Solubility

Solubility sits at the intersection of equilibrium chemistry, acid–base theory, and coordination chemistry. The central idea: when an ionic solid dissolves, it reaches a dynamic equilibrium with its ions in solution. Every topic here is a variation on "how much dissolves?" or "what shifts that equilibrium?" — and along the way you reinforce Le Châtelier's principle and Lewis acid–base theory.

Priority labels: Must know = cold; Know the logic = mechanism not names; Passage-level = recognize, don't memorize; Optional = skippable.

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Units of Concentration

You need to express how much solute is in solution before reasoning about equilibrium.

Molarity (M)

Must know

Molarity is moles of solute per liter of solution:

$M = \frac{n_{\text{solute}}}{V_{\text{solution (L)}}}$

This is the unit for every equilibrium expression and Ksp calculation. In an equilibrium context, concentration always means molarity.

Dilution (moles conserved) — the single most-tested concentration manipulation:

$M_1 V_1 = M_2 V_2$

Molality (m)

Must know

Molality is moles of solute per kilogram of solvent:

$m = \frac{n_{\text{solute}}}{m_{\text{solvent (kg)}}}$

It appears in colligative property problems because, unlike molarity, it does not change with temperature (mass of solvent is temperature-independent; volume is not).

Mole Fraction ($\chi$), Mass Percent, ppm/ppb

Must know

$\chi_A = \frac{n_A}{n_{\text{total}}} \qquad \text{mass \%} = \frac{m_{\text{solute}}}{m_{\text{solution}}} \times 100\%$

Mole fractions are dimensionless and sum to 1 (they appear in Raoult's Law). Passage-level ppm/ppb are mass percent scaled further (1 ppm ≈ 1 mg/L for dilute aqueous solutions); recognize them in environmental/clinical contexts.

Fully Worked Example: Dilution

Must know

A stock solution of $\ce{NaCl}$ is 3.0 M. You need 250 mL of a 0.60 M solution. What volume of stock do you take?

$V_1 = \frac{M_2 V_2}{M_1} = \frac{(0.60 \text{ M})(0.250 \text{ L})}{3.0 \text{ M}} = 0.050 \text{ L} = 50 \text{ mL}$

Sanity check: diluting 5-fold (3.0 ÷ 0.60 = 5), so take 1/5 of final volume: 250/5 = 50 mL. ✓

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Quick check: A student dissolves 0.10 mol of glucose in 500 g of water. What is the molality?

Answer: $m = 0.10 \text{ mol} / 0.500 \text{ kg} = 0.20 \text{ m}$. Molarity would require the total solution volume (not given) — this is why molality and molarity aren't interchangeable.

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The Solubility Product Constant, $K_{sp}$

Qualitative Solubility Rules

Must know

The MCAT often asks whether a precipitate forms without giving $K_{sp}$ values. Must know the broad categories (top entries win when rules conflict):

• Always soluble: $\ce{Na+}$, $\ce{K+}$, $\ce{NH4+}$, $\ce{NO3-}$ salts (also most acetates).
• Usually soluble: halides and sulfates — except halides with $\ce{Ag+}$, $\ce{Pb^{2+}}$, $\ce{Hg2^{2+}}$, and sulfates with $\ce{Ba^{2+}}$, $\ce{Pb^{2+}}$, $\ce{Ca^{2+}}$, $\ce{Sr^{2+}}$.
• Usually insoluble: carbonates, phosphates, sulfides, hydroxides — except with Group I cations or $\ce{NH4+}$.

A double-displacement reaction precipitates only if a possible product is insoluble.

The Equilibrium Expression

Must know

For a slightly soluble salt $\ce{M_aX_b}$:

$\ce{M_aX_b(s) <=> a M^{n+}(aq) + b X^{m-}(aq)}$

Pure solids are omitted (their "concentration" is constant), so the solubility product is:

$K_{sp} = [\text{M}^{n+}]^a[\text{X}^{m-}]^b$

A small $K_{sp}$ means little dissolves; a large $K_{sp}$ means relatively soluble. Be comfortable working in both directions.

From $K_{sp}$ to Molar Solubility

Must know

Molar solubility ($s$) is the moles of salt that dissolve per liter to reach equilibrium. Use an ICE table.

Fully Worked Example: Molar Solubility of PbI₂

$\ce{PbI2(s) <=> Pb^{2+}(aq) + 2 I^-(aq)} \quad K_{sp} = 9.8 \times 10^{-9}$

At equilibrium $[\ce{Pb^{2+}}] = s$ and $[\ce{I^-}] = 2s$, so:

$K_{sp} = (s)(2s)^2 = 4s^3 \;\Rightarrow\; s^3 = \frac{9.8 \times 10^{-9}}{4} = 2.45 \times 10^{-9}$

$s = \sqrt[3]{2.45 \times 10^{-9}} \approx 1.35 \times 10^{-3} \text{ M}$

The Reaction Quotient Q vs. $K_{sp}$

Must know

The ion product $Q$ is calculated like $K_{sp}$ but using actual concentrations at a given moment:

• $Q < K_{sp}$: unsaturated; more solid can dissolve.
• $Q = K_{sp}$: saturated (at equilibrium).
• $Q > K_{sp}$: supersaturated; a precipitate forms.

Quick check: You mix equal volumes of $1.0 \times 10^{-4}$ M $\ce{Pb^{2+}}$ and $1.0 \times 10^{-2}$ M $\ce{I^-}$. Will $\ce{PbI2}$ precipitate? ($K_{sp} = 9.8 \times 10^{-9}$)

Answer: After mixing, concentrations halve: $[\ce{Pb^{2+}}] = 5.0 \times 10^{-5}$ M, $[\ce{I^-}] = 5.0 \times 10^{-3}$ M.
$Q = (5.0 \times 10^{-5})(5.0 \times 10^{-3})^2 = 1.25 \times 10^{-9}$.
Since $Q < K_{sp}$ → no precipitate forms.

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Temperature and Gas Solubility

Know the logic

Dissolution has a heat of solution, and Le Châtelier predicts the temperature response:

• For most solids, dissolving is endothermic, so higher temperature increases solubility (underlies recrystallization: dissolve hot, cool slowly to drop a pure crystal).
• For gases, dissolving is exothermic, so solubility decreases with rising temperature — warm soda goes flat; warm water holds less $\ce{O2}$.

Henry's Law (Gas Solubility vs. Pressure)

Know the logic

Gas solubility is proportional to its partial pressure above the liquid:

$C = k_H P$

This explains carbonation, blood-gas exchange tracking alveolar partial pressures, and decompression sickness ("the bends" — dissolved gas bubbles out when a diver ascends too fast).

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Common-Ion Effect and Laboratory Separations

The Core Idea

Must know

Add $\ce{NaCl}$ to saturated $\ce{AgCl}$: the extra $\ce{Cl^-}$ raises $Q$ above $K_{sp}$, shifting equilibrium left (Le Châtelier) so more $\ce{AgCl}$ precipitates. This is the common-ion effect — adding an ion already in the equilibrium expression suppresses solubility.

$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}$

Quantitative Treatment

Must know

Worked Example: Molar solubility of $\ce{AgCl}$ ($K_{sp} = 1.8 \times 10^{-10}$) in 0.10 M $\ce{NaCl}$?

With $[\ce{Cl^-}] \approx 0.10$ (since $s \ll 0.10$) and $[\ce{Ag^+}] = s$:

$K_{sp} = (s)(0.10) \;\Rightarrow\; s = \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9} \text{ M}$

Compare to pure water, where $s = \sqrt{1.8 \times 10^{-10}} \approx 1.3 \times 10^{-5}$ M — the common ion cuts solubility ~10,000-fold.

Selective Precipitation

Know the logic

A mixture of cations can be separated by adding an anion whose $K_{sp}$ differs greatly with each. With $\ce{Cl^-}$, $\ce{AgCl}$ ($K_{sp} = 1.8 \times 10^{-10}$) precipitates at far lower $[\ce{Cl^-}]$ than $\ce{PbCl2}$ ($K_{sp} = 1.7 \times 10^{-5}$). Adding $\ce{Cl^-}$ to a concentration between the two thresholds precipitates $\ce{Ag^+}$ while leaving $\ce{Pb^{2+}}$ in solution — selective precipitation.

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Quick check: Two cations, $\ce{Ba^{2+}}$ and $\ce{Ca^{2+}}$, are in solution. $K_{sp}(\ce{BaSO4}) = 1.1 \times 10^{-10}$; $K_{sp}(\ce{CaSO4}) = 4.9 \times 10^{-5}$. Adding $\ce{Na2SO4}$ slowly, which precipitates first?

Answer: $\ce{Ba^{2+}}$ — $\ce{BaSO4}$ has the smaller $K_{sp}$, so it becomes insoluble at a much lower $[\ce{SO4^{2-}}]$.

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Complex Ion Formation

Lewis Acid–Base Chemistry

Know the logic

A complex ion is a central metal cation (Lewis acid — accepts electron pairs) surrounded by ligands (Lewis bases — donate electron pairs). The coordination number is the number of ligand-to-metal bonds (commonly 4 or 6).

Optional

Reference: common ligands are mostly monodentate ($\ce{H2O}$, $\ce{NH3}$, $\ce{OH^-}$, $\ce{CN^-}$); a few are polydentate ($\ce{S2O3^{2-}}$ bidentate, EDTA hexadentate) — don't memorize denticities.

Formation Constant $K_f$

Must know

Complex ion formation is itself an equilibrium with a formation constant $K_f$:

$\ce{Ag+(aq) + 2 NH3(aq) <=> [Ag(NH3)2]+(aq)} \quad K_f = 1.7 \times 10^7$

A large $K_f$ means the complex is stable and forms readily; common $K_f$ values are $\gg 1$, so the reaction lies far right.

Passage-level

Recognize classic complexes if a passage uses them — e.g. $\ce{[Ag(NH3)2]+}$ (Tollens' reagent), $\ce{[Cu(NH3)4]^{2+}}$ (deep blue), $\ce{[Al(OH)4]-}$ and $\ce{[Zn(OH)4]^{2-}}$ (amphoteric hydroxides). Don't memorize the list.

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Quick check: In $\ce{[Cu(NH3)4]^{2+}}$, identify the Lewis acid and Lewis base.

Answer: $\ce{Cu^{2+}}$ is the Lewis acid (accepts lone pairs); $\ce{NH3}$ are the Lewis bases (donate N lone pairs). Coordination number is 4.

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Complex Ions and Solubility

The Coupled-Equilibria Concept

Must know

Complex ion formation can dramatically increase the solubility of an otherwise insoluble salt. Two equilibria combine:

1. Dissolution: $\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}$ (small $K_{sp}$)
2. Complexation: $\ce{Ag+(aq) + 2 NH3(aq) <=> [Ag(NH3)2]+(aq)}$ (large $K_f$)

Adding them gives the net reaction, with $K_{\text{net}} = K_{sp} \times K_f$:

$\ce{AgCl(s) + 2 NH3(aq) <=> [Ag(NH3)2]+(aq) + Cl-(aq)}$

$K_{\text{net}}$ is much larger than $K_{sp}$ alone: complexation constantly removes free $\ce{Ag^+}$ (Le Châtelier), pulling dissolution right. So $\ce{AgCl}$ dissolves far more readily in ammonia than in water.

Passage-level

The body uses complex-ion chemistry constantly — iron in hemoglobin (porphyrin–iron), zinc–histidine in enzyme active sites, the zinc–hydroxide of carbonic anhydrase.

Amphoteric Hydroxides

Must know

Amphoteric hydroxides (e.g. $\ce{Al(OH)3}$, $\ce{Zn(OH)2}$, $\ce{Cr(OH)3}$) are insoluble at neutral pH but dissolve in both acid and strong base:

In acid: $\ce{Al(OH)3(s) + 3 H+(aq) -> Al^{3+}(aq) + 3 H2O(l)}$

In excess base (complex formation): $\ce{Al(OH)3(s) + OH-(aq) -> [Al(OH)4]-(aq)}$

The excess base case is the same principle as the ammonia–silver example, with $\ce{OH^-}$ as the ligand.

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Quick check: A student adds excess $\ce{NaOH}$ to a precipitate of $\ce{Zn(OH)2}$. Will it dissolve or persist? Why?

Answer: It dissolves. $\ce{Zn(OH)2}$ is amphoteric; excess $\ce{OH^-}$ forms $\ce{[Zn(OH)4]^{2-}}$, shifting dissolution right.

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Solubility and pH

Why pH Matters for Solubility

Must know

If a salt's anion is the conjugate base of a weak acid, it reacts with $\ce{H+}$. Lowering pH consumes the anion, which (Le Châtelier) pulls dissolution right and increases solubility:

$\ce{MA(s) <=> M+(aq) + A-(aq)} \qquad \ce{A-(aq) + H+(aq) -> HA(aq)}$

Key Examples

Know the logic

One line each; don't memorize the $K_{sp}$ values:

• $\ce{CaF2}$ (tooth enamel): $\ce{F^-}$ is the conjugate base of weak HF, so acid converts it to HF and erodes enamel — why acidic foods damage teeth.
• $\ce{CaCO3}$ (limestone, shells, antacids): in acid, $\ce{CO3^{2-} -> HCO3^- -> CO2}$ gas, so solubility rises sharply — acid rain dissolves marble; acidic groundwater carves limestone caves.
• $\ce{Ca3(PO4)2}$ (bone): $\ce{PO4^{3-}}$ is tribasic and very reactive with $\ce{H^+}$; bone mineral dissolves more at low pH (osteoclast resorption).
• $\ce{Fe(OH)3}$: essentially insoluble at physiological pH; dissolves in acid ($\ce{+3H+ -> Fe^{3+} + 3H2O}$), which is why gut iron absorption needs reduction to $\ce{Fe^{2+}}$ or chelation by transferrin.

Salts with Acidic Cations

Optional

The reverse case — if the cation is the conjugate acid of a weak base ($\ce{NH4^+}$), raising pH favors dissolution. Rarely tested.

When pH Does Not Affect Solubility

Must know

If the anion is the conjugate base of a strong acid ($\ce{Cl^-}$, $\ce{SO4^{2-}}$, $\ce{NO3^-}$), it has no affinity for $\ce{H^+}$, so pH is irrelevant. $\ce{AgCl}$ is pH-independent; $\ce{CaF2}$ is pH-dependent.

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Quick check: Which salt's solubility is more affected by a decrease in pH: $\ce{AgCl}$ or $\ce{Ag2CO3}$? Explain.

Answer: $\ce{Ag2CO3}$. Carbonate is the conjugate base of weak $\ce{HCO3^-}$, so $\ce{H^+}$ consumes $\ce{CO3^{2-}}$ and pulls dissolution right. Chloride is the conjugate base of strong HCl and is unaffected by pH.

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Common Confusions & Tricks

1. Molar solubility ≠ $K_{sp}$ directly.
For 1:1 $\ce{AgCl}$, $s = \sqrt{K_{sp}}$. For $\ce{PbI2}$ (1:2), $K_{sp} = 4s^3$ so $s = (K_{sp}/4)^{1/3}$. The stoichiometric coefficients become exponents and multipliers — students forget the "4" in $4s^3$.

2. Comparing $K_{sp}$ values only works for the same formula type.
A salt with a larger $K_{sp}$ can have lower molar solubility if its formula produces more ions. For different formula types, calculate $s$ from each $K_{sp}$ and compare.

3. Common-ion effect ↓ solubility; complex ion formation ↑ solubility.
Adding $\ce{Cl^-}$ to $\ce{AgCl}$: solubility ↓. Adding $\ce{NH3}$ to $\ce{AgCl}$: solubility ↑. A passage may mix both.

4. pH affects solubility only with weak-acid anions (or weak-base cations).
$\ce{Cl^-}$, $\ce{NO3^-}$, $\ce{SO4^{2-}}$ → pH irrelevant. $\ce{F^-}$, $\ce{CO3^{2-}}$, $\ce{PO4^{3-}}$, $\ce{OH^-}$, $\ce{S^{2-}}$ → pH matters.

5. Amphoteric hydroxides dissolve in excess base — they don't just neutralize it.
The word "excess" signals complexation, not simple neutralization.

6. $K_{sp}$ expressions never include the solid.
Pure solids and pure liquids are always omitted.

7. Molarity vs. molality.
For dilute aqueous solutions M ≈ m (density ≈ 1 g/mL); they diverge for concentrated or non-aqueous solutions. Molality → colligative properties; molarity → equilibrium.

8. Tollens' reagent / silver mirror test → think $\ce{[Ag(NH3)2]+}$.
This complex is the active oxidizer, used to distinguish aldehydes from ketones.

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Key Equations

Equation	Variables & Use
$M = \dfrac{n_{\text{solute}}}{V_{\text{solution (L)}}}$	Molarity: universal concentration unit for equilibrium.
$M_1 V_1 = M_2 V_2$	Dilution: moles conserved. Solve for any unknown M or V.
$m = \dfrac{n_{\text{solute}}}{m_{\text{solvent (kg)}}}$	Molality: use for colligative properties (not equilibrium).
$\chi_A = \dfrac{n_A}{n_{\text{total}}}$	Mole fraction; appears in Raoult's Law.
$K_{sp} = [\text{M}^{n+}]^a[\text{X}^{m-}]^b$	Solubility product for $\ce{M_aX_b}$. Solid not included.
$K_{sp} = s^2$	Molar solubility for 1:1 salts (e.g., $\ce{AgCl}$).
$K_{sp} = 4s^3$	Molar solubility for 1:2 / 2:1 salts (e.g., $\ce{PbI2}$).
$K_{sp} = 27s^4$	Molar solubility for 1:3 / 3:1 salts (e.g., $\ce{AlF3}$).
$Q = [\text{M}^{n+}]^a[\text{X}^{m-}]^b \text{ (actual)}$	Ion product: $Q > K_{sp}$ → precipitate; $Q < K_{sp}$ → unsaturated.
$K_{\text{net}} = K_{sp} \times K_f$	Net constant for dissolving a salt in a complexing agent.
$\ce{M^{n+}(aq) + n L(aq) <=> [ML_n]^{n+}(aq)}$	Complex ion formation with ligand L; constant is $K_f$.