The task “molar solubility from ksp” means converting a solubility product constant \(K_{sp}\) into a molar solubility \(s\) (in \(\text{mol}\cdot\text{L}^{-1}\)) by using the dissolution stoichiometry of the salt.
Core idea: express every equilibrium concentration in terms of \(s\)
For a sparingly soluble ionic solid \(M_aX_b(s)\) dissolving in pure water (no added common ions, no complexing agents, no significant hydrolysis), write the dissolution reaction:
\[ M_aX_b(s) \rightleftharpoons a\,M^{\,\text{(charge)}}(aq) + b\,X^{\,\text{(charge)}}(aq) \]
Define \(s\) as the molar solubility of the solid: the number of moles of \(M_aX_b\) that dissolve per liter at equilibrium. Then the ion concentrations created by dissolution are:
\[ [M^{\,\cdot}] = a\,s,\qquad [X^{\,\cdot}] = b\,s \]
Substitute into the solubility product expression:
\[ K_{sp} = [M^{\,\cdot}]^{a}[X^{\,\cdot}]^{b} = (a\,s)^{a}(b\,s)^{b} = a^{a}b^{b}\,s^{a+b} \]
Solve for \(s\):
\[ s = \left(\frac{K_{sp}}{a^{a}b^{b}}\right)^{\!\frac{1}{a+b}} \]
This closed-form result applies cleanly when dissolution is the only source of ions (pure water) and the equilibrium expression contains only \(a\,s\) and \(b\,s\). If a common ion is present, or if pH/complex ions affect the equilibrium, then concentrations are not simply \(a\,s\) and \(b\,s\), and an ICE-table setup is required.
Quick reference table: common stoichiometries
| Salt type | Dissolution | \(K_{sp}\) in terms of \(s\) | \(s\) in terms of \(K_{sp}\) |
|---|---|---|---|
| \(MX\) (1:1) | \(MX(s)\rightleftharpoons M^{+}+X^{-}\) | \(K_{sp}=s^2\) | \(s=\sqrt{K_{sp}}\) |
| \(MX_2\) (1:2) | \(MX_2(s)\rightleftharpoons M^{2+}+2X^{-}\) | \(K_{sp}=s(2s)^2=4s^3\) | \(s=\left(\frac{K_{sp}}{4}\right)^{\!\frac{1}{3}}\) |
| \(M_2X_3\) (2:3) | \(M_2X_3(s)\rightleftharpoons 2M^{3+}+3X^{2-}\) | \(K_{sp}=(2s)^2(3s)^3=108s^5\) | \(s=\left(\frac{K_{sp}}{108}\right)^{\!\frac{1}{5}}\) |
Worked examples of molar solubility from \(K_{sp}\)
Example 1 (1:1 salt): For \(AgCl(s)\rightleftharpoons Ag^{+}+Cl^{-}\), \(K_{sp}=[Ag^{+}][Cl^{-}]=s\cdot s=s^2\). Therefore:
\[ s=\sqrt{K_{sp}} \]
Example 2 (1:2 salt): For \(CaF_2(s)\rightleftharpoons Ca^{2+}+2F^{-}\), let \(s\) be the molar solubility. Then \([Ca^{2+}]=s\) and \([F^{-}]=2s\). So:
\[ K_{sp}=[Ca^{2+}][F^{-}]^2=s(2s)^2=4s^3 \quad\Rightarrow\quad s=\left(\frac{K_{sp}}{4}\right)^{\!\frac{1}{3}} \]
Visualization: path from stoichiometry to molar solubility
Common pitfalls (and how to avoid them)
- Forgetting exponents: coefficients in the dissolution reaction become exponents in \(K_{sp}\) (e.g., \(2X^{-}\) leads to \([X^-]^2\)).
- Mixing “\(s\)” with ion concentrations: \(s\) is for the solid; ion concentrations are multiples like \(a\,s\), \(b\,s\).
- Using the pure-water shortcut when a common ion exists: if initial \([X^-]_0\neq 0\), then \([X^-]=[X^-]_0+b\,s\) and the equation changes.