. This is the acid-enhanced solubility effect.
| pH | [H+] (M) | Free fluoride fraction α for HF, pKa = 3.17 | Calculated s for CaF2, Ksp = 3.45 × 10-11 (M) |
|---|---|---|---|
| 7 | 1.0 × 10-7 | 0.9993 | 2.05 × 10-4 |
| 4 | 1.0 × 10-4 | 0.426 | 3.61 × 10-4 |
| 3 | 1.0 × 10-3 | 0.0634 | 1.28 × 10-3 |
| 2 | 1.0 × 10-2 | 0.00672 | 5.82 × 10-3 |
The table illustrates a real and important phenomenon: the solid appears much more soluble in acid because free F– is consumed by protonation. In environmental chemistry, mineral dissolution, and analytical chemistry, this kind of pH dependence can strongly affect observed concentrations.
Step-by-step method for solving by hand
- Write the balanced dissolution equation. Identify how many ions are produced per mole of solid.
- Write the Ksp expression. Include only dissolved species, never solids.
- Decide whether pH matters. Ask whether H+ or OH– changes the concentration of a Ksp ion.
- Convert pH into [H+] or [OH–]. At 25 C, [H+] = 10-pH and [OH–] = 10-(14-pH).
- Relate ion concentrations to molar solubility. Use stoichiometric coefficients carefully.
- Solve algebraically for s. Check whether the units and magnitude are reasonable.
- Interpret the chemistry. Low pH usually increases solubility for salts with basic anions, while high pH usually decreases solubility for hydroxides.
Common mistakes to avoid
- Ignoring stoichiometry. A 1:2 salt does not dissolve with equal ion concentrations.
- Using pH directly in Ksp. Ksp contains concentrations of ions, not pH itself. You must convert pH to [H+] or [OH–].
- Forgetting common-ion effects. If OH– is already present, hydroxide solids become less soluble.
- Using the wrong acid-base constant. For a protonated anion model, use the pKa of the conjugate acid of the anion.
- Applying ideal formulas at high ionic strength. In real systems, activities can differ significantly from concentrations.
Why this matters in real chemistry
Molar solubility from Ksp and pH is not just a textbook topic. It appears in water treatment, geochemistry, drug formulation, corrosion science, and biological buffering systems. For example, metal hydroxide precipitation is used to remove contaminants from wastewater. Fluoride, phosphate, and carbonate chemistry all depend strongly on pH. In laboratory analysis, the pH chosen for a separation can determine whether a precipitate forms or dissolves. A chemist who understands Ksp alone is useful; a chemist who understands Ksp together with pH can predict what actually happens in solution.
For further reading, consult authoritative chemistry and water-quality references such as the U.S. Environmental Protection Agency water quality resources, the NIST Chemistry WebBook, and the Purdue University general chemistry review on Ksp.
Using the calculator effectively
If your compound does not participate in acid-base chemistry under the stated conditions, select the generic pH-independent model and enter Ksp plus stoichiometric coefficients. If you are working with a metal hydroxide and the problem gives pH, choose the hydroxide model. The calculator will convert pH to [OH–] and estimate solubility under the fixed-pH assumption. If your anion is basic and becomes protonated in acid, choose the protonated-anion model and enter the pKa of its conjugate acid. The graph will then show how the estimated molar solubility changes across the full pH range.
Remember that all such calculations depend on the quality of the assumptions. In buffered systems, fixed-pH approximations are usually reasonable. In pure water, however, the dissolved salt itself may contribute significantly to pH, and a more exact equilibrium treatment may be needed. Even so, the methods here capture the main chemistry and are the standard approach for most instructional and many practical calculations.