1. $K_{sp}$ Expressions — the foundation

For a generic salt $A_p B_q(s) \rightleftharpoons p\,A^{m+}(aq) + q\,B^{n-}(aq)$,

$K_{sp} = [A^{m+}]^{p}\,[B^{n-}]^{q}.$

Pure solids and pure liquids do not appear in the expression. Examples:

• $\text{AgCl}(s) \rightleftharpoons \text{Ag}^{+} + \text{Cl}^{-}$: $K_{sp} = [\text{Ag}^{+}][\text{Cl}^{-}]$
• $\text{PbCl}_{2}(s) \rightleftharpoons \text{Pb}^{2+} + 2\,\text{Cl}^{-}$:
• $\text{Ca}_{3}(\text{PO}_{4})_{2}(s) \rightleftharpoons 3\,\text{Ca}^{2+} + 2\,\text{PO}_{4}^{3-}$:

The smaller the $K_{sp}$, the less soluble the salt. But you can only directly compare solubilities of two salts by $K_{sp}$ if they have the same stoichiometry — otherwise compute molar solubility first.

2. Molar Solubility from $K_{sp}$

Let $s$ = molar solubility (mol/L of salt that dissolves to saturate the solution).

For $\text{AgCl}$: $[\text{Ag}^{+}] = s$, $[\text{Cl}^{-}] = s$ → $K_{sp} = s^{2}$, so $s = \sqrt{K_{sp}}$.

For $\text{PbCl}_{2}$: $[\text{Pb}^{2+}] = s$, $[\text{Cl}^{-}] = 2s$ → $K_{sp} = s(2s)^{2} = 4s^{3}$, so $s = \sqrt[3]{K_{sp}/4}$.

For $\text{Ca}_{3}(\text{PO}_{4})_{2}$: $K_{sp} = (3s)^{3}(2s)^{2} = 108\,s^{5}$.

AP tip. Always start with the dissolution equation, then read off the stoichiometric coefficients to set the ion concentrations in terms of $s$. Most "$K_{sp}$ vs. solubility" wrong answers come from forgetting the coefficient ($2s$, not $s$) or failing to raise to the right power.

3. The Common-Ion Effect

By Le Châtelier's principle, adding an ion already present in the dissolution equilibrium shifts the equilibrium back toward the solid, decreasing solubility.

Example. $K_{sp}(\text{AgCl}) = 1.8 \times 10^{-10}$. In pure water, $s = \sqrt{K_{sp}} = 1.3 \times 10^{-5}$ M.

In 0.10 M NaCl, $[\text{Cl}^{-}]$ is essentially fixed at 0.10 M (the trace from AgCl is negligible). So $[\text{Ag}^{+}] = K_{sp}/[\text{Cl}^{-}] = (1.8 \times 10^{-10})/(0.10) = 1.8 \times 10^{-9}\,\text{M}.$ The molar solubility crashes from $1.3 \times 10^{-5}$ M to $1.8 \times 10^{-9}$ M — about four orders of magnitude lower.

The common-ion effect underlies why "salting out" works in many separations.

4. $Q$ vs. $K_{sp}$ — Will a Precipitate Form?

The reaction quotient for a dissolution uses the same expression as $K_{sp}$ but plugged with current (not equilibrium) ion concentrations.

Worked example. Mix 50 mL of 0.0010 M $\text{Pb(NO}_{3})_{2}$ with 50 mL of 0.020 M NaI. $K_{sp}(\text{PbI}_{2}) = 9.8 \times 10^{-9}$.

After mixing (volumes add → each is diluted by ½):

• $[\text{Pb}^{2+}] = 5.0 \times 10^{-4}$ M
• $[\text{I}^{-}] = 0.010$ M
• $Q = (5.0 \times 10^{-4})(0.010)^{2} = 5.0 \times 10^{-8}$

Since $Q\;(5.0 \times 10^{-8}) > K_{sp}\;(9.8 \times 10^{-9})$, PbI₂ precipitates.

5. pH Effects on Solubility

A salt becomes more soluble in acidic solution if its anion is a conjugate base of a weak acid (or if its cation reacts with OH⁻ in basic solution).

Examples.

• $\text{CaF}_{2}$ is more soluble in acidic solution. The $\text{F}^{-}$ ion reacts with $\text{H}^{+}$ to form HF (a weak acid), removing $\text{F}^{-}$ from solution and pulling the dissolution equilibrium right.
• $\text{Mg(OH)}_{2}$ is more soluble in acidic solution (acid neutralizes OH⁻) and less soluble in basic solution (added OH⁻ is a common ion).
• $\text{AgCl}$ solubility is not noticeably affected by pH because $\text{Cl}^{-}$ is the conjugate base of a strong acid (HCl) and doesn't react with $\text{H}^{+}$.

Rule of thumb. Salts of weak-acid anions (F⁻, OH⁻, S²⁻, $\text{CO}_{3}^{2-}$, $\text{PO}_{4}^{3-}$, …) become more soluble at low pH. Salts of strong-acid anions (Cl⁻, Br⁻, I⁻, $\text{NO}_{3}^{-}$, $\text{ClO}_{4}^{-}$) are essentially pH-independent.

6. Selective Precipitation

When a solution contains two ions that both form an insoluble salt with a common reagent, the one with the smaller $K_{sp}$ precipitates first (assuming similar stoichiometry).

Example. A solution is 0.10 M in both $\text{Cl}^{-}$ and $\text{I}^{-}$. Slowly add AgNO₃. Use $K_{sp}(\text{AgCl}) = 1.8 \times 10^{-10}$ and $K_{sp}(\text{AgI}) = 8.5 \times 10^{-17}$.

• AgI starts to precipitate when $[\text{Ag}^{+}] = K_{sp}/[\text{I}^{-}] = 8.5 \times 10^{-17}/0.10 = 8.5 \times 10^{-16}$ M.
• AgCl starts when $[\text{Ag}^{+}] = 1.8 \times 10^{-10}/0.10 = 1.8 \times 10^{-9}$ M.

AgI begins to precipitate at a much lower [Ag⁺]. By the time AgCl just begins to precipitate, $[\text{I}^{-}]$ has been reduced to $K_{sp}(\text{AgI})/(1.8 \times 10^{-9}) = 4.7 \times 10^{-8}$ M — i.e., over 99.99999% of the iodide has been removed before any chloride precipitates. This is the basis of qualitative-analysis cation/anion separation schemes.

7. Synthesis & AP Review

Big ideas to leave with:

• $K_{sp}$ is just an equilibrium constant for the dissolution of a sparingly soluble salt. Solid omitted, ions to their stoichiometric powers.
• Always relate molar solubility $s$ to $K_{sp}$ via the dissolution stoichiometry (e.g., $K_{sp} = 4s^{3}$ for $A B_{2}$).
• The common-ion effect dramatically lowers solubility (Le Châtelier on the dissolution equilibrium).
• Compare $Q$ to $K_{sp}$ to decide if a precipitate forms. Don't forget to dilute ion concentrations after mixing.
• Solubility of salts with weak-acid anions increases at low pH; salts of strong-acid anions are essentially pH-independent.
• Selective precipitation exploits differences in $K_{sp}$ to separate ions.

Common mistakes

• Forgetting to raise ion concentrations to the stoichiometric power in $K_{sp}$ ($[\text{Cl}^{-}]^{2}$, not $[\text{Cl}^{-}]$, for $\text{PbCl}_{2}$).
• Using molar solubility $s$ as a $K_{sp}$ value, or vice versa.
• Forgetting to dilute when mixing two solutions before computing $Q$.
• Comparing $K_{sp}$ values of salts with different stoichiometries to rank solubilities. Compute $s$ first.
• Saying that pH affects all salts. It only matters when the cation reacts with OH⁻ or the anion reacts with H⁺.

Quick reference card

• $K_{sp}$ for $A B$: $K_{sp} = s^{2}$
• $K_{sp}$ for $A B_{2}$ or $A_{2} B$:
• $K_{sp}$ for $A B_{3}$ or $A_{3} B$:
• $K_{sp}$ for $A_{3} B_{2}$:
• Common ion → solubility ↓ (Le Châtelier)
• $Q > K_{sp}$ → precipitate forms
• Weak-acid anion (F⁻, OH⁻, $\text{CO}_{3}^{2-}$, …) → solubility ↑ at low pH

$\boxed{K_{sp} = [\text{Ca}^{2+}]^{3}[\text{PO}_{4}^{3-}]^{2}}.$

Note both the coefficients (3 cations, 2 anions per formula unit) and the corresponding exponents.

Let $s$ = molar solubility. Then $[\text{Ba}^{2+}] = s$ and $[\text{SO}_{4}^{2-}] = s$:

$K_{sp} = s^{2} = 1.1 \times 10^{-10}$ → $s = \sqrt{1.1 \times 10^{-10}} = \boxed{1.05 \times 10^{-5}\,\text{M}}$.

ICE: $[\text{Pb}^{2+}] = s$, $[\text{Cl}^{-}] = 2s$.

$K_{sp} = s(2s)^{2} = 4s^{3} = 1.7 \times 10^{-5}$

$s^{3} = 4.25 \times 10^{-6}$ → $s = \boxed{1.6 \times 10^{-2}\,\text{M}}$.

A notably more-soluble salt than the silver halides above — consistent with $K_{sp}$ being roughly $10^{5}$ times larger.

In 0.10 M NaCl: $[\text{Cl}^{-}] \approx 0.10$ M (the trace from AgCl is negligible). Then

$[\text{Ag}^{+}] = K_{sp}/[\text{Cl}^{-}] = (1.8 \times 10^{-10})/(0.10) = 1.8 \times 10^{-9}$ M.

$[\text{Ag}^{+}]$ now equals the molar solubility, so $s' = 1.8 \times 10^{-9}$ M.

Factor: $s/s' = 1.34 \times 10^{-5}/1.8 \times 10^{-9} = \boxed{\sim 7400 \times \text{ less soluble}}$.

Step 2. $Q = [\text{Ag}^{+}][\text{Cl}^{-}] = (1.0 \times 10^{-3})(1.0 \times 10^{-3}) = 1.0 \times 10^{-6}$.

Step 3. $Q\;(1.0 \times 10^{-6}) \gg K_{sp}\;(1.8 \times 10^{-10})$, so $\boxed{\text{yes, AgCl precipitates}}$ until $Q$ falls back to $K_{sp}$.

AgCl starts when $[\text{Ag}^{+}] = (1.8 \times 10^{-10})/(0.010) = 1.8 \times 10^{-8}$ M.

AgI requires far less $[\text{Ag}^{+}]$, so $\boxed{\text{AgI precipitates first}}$.

(b) When AgCl just begins to precipitate, $[\text{Ag}^{+}] = 1.8 \times 10^{-8}$ M. At that point

$[\text{I}^{-}] = K_{sp}(\text{AgI})/[\text{Ag}^{+}] = (8.5 \times 10^{-17})/(1.8 \times 10^{-8}) = \boxed{4.7 \times 10^{-9}\,\text{M}}$.

Fraction remaining: $4.7 \times 10^{-9}/0.010 = 4.7 \times 10^{-7}$, i.e. >99.99995% of the iodide has been removed before any chloride precipitates. Excellent separation.

(a) Pure water. Let $s$ = molar solubility. Then $[\text{Ag}^{+}] = 2s$, $[\text{CrO}_{4}^{2-}] = s$.

$K_{sp} = (2s)^{2}(s) = 4s^{3} = 1.1 \times 10^{-12}$ → $s^{3} = 2.75 \times 10^{-13}$ → $\boxed{s = 6.5 \times 10^{-5}\,\text{M}}$.

(b) 0.010 M $\text{K}_{2}\text{CrO}_{4}$ (common ion = $\text{CrO}_{4}^{2-}$). Now $[\text{CrO}_{4}^{2-}] \approx 0.010$ M (the trace from $\text{Ag}_{2}\text{CrO}_{4}$ is negligible). Let $s' =$ new molar solubility, so $[\text{Ag}^{+}] = 2s'$.

$K_{sp} = (2s')^{2}(0.010) = 4s'^{2}(0.010) = 1.1 \times 10^{-12}$

$s'^{2} = 2.75 \times 10^{-11}$ → $\boxed{s' = 5.2 \times 10^{-6}\,\text{M}}$.

The common ion drops solubility by ~12×.

$K_{sp} = s(2s)^{2} = 4s^{3} = 3.9 \times 10^{-11}$ → $s^{3} = 9.75 \times 10^{-12}$ → $\boxed{s = 2.1 \times 10^{-4}\,\text{M}}$.

(b) $\text{F}^{-}$ is the conjugate base of the weak acid HF (pKa ≈ 3.2). In 0.10 M HCl, the high $[\text{H}^{+}]$ converts much of the dissolved $\text{F}^{-}$ to undissociated HF, removing $\text{F}^{-}$ from the equilibrium. By Le Châtelier, dissolution shifts right and solubility increases substantially.

For $\text{CaCl}_{2}$, the anion is $\text{Cl}^{-}$, the conjugate base of the strong acid HCl — it has essentially no tendency to combine with $\text{H}^{+}$. So acid does not consume $\text{Cl}^{-}$ and the dissolution equilibrium is unaffected by pH. ($\text{CaCl}_{2}$ is also extremely soluble in water for this kind of analysis to apply, but the key point is the lack of pH coupling.)