Why is Ksp temperature-dependent?

Dissolution has a standard-state ΔG° that depends on T through ΔH° and ΔS°: ΔG° = ΔH° − TΔS°, and Ksp = exp(−ΔG°/RT). For endothermic dissolutions (most simple salts), Ksp grows with T; for exothermic ones (e.g. Ca(OH)₂, Li₂CO₃), Ksp shrinks. All the values in this tool are quoted at 25 °C. To predict Ksp at another temperature, use the van't Hoff relation ln(Ksp,T₂/Ksp,T₁) = −(ΔH°/R)(1/T₂ − 1/T₁), or look it up in Table 4-104 of CRC 97e which lists many salts across 0–100 °C.

When does the pure-Ksp calculation break down?

Two big cases. First, activity coefficients: Ksp is defined for activities, and this tool uses concentrations. Below ionic strength μ ≈ 0.01 mol/L the two agree to a few percent, but at μ = 0.1 mol/L, γ± drops to ~0.75 for a 1:1 salt and the concentration-Ksp underestimates true solubility by ~30 %. Second, ion pairing: polyvalent salts (Ca₃(PO₄)₂, La₂(SO₄)₃) form soluble ion pairs whose concentrations aren't in the Ksp equation, so real solubility can be 10× the naive answer. Correct with Davies-equation activity coefficients or a formal-charge speciation code.

How does the common-ion effect change s?

Adding a soluble source of one of the salt's ions shifts Ksp = [M]ᵃ[X]ᵇ so the *other* species has to drop to hold the product constant. For AgCl (Ksp = 1.77×10⁻¹⁰) in pure water, s = 1.33×10⁻⁵ M; in 0.10 M NaCl the Cl⁻ term is dominated by NaCl and s falls to Ksp/0.10 = 1.77×10⁻⁹ M — 7,500× less soluble. This is why gravimetric silver-chloride analyses wash the precipitate with dilute HNO₃, not pure water: minimising solubility losses.

Ksp vs Q — will a precipitate form?

Compute the reaction quotient Q = [M]ᵃ[X]ᵇ using the actual (not equilibrium) concentrations after mixing. If Q > Ksp the solution is supersaturated and a precipitate forms until Q falls to Ksp; if Q < Ksp everything stays dissolved; if Q = Ksp you are exactly at the saturation boundary. This is separate from the Ksp calc but uses the same product expression.

What is the difference between Ksp and solubility (g/L)?

Molar solubility s is mol of formula unit dissolved per litre. Solubility in g/L is s multiplied by the salt's molar mass. Ksp is neither — it's the equilibrium constant for the dissolution product, dimensionless in the modern IUPAC convention. Tables sometimes quote log Ksp or pKsp; pKsp = −log₁₀(Ksp), so pKsp(AgCl) = 9.75 corresponds to Ksp = 1.77×10⁻¹⁰.

Ksp Calculator — Molar Solubility & Common Ion

The solubility product Ksp is the equilibrium constant for the dissolution of a sparingly-soluble ionic salt. It quantifies the extent to which the solid dissolves in pure water: a smaller Ksp means less soluble. Understanding it is the foundation of gravimetric analysis, hard-water chemistry, mineral geochemistry, and the design of every precipitation-based separation you’ll meet in an analytical lab.

Ksp / Molar Solubility Calculator

Enter the stoichiometry (a for cation, b for anion) and either Ksp or the molar solubility s. Presets prefill both from CRC 97e. Optional common-ion field solves the depression numerically.

Solve for Molar solubility (s) Solubility product (Ksp)

Preset (fills Ksp, a, b)

Cation stoichiometry a

Anion stoichiometry b

Ksp

Molar solubility s (mol/L)

Common ion (optional) None Add cation Add anion

Result

Equations used

K_{s p} = a^{a} b^{b} s^{a + b}

s = (\frac{K _{s p}}{a ^{a} b ^{b}})^{1/ (a + b)}

Ksp values from CRC Handbook of Chemistry and Physics 97e (Haynes 2017), Table 4-104, at 25 °C. Values assume unit activity coefficients and no ion pairing — good to 10 % for μ < 0.01 M; polyvalent salts and concentrated solutions need activity corrections.

The equilibrium expression

For a general salt M_aX_b that dissociates as

$M_aX_b(s) \rightleftharpoons a\,M^{y+}(aq) + b\,X^{z-}(aq)$

the solubility product is

$K_{sp} = [M^{y+}]^a\,[X^{z-}]^b$

The solid is not in the expression (pure phases have unit activity). Ksp is temperature-dependent — every value here is at 25 °C — but not concentration-dependent, which is what makes it a useful equilibrium constant in the first place.

Pure water: getting s from Ksp

Let s = mol/L of the salt that dissolves. Then [M] = a·s and [X] = b·s, so

$K_{sp} = (a s)^a (b s)^b = a^a\,b^b\,s^{a+b}$

which rearranges to

$s = \left(\frac{K_{sp}}{a^a\,b^b}\right)^{1/(a+b)}$

For 1:1 salts (AgCl, BaSO₄), that simplifies to $s = \sqrt{K_{sp}}$ . For 1:2 salts (CaF₂, Mg(OH)₂), $s = (K_{sp}/4)^{1/3}$ . Get the exponent wrong on a 1:3 salt and you can be a factor of 10 off — the calculator handles the general case for any integer a, b.

Worked example — calcium fluoride

CaF₂ has Ksp = 3.45×10⁻¹¹ at 25 °C (CRC 97e).

Worked example — molar solubility of CaF₂

Salt: CaF₂ (a = 1, b = 2)
Ksp (CRC 97e): 3.45 × 10⁻¹¹

Formula
$s = (K_{s p} / (a^{a} b^{b}))^{1/ (a + b)}$
Substitute
$s = (3.45 \times 1 0^{- 11} /4)^{1/3}$
Note
That's 0.0000205 mol Ca²⁺ per litre — about 16 ppm Ca as CaF₂, which explains why fluoride precipitation is a workable water-hardness treatment. Notice a²·b² = 1·4 = 4 in the denominator; forgetting the 4 gives 2.4×10⁻⁴ M, about 20% high.
Result
$s = 2.05 \times 1 0^{- 4} mol/L$

Common-ion effect

Add a source of the cation or the anion to the solvent, and Ksp forces the other species down. For AgCl (Ksp = 1.77×10⁻¹⁰) in 0.10 M NaCl:

$K_{sp} = [Ag^+][Cl^-] \approx [Ag^+] \cdot 0.10 \implies [Ag^+] = 1.77 \times 10^{-9}\ \text{M}$

Silver solubility drops from 1.33×10⁻⁵ M (pure water) to 1.77×10⁻⁹ M — a 7,500× reduction. The approximation [Cl⁻] ≈ 0.10 works because s ≪ 0.10; when the added common ion is only ~10× the pure-water s, you need to solve $(s + C)(s) = K_{sp}$ numerically (which the calculator does automatically via bisection).

Reference values (CRC 97e, 25 °C)

Salt	Ksp	Stoichiometry	s in pure water (mol/L)
AgCl	1.77 × 10⁻¹⁰	1 : 1	1.33 × 10⁻⁵
AgBr	5.35 × 10⁻¹³	1 : 1	7.31 × 10⁻⁷
AgI	8.52 × 10⁻¹⁷	1 : 1	9.23 × 10⁻⁹
BaSO₄	1.08 × 10⁻¹⁰	1 : 1	1.04 × 10⁻⁵
CaF₂	3.45 × 10⁻¹¹	1 : 2	2.05 × 10⁻⁴
Mg(OH)₂	5.61 × 10⁻¹²	1 : 2	1.12 × 10⁻⁴
PbCl₂	1.70 × 10⁻⁵	1 : 2	1.62 × 10⁻²
Ag₂SO₄	1.20 × 10⁻⁵	2 : 1	1.44 × 10⁻²

Common pitfalls

Confusing solubility (g/L) with molar solubility (mol/L). Multiply s by the molar mass to convert; use the molar-mass calculator if in doubt.
Ignoring activity coefficients above μ ≈ 0.01 M. The calc uses concentrations; when the ionic strength is high, the “true” Ksp expression uses activities and this tool will underestimate solubility.
Applying the pure-water formula to a polyvalent salt with ion pairing. Ca₃(PO₄)₂ and rare-earth phosphates form soluble intermediate species (CaHPO₄⁺, LaPO₄⁰) whose concentrations are not in the Ksp expression. The calc’s answer is a lower bound; real solubility can be 10–100× higher.
Wrong stoichiometry. Ag₂CrO₄ is 2:1 (a = 2, b = 1), not 1:1. The stoichiometry exponents matter — an off-by-one error changes s by an order of magnitude.

Frequently asked questions

Why is Ksp temperature-dependent?: Dissolution has a standard-state ΔG° that depends on T through ΔH° and ΔS°: ΔG° = ΔH° − TΔS°, and Ksp = exp(−ΔG°/RT). For endothermic dissolutions (most simple salts), Ksp grows with T; for exothermic ones (e.g. Ca(OH)₂, Li₂CO₃), Ksp shrinks. All the values in this tool are quoted at 25 °C. To predict Ksp at another temperature, use the van't Hoff relation ln(Ksp,T₂/Ksp,T₁) = −(ΔH°/R)(1/T₂ − 1/T₁), or look it up in Table 4-104 of CRC 97e which lists many salts across 0–100 °C.
When does the pure-Ksp calculation break down?: Two big cases. First, activity coefficients: Ksp is defined for activities, and this tool uses concentrations. Below ionic strength μ ≈ 0.01 mol/L the two agree to a few percent, but at μ = 0.1 mol/L, γ± drops to ~0.75 for a 1:1 salt and the concentration-Ksp underestimates true solubility by ~30 %. Second, ion pairing: polyvalent salts (Ca₃(PO₄)₂, La₂(SO₄)₃) form soluble ion pairs whose concentrations aren't in the Ksp equation, so real solubility can be 10× the naive answer. Correct with Davies-equation activity coefficients or a formal-charge speciation code.
How does the common-ion effect change s?: Adding a soluble source of one of the salt's ions shifts Ksp = [M]ᵃ[X]ᵇ so the *other* species has to drop to hold the product constant. For AgCl (Ksp = 1.77×10⁻¹⁰) in pure water, s = 1.33×10⁻⁵ M; in 0.10 M NaCl the Cl⁻ term is dominated by NaCl and s falls to Ksp/0.10 = 1.77×10⁻⁹ M — 7,500× less soluble. This is why gravimetric silver-chloride analyses wash the precipitate with dilute HNO₃, not pure water: minimising solubility losses.
Ksp vs Q — will a precipitate form?: Compute the reaction quotient Q = [M]ᵃ[X]ᵇ using the actual (not equilibrium) concentrations after mixing. If Q > Ksp the solution is supersaturated and a precipitate forms until Q falls to Ksp; if Q < Ksp everything stays dissolved; if Q = Ksp you are exactly at the saturation boundary. This is separate from the Ksp calc but uses the same product expression.
What is the difference between Ksp and solubility (g/L)?: Molar solubility s is mol of formula unit dissolved per litre. Solubility in g/L is s multiplied by the salt's molar mass. Ksp is neither — it's the equilibrium constant for the dissolution product, dimensionless in the modern IUPAC convention. Tables sometimes quote log Ksp or pKsp; pKsp = −log₁₀(Ksp), so pKsp(AgCl) = 9.75 corresponds to Ksp = 1.77×10⁻¹⁰.

Sources

CRC Handbook of Chemistry and Physics, 97th ed. (Haynes 2017), Table 4-104 — Aqueous solubility of inorganic compounds at various temperatures
Zumdahl, Chemistry, 10th ed. (2018), §15.6 — Solubility equilibria
Harris, Quantitative Chemical Analysis, 10th ed. (2020), §6-4 & Appendix F — Solubility products
IUPAC Gold Book — solubility product
Skoog, West, Holler & Crouch, Fundamentals of Analytical Chemistry, 10th ed. (2022), Ch. 11

Last updated: 2026-09-18.