Relating Ksp and molar solubility s
The solubility product constant, \(K_{sp}\), quantifies the equilibrium between a sparingly soluble ionic solid
and the ions it produces in water. The “why” behind \(K_{sp}\) is the law of mass action: at equilibrium, the
product of the dissolved ion concentrations (each raised to its stoichiometric power) is constant at a fixed
temperature. This makes \(K_{sp}\) a powerful way to compare how soluble different salts are and to compute a
salt’s molar solubility \(s\) (mol·L\(^{-1}\)) in pure water.
\(\mathrm{A}_m\mathrm{B}_n(s) \;\rightleftharpoons\; m\,\mathrm{A}^{\,z_A+}(aq) \;+\; n\,\mathrm{B}^{\,z_B-}(aq)\)
Because the solid is a pure phase, its activity is treated as 1 and does not appear in the equilibrium expression.
Under the common approximation that activities \(\approx\) concentrations in dilute solutions, the equilibrium
constant becomes:
\(K_{sp} \;=\; \big[\mathrm{A}^{\,z_A+}\big]^{m}\big[\mathrm{B}^{\,z_B-}\big]^{n}\)
How the “s” relationship is built (the stoichiometry step)
The “how” is a stoichiometry-to-equilibrium bridge. If the molar solubility in pure water is \(s\), then dissolving
\(s\) moles per liter of \(\mathrm{A}_m\mathrm{B}_n\) produces \(m\,s\) moles per liter of \(\mathrm{A}^{\,z_A+}\) and
\(n\,s\) moles per liter of \(\mathrm{B}^{\,z_B-}\). This is true because the coefficients \(m\) and \(n\) are the
mole ratios in the dissolution equation.
\(\big[\mathrm{A}^{\,z_A+}\big]=m\,s \quad\text{and}\quad \big[\mathrm{B}^{\,z_B-}\big]=n\,s\)
Substituting into the \(K_{sp}\) expression gives a single equation in \(s\). Notice how the total power on \(s\)
becomes \(m+n\), because \((m\,s)^m(n\,s)^n = m^m n^n s^{m+n}\). This is the reason salts with larger ion counts
(like AB3) often have much smaller \(s\) than a 1:1 salt with a similar \(K_{sp}\): the exponent on \(s\)
is larger, so \(s\) must be smaller to keep the product constant.
\(K_{sp} \;=\; (m\,s)^{m}(n\,s)^{n} \;=\; m^{m}n^{n}s^{\,m+n}\)
- \(\displaystyle s \;=\;\Bigg(\dfrac{K_{sp}}{m^{m}n^{n}}\Bigg)^{\!1/(m+n)}\)
- \(\displaystyle K_{sp}\;=\;m^{m}n^{n}s^{\,m+n}\)
Standard shortcuts (pure water)
The following are common “memorize-friendly” special cases obtained by plugging the stoichiometric coefficients
into the general formula. They are widely used in general chemistry and analytical chemistry problems.
-
AB (1:1)
\(K_{sp}=s^{2}\), \(\displaystyle s=\sqrt{K_{sp}}\)
Example pattern: \(\mathrm{AgCl}\), \(\mathrm{BaSO_4}\) (1:1 ions).
-
AB2 / A2B (1:2)
\(K_{sp}=4\,s^{3}\), \(\displaystyle s=\sqrt[3]{\dfrac{K_{sp}}{4}}\)
Example pattern: \(\mathrm{CaF_2}\) (1 Ca2+, 2 F-).
-
AB3 / A3B (1:3)
\(K_{sp}=27\,s^{4}\), \(\displaystyle s=\sqrt[4]{\dfrac{K_{sp}}{27}}\)
Example pattern: \(\mathrm{Fe(OH)_3}\) treated as 1:3 for ion count.
-
A2B3 / A3B2 (2:3)
\(K_{sp}=108\,s^{5}\), \(\displaystyle s=\sqrt[5]{\dfrac{K_{sp}}{108}}\)
Example pattern: salts producing 2 cations and 3 anions per formula unit.
When these relations do and do not apply
These \(K_{sp}\)–\(s\) conversions assume pure water and that the only source of the ions is the
dissolving solid. In the presence of a common ion (for example, dissolving \(\mathrm{AgCl}\) in a
solution that already contains \(\mathrm{Cl^-}\)), the equilibrium shifts left and the solubility decreases, so
\([\text{ion}] \neq\) simple multiples of \(s\). Likewise, at higher ionic strength the activity coefficients can deviate
from 1, meaning \(K_{sp}\) relates to activities rather than raw concentrations. For most dilute textbook and lab
settings, the concentration approximation gives excellent results and is the standard way to connect
solubility product and molar solubility.
Assumes pure water, no common ions, and activities ≈ concentrations. The calculator uses these relations to
convert between \(K_{sp}\) and molar solubility \(s\) at a fixed temperature.
