Solubility Product (Ksp) Calculator
Inputs
| Solve for | Find molar solubility |
|---|---|
| Salt type | AB type (e.g. AgCl) |
| Ksp | 1.8e-10 |
| Molar solubility | 1e-5 M |
Solubility Product (Ksp) Calculator
Convert between the solubility product Ksp and molar solubility s for a sparingly soluble salt. Pick the dissolution stoichiometry (AB, AB₂, AB₃, or A₂B₃) and solve in either direction.
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Enter a value to see results.
Understanding the solubility product
When a sparingly soluble salt sits in water, a little of it dissolves into ions until the solution is saturated. At that point the rate of dissolving equals the rate of crystallising, and the system is at equilibrium. The solubility product Ksp is the equilibrium constant for that process. For a salt dissolving as
AxBy⇌xAy++yBx−the solubility product is the product of the ion concentrations, each raised to its stoichiometric coefficient:
Ksp=[Ay+]x[Bx−]yKsp is written as a dimensionless number here because it is built from ion activities — ratios to a standard 1 mol/L state — rather than raw concentrations.
The molar solubility s is the number of moles of the salt that dissolve per litre of saturated solution. Because the ions come from a single source, every ion concentration can be written in terms of s, which collapses Ksp into one tidy expression per salt type.
| Salt type | Dissolution | Ksp in terms of s | Solubility s |
|---|---|---|---|
| AB | A⁺ + B⁻ | ||
| AB₂ / A₂B | A²⁺ + 2 B⁻ | ||
| AB₃ / A₃B | A³⁺ + 3 B⁻ | ||
| A₂B₃ / A₃B₂ | 2 A³⁺ + 3 B²⁻ |
Worked example
Silver chloride is a classic 1:1 (AB) salt:
AgCl⇌Ag++Cl−At 25 °C its solubility product is . Because one Ag⁺ and one Cl⁻ are released per formula unit, both ion concentrations equal the molar solubility s, so . Solving for s:
s=Ksp=1.8×10−10≈1.34×10−5 mol/LSo a saturated solution of AgCl holds only about 13 micromoles of dissolved salt per litre — which is why silver chloride looks insoluble.
Why the coefficient grows with stoichiometry
Salts that release more ions get larger coefficients. Calcium fluoride dissolves as CaF₂ → Ca²⁺ + 2 F⁻, so the fluoride concentration is twice the calcium concentration: and . Substituting gives
Ksp=[Ca2+][F−]2=s(2s)2=4s3The same logic produces for an AB₃ salt and for an A₂B₃ salt. Because the exponent on s changes with the salt type, two salts with identical Ksp values can have very different molar solubilities — you cannot compare solubilities directly from Ksp unless the stoichiometries match.
The common-ion effect
The conversions above assume the salt dissolves in pure water. If the water already contains one of the ions — say chloride from dissolved NaCl — the dissolution equilibrium is pushed back toward the solid and less salt dissolves. This is the common-ion effect. Ksp itself is unchanged, but the molar solubility drops, sometimes by orders of magnitude. To handle it you keep the Ksp expression but substitute the elevated background concentration of the shared ion instead of assuming both ions come only from the dissolving salt.
Frequently Asked Questions (FAQ)
How is Ksp related to molar solubility?
For a salt that dissolves as AₓBᵧ, the solubility product equals the product of the ion concentrations, each raised to its stoichiometric coefficient. Writing every ion concentration in terms of the molar solubility s gives a single relationship that depends only on the stoichiometry: Ksp = s² for an AB salt, Ksp = 4s³ for AB₂ or A₂B, Ksp = 27s⁴ for AB₃ or A₃B, and Ksp = 108s⁵ for A₂B₃. Rearranging each one solves for s — for example s = √Ksp for an AB salt.
What is molar solubility?
Molar solubility is the number of moles of a solid that dissolve in one litre of solution to reach a saturated equilibrium, expressed in mol/L. It is different from the bare Ksp value: two salts can have similar Ksp values but very different molar solubilities if their stoichiometries differ, because the ions are released in different ratios. Molar solubility is the quantity you would actually measure by saturating water with the salt and analysing the dissolved ions.
Why does the formula change with the salt type?
The exponents and coefficients come directly from the dissolution equation. AgCl → Ag⁺ + Cl⁻ gives Ksp = [Ag⁺][Cl⁻] = s·s = s². CaF₂ → Ca²⁺ + 2F⁻ gives Ksp = [Ca²⁺][F⁻]² = s·(2s)² = 4s³. Fe(OH)₃ → Fe³⁺ + 3OH⁻ gives Ksp = s·(3s)³ = 27s⁴, and a 2:3 salt such as A₂B₃ gives Ksp = (2s)²·(3s)³ = 108s⁵. Because the relationship is non-linear, you must match the salt type to its actual dissolution before converting.
What is the common-ion effect?
Adding a soluble salt that shares an ion with a sparingly soluble salt lowers that salt’s solubility. For example, AgCl is far less soluble in a solution that already contains chloride ions (from NaCl) than in pure water. The dissolved chloride pushes the dissolution equilibrium back toward the solid, so less AgCl dissolves — even though Ksp itself does not change with concentration. The simple Ksp = s² conversion here assumes pure water with no common ions present; in their presence you would solve the equilibrium with the extra ion included.
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