Solubility Product (Ksp) Calculator
Calculate the solubility product (Ksp) of a sparingly soluble salt from its molar solubility. Enter the solubility, its unit, and the cation and anion coefficients from the formula AaBb.
Find Ksp from Molar Solubility
Solubility Product
Where:
- \(s\) — molar solubility of the salt in pure water (mol/L)
- \(a\) — number of cations released by one formula unit
- \(b\) — number of anions released by one formula unit
- \([A]\), \([B]\) — equilibrium ion concentrations (mol/L)
For a salt \(A_aB_b\) that dissolves as \(A_aB_b(s) \rightleftharpoons aA^{}+bB\), each dissolved formula unit releases \(a\) cations and \(b\) anions, so \([A] = a\,s\) and \([B] = b\,s\). Substituting these into the equilibrium expression gives \(K_{sp}\) directly from the molar solubility.
Solubility Product – Calculation Example
Example: calcium fluoride, CaF\(_2\), has a molar solubility of \(2.1\times10^{-4}\) mol/L in pure water at 25 \(^\circ\)C. Each formula unit releases one Ca\(^{2+}\) and two F\(^{-}\), so \(a = 1\) and \(b = 2\).
\[ [\mathrm{Ca^{2+}}] = s = 2.1\times10^{-4}\ \mathrm{mol/L}, \quad [\mathrm{F^-}] = 2s = 4.2\times10^{-4}\ \mathrm{mol/L} \]
\[ K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{F^-}]^2 = (s)(2s)^2 = 4s^3 = 4\,(2.1\times10^{-4})^3 \]
Result: \(K_{sp} \approx 3.7\times10^{-11}\), which matches the tabulated literature value of \(3.9\times10^{-11}\) for CaF\(_2\) at 25 \(^\circ\)C.
How to Use the Calculator
- Enter the solubility of the salt in the Molar Solubility (s) field and choose its unit (mol/L, mmol/L, g/L, or mg/L) from the dropdown.
- If you selected g/L or mg/L, the Molar Mass of the Salt (M) field appears — enter the salt's molar mass in g/mol so the value can be converted to mol/L. If you're not sure, you can find it with the percent composition from chemical formula calculator.
- Set the Cation Coefficient (a) to the number of cations in one formula unit (1 for Ca\(^{2+}\) in CaF\(_2\), 2 for Ag\(^+\) in Ag\(_2\)CrO\(_4\)).
- Set the Anion Coefficient (b) to the number of anions in one formula unit (2 for F\(^{-\) in CaF\(_2\), 1 for CrO\(_4^{2-}\)).
- Read the Solubility Product, the equilibrium cation and anion concentrations, the full calculation details, and a table of common literature \(K_{sp}\) values for comparison.
What Is the Solubility Product?
The solubility product, \(K_{sp}\), is the equilibrium constant for the dissolution of a sparingly soluble ionic solid in water. When a salt such as AgCl or CaF\(_2\) is added to water, it dissolves until the solution is saturated: the solid and its dissolved ions are then in dynamic equilibrium. Unlike soluble salts, these compounds dissolve only to a very small extent, and \(K_{sp}\) quantifies exactly how small that extent is.
\(K_{sp}\) depends only on temperature (typically tabulated at 25 \(^\circ\)C) and on the identity of the salt — not on how much solid is present, as long as some undissolved solid remains. A smaller \(K_{sp}\) means a less soluble salt. For example, AgBr (\(K_{sp} = 5.4\times10^{-13}\)) is far less soluble than CaSO\(_4\) (\(K_{sp} = 2.4\times10^{-5}\)).
An important subtlety is the stoichiometry. You cannot compare \(K_{sp}\) values directly across salts with different ion ratios: for CaF\(_2\), \(K_{sp} = 4s^3\), while for AgCl, \(K_{sp} = s^2\). The relationship between \(K_{sp}\) and molar solubility always passes through the coefficients \(a\) and \(b\), which is why the calculator asks for them.
\(K_{sp}\) is also the basis for predicting whether a precipitate will form when two solutions are mixed — compare the ion product with \(K_{sp}\) using the precipitation prediction calculator. The same equilibrium governs the common-ion effect, qualitative analysis separations, and the behavior of scale-forming minerals measured by the water hardness calculator.
Assumptions, Limitations and Practical Tips
- Pure water, no other ion sources. The formula assumes the dissolving salt is the only source of both ions. If the solution already contains a common ion (e.g. F\(^{-\) from NaF), the solubility drops and the calculated \(K_{sp}\) from simple \(s\) no longer applies.
- No side reactions. Ions that hydrolyze, complex, or react with water (carbonates, sulfides, salts of weak acids or bases) have apparent solubilities that differ from the simple \(K_{sp}\) model. Mg(OH)\(_2\), for instance, is affected by the pH-dependent OH\(^{-\) concentration.
- Concentrations approximate activities. Strictly, \(K_{sp}\) is written in terms of ion activities. For dilute saturated solutions of sparingly soluble salts this approximation is excellent, which is why the calculator rejects solubilities above roughly 0.1 mol/L.
- Temperature matters. All results assume 25 \(^\circ\)C. Solubility usually increases with temperature for endothermic dissolution, so \(K_{sp}\) changes accordingly — see the Van 't Hoff equation calculator for the temperature dependence of equilibrium constants.
- Check against literature values. Comparing your result with the reference table shown in the output is a quick sanity check; small deviations reflect rounding in the solubility data and ionic-strength effects.
Common Ksp Values at 25 \(^\circ\)C
| Salt | Dissolution equilibrium | \(K_{sp}\) |
|---|---|---|
| AgCl | \(\mathrm{AgCl \rightleftharpoons Ag^+ + Cl^-}\) | \(1.8\times10^{-10}\) |
| AgBr | \(\mathrm{AgBr \rightleftharpoons Ag^+ + Br^-}\) | \(5.4\times10^{-13}\) |
| Ag\(_2\)CrO\(_4\) | \(\mathrm{Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-}}\) | \(1.1\times10^{-12}\) |
| BaSO\(_4\) | \(\mathrm{BaSO_4 \rightleftharpoons Ba^{2+} + SO_4^{2-}}\) | \(1.1\times10^{-10}\) |
| CaCO\(_3\) (calcite) | \(\mathrm{CaCO_3 \rightleftharpoons Ca^{2+} + CO_3^{2-}}\) | \(3.3\times10^{-9}\) |
| CaF\(_2\) | \(\mathrm{CaF_2 \rightleftharpoons Ca^{2+} + 2F^-}\) | \(3.9\times10^{-11}\) |
| CaSO\(_4\) | \(\mathrm{CaSO_4 \rightleftharpoons Ca^{2+} + SO_4^{2-}}\) | \(2.4\times10^{-5}\) |
| Mg(OH)\(_2\) | \(\mathrm{Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^-}\) | \(5.6\times10^{-12}\) |
| PbI\(_2\) | \(\mathrm{PbI_2 \rightleftharpoons Pb^{2+} + 2I^-}\) | \(7.1\times10^{-9}\) |
These values are useful for verifying a calculation: entering the literature solubility of any of these salts with the correct coefficients should reproduce the tabulated \(K_{sp}\) to within rounding error.
Frequently Asked Questions
How do I calculate Ksp by hand without a calculator?
Write the balanced dissolution equation, then raise each ion concentration to the power of its coefficient. Since \([A] = a\,s\) and \([B] = b\,s\), you get \(K_{sp} = a^a b^b s^{a+b}\). For AgCl (a = b = 1): \(K_{sp} = s^2\). For PbI\(_2\) (a = 1, b = 2): \(K_{sp} = (s)(2s)^2 = 4s^3\). Plug in \(s\) in mol/L and evaluate.
What is the difference between solubility and the solubility product?
Solubility (\(s\)) is how many moles of the salt dissolve per liter of solution — it has units of mol/L (or g/L). \(K_{sp}\) is an equilibrium constant: a dimensionless number describing the saturated state of the dissociation reaction. Solubility changes when a common ion is added or the pH shifts, but \(K_{sp}\) stays constant at a given temperature — that is the key distinction.
Does Ksp have units?
Formally, no. \(K_{sp}\) is defined in terms of ion activities relative to the standard state of 1 mol/L, which makes each term dimensionless, so \(K_{sp}\) is a pure number. Older textbooks sometimes attach units like mol\(^3\)/L\(^3\) (for salts releasing three ions), but these are best avoided. Always report \(K_{sp}\) as a dimensionless value, e.g. \(3.9\times10^{-11}\) for CaF\(_2\).
How do I convert solubility in g/L to molar solubility?
Divide the mass solubility by the salt's molar mass: \(s = \frac{\text{g/L}}{M}\). For example, CaF\(_2\) has \(M = 78.07\) g/mol, so a solubility of 0.0164 g/L equals \(0.0164 / 78.07 \approx 2.1\times10^{-4}\) mol/L. If your value is in mg/L, first divide by 1000 to get g/L. The calculator handles this conversion automatically when you enter the molar mass.
What is a typical Ksp value, and what counts as 'low'?
Sparingly soluble salts typically have \(K_{sp}\) values between about \(10^{-3}\) and \(10^{-40}\). Anything below roughly \(10^{-10}\) is considered very insoluble — AgBr at \(5.4\times10^{-13}\) dissolves only micrograms per liter. But never compare \(K_{sp}\) values across salts with different stoichiometries: Ag\(_2\)CrO\(_4\) (\(K_{sp} = 1.1\times10^{-12}\)) is actually more soluble in mol/L than AgCl (\(K_{sp} = 1.8\times10^{-10}\)) because of its ion ratio.
Why is my calculated Ksp different from the literature value?
Small mismatches usually come from rounding in the solubility data, experimental ionic-strength effects, or tabulated values measured at slightly different conditions. If your result is off by a factor near your coefficients (e.g. 4×, 8×, 27×), check the stoichiometry — forgetting that \([B] = b\,s\) must be raised to the power \(b\) is the most common error. Large discrepancies may mean hydrolysis or complexation invalidates the simple model.
Can I use this calculator to find molar solubility from Ksp instead?
Not directly — this tool works in one direction (solubility → \(K_{sp}\)). For the reverse, rearrange \(K_{sp} = a^a b^b s^{a+b}\) to get \(s = \sqrt[a+b]{K_{sp} / (a^a b^b)}\). For example, AgCl with \(K_{sp} = 1.8\times10^{-10}\) gives \(s = \sqrt{1.8\times10^{-10}} \approx 1.3\times10^{-5}\) mol/L. You can verify your answer by entering it back into this calculator.