Solubility Equilibria

Setting up the equilibrium. A "sparingly soluble" salt is not insoluble — it dissolves just enough to reach a genuine dynamic equilibrium between the undissolved solid and its dissolved ions. Write the general dissolution as

Because the solid's activity is taken as 1 (the same convention used for pure solids and liquids in every $K_c$ expression you have met), the equilibrium constant involves only the dissolved ions:

where: $[\text{A}^{y+}]$ and $[\text{B}^{x-}]$ are equilibrium molar concentrations of the two ions, and $x, y$ are their stoichiometric coefficients in the balanced dissolution. $K_{sp}$ is fixed at a given temperature, and a smaller $K_{sp}$ means a less soluble salt.

From $K_{sp}$ to molar solubility. Let $s$ be the molar solubility of $\text{A}_x\text{B}_y$ in pure water — the moles of salt that dissolve per litre. Then $[\text{A}^{y+}] = xs$ and $[\text{B}^{x-}] = ys$, so

For a 1:1 salt like $\text{AgCl}$ this collapses to $K_{sp} = s^2$, so $s = \sqrt{K_{sp}}$. For a 1:2 salt like $\text{Ag}_2\text{CrO}_4$ or $\text{Mg(OH)}_2$, $K_{sp} = 4s^3$, so $s = \sqrt[3]{K_{sp}/4}$. The exponent on $s$ always equals the total number of ions released per formula unit — never assume a blanket formula without checking the stoichiometry.

Predicting precipitation with $Q_{sp}$. The ionic product $Q_{sp}$ has exactly the same algebraic form as $K_{sp}$ but is evaluated from whatever concentrations happen to be present at any instant, equilibrium or not.

Common-ion effect on solubility. Adding a soluble salt that shares an ion with a sparingly soluble one suppresses the latter's solubility. Dissolving $\text{NaCl}$ into a saturated $\text{AgCl}$ solution raises $[\text{Cl}^-]$; since $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$ must stay fixed, $[\text{Ag}^+]$ — and therefore how much solid $\text{AgCl}$ can stay dissolved — must fall. Le Chatelier's principle gives the same conclusion qualitatively: the extra $\text{Cl}^-$ pushes the dissolution equilibrium backward, toward more solid.

Common-ion effect on weak-acid/base ionisation. The identical mechanism governs weak electrolytes. Adding sodium acetate (a source of $\text{CH}_3\text{COO}^-$) to acetic acid pushes $\text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{COO}^-$ backward, keeping $K_a$ constant while lowering the degree of ionisation, $\alpha$. This is the chemistry underlying every acetic acid/acetate buffer.

The goal is to derive, purely from the equilibrium expression and stoichiometry, a general formula linking $K_{sp}$ to molar solubility $s$ for a salt of arbitrary formula $\text{A}_x\text{B}_y$, then specialise it to the 1:1 and 1:2 cases used throughout this article and in numerical problems.

Step 1. Write the dissolution equilibrium and let molar solubility be $s$ (mol of salt dissolved per litre). By definition of $s$ and the stoichiometry of $\text{A}_x\text{B}_y\text{(s)} \rightleftharpoons x\text{A}^{y+} + y\text{B}^{x-}$, each mole of salt that dissolves releases $x$ moles of $\text{A}^{y+}$ and $y$ moles of $\text{B}^{x-}$, so at equilibrium $[\text{A}^{y+}] = xs$ and $[\text{B}^{x-}] = ys$.

Step 2. Substitute directly into the solubility-product expression. $K_{sp} = [\text{A}^{y+}]^x[\text{B}^{x-}]^y = (xs)^x(ys)^y$.

Step 3. Separate the constants from the variable. Expanding, $(xs)^x(ys)^y = x^x s^x \cdot y^y s^y = x^x y^y s^{(x+y)}$. This is the general relation: $K_{sp} = x^x y^y s^{(x+y)}$, valid for any sparingly soluble salt once $x$ and $y$ are read off the balanced equation.

Step 4. Specialise. For a 1:1 salt ($x = y = 1$, e.g. AgCl), $K_{sp} = 1^1 \cdot 1^1 \cdot s^2 = s^2$, so $s = \sqrt{K_{sp}}$. For a 1:2 salt ($x=1, y=2$, e.g. $\text{Ag}_2\text{CrO}_4$ or $\text{Mg(OH)}_2$), $K_{sp} = 1^1 \cdot 2^2 \cdot s^3 = 4s^3$, so $s = \sqrt[3]{K_{sp}/4}$. The exponent on $s$ is always $x+y$, the total ion count per formula unit — this is the check to run before plugging numbers into any solubility problem.

Problem. The solubility product of $\text{AgCl}$ is $K_{sp} = 1.8\times10^{-10}$ at 298 K (molar mass of AgCl = 143.5 g mol$^{-1}$). Find (a) the molar solubility of AgCl in pure water and in g L$^{-1}$, and (b) the molar solubility of AgCl in $0.10\ \text{mol L}^{-1}\ \text{NaCl}$ solution, comparing the two.

Step 1. In pure water, let molar solubility be $s$. Since AgCl is a 1:1 salt, $[\text{Ag}^+] = [\text{Cl}^-] = s$, so $K_{sp} = s^2 = 1.8\times10^{-10}$, giving $s = \sqrt{1.8\times10^{-10}} = 1.34\times10^{-5}\ \text{mol L}^{-1}$.

Step 2. Convert to g L$^{-1}$: mass $= s \times M = 1.34\times10^{-5} \times 143.5 = 1.92\times10^{-3}\ \text{g L}^{-1} \approx 1.9\ \text{mg}$ per litre — confirming AgCl is only very sparingly soluble.

Step 3. Now dissolve AgCl in $0.10\ \text{mol L}^{-1}$ NaCl. $[\text{Cl}^-]$ has two sources: the tiny amount $s'$ from dissolved AgCl, and $0.10\ \text{mol L}^{-1}$ from the fully dissociated NaCl. Since $s'$ will turn out negligible, $[\text{Cl}^-] \approx 0.10\ \text{mol L}^{-1}$, while $[\text{Ag}^+] = s'$ (its only source is the dissolving AgCl).

Step 4. $K_{sp} = [\text{Ag}^+][\text{Cl}^-] = s' \times 0.10 = 1.8\times10^{-10}$, so $s' = 1.8\times10^{-9}\ \text{mol L}^{-1}$ — indeed negligible next to 0.10, so the approximation in Step 3 holds.

Step 5. Compare: solubility falls from $1.34\times10^{-5}\ \text{mol L}^{-1}$ in pure water to $1.8\times10^{-9}\ \text{mol L}^{-1}$ in 0.10 M NaCl, roughly a 7,400-fold decrease. This is the common-ion effect: added Cl⁻ drives $\text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$ backward, keeping $K_{sp}$ fixed while $[\text{Ag}^+]$ — and hence solubility — collapses.

Competitive stretch

If instead $0.10\ \text{mol L}^{-1}\ \text{AgNO}_3$ were added rather than NaCl, the roles reverse: $[\text{Ag}^+] \approx 0.10\ \text{mol L}^{-1}$ dominates, and solving $K_{sp} = 0.10 \times [\text{Cl}^-]$ gives the same suppressed solubility, $1.8\times10^{-9}\ \text{mol L}^{-1}$ — the common-ion effect works symmetrically on whichever ion is added in excess.

• Writing $K_{sp} = s^2$ for every salt regardless of formula — the correct power depends on stoichiometry ($K_{sp} = 4s^3$ for a 1:2 salt); always derive it from the balanced dissolution equation rather than memorising one shape.
• Confusing $K_{sp}$ (a fixed equilibrium constant at a given temperature) with the ionic product $Q_{sp}$ (computed from whatever concentrations are actually present); precipitation is predicted by comparing $Q_{sp}$ to $K_{sp}$, not by inspecting $K_{sp}$ alone.
• Ignoring a common ion already present from another source — forgetting the extra $\text{Cl}^-$ contributed by NaCl and computing solubility as if in pure water gives an answer far too high.
• Treating "sparingly soluble" as "insoluble" — a small $K_{sp}$ still means some ions are present at equilibrium; it is never exactly zero.

Ask the tutor: want to see how a $Q_{sp}$-versus-$K_{sp}$ comparison predicts whether mixing two salt solutions produces a precipitate — ask the AI tutor to walk through that calculation with you.