AP Chemistry Flashcards: Introduction To Solubility Equilibria

What this deck covers

This deck focuses on Introduction To Solubility Equilibria, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Identify the $K_{sp}$ expression for $BaSO_4$.

Answer: $K_{sp} = [Ba^{2+}][SO_4^{2-}]$. 1:1 stoichiometry gives simple product of ion concentrations.

Flashcard 2: Identify the $K_{sp}$ expression for $CuS$.

Answer: $K_{sp} = [Cu^{2+}][S^{2-}]$. 1:1 stoichiometry for copper(II) sulfide dissolution.

Flashcard 3: Identify the $K_{sp}$ expression for $PbSO_4$.

Answer: $K_{sp} = [Pb^{2+}][SO_4^{2-}]$. 1:1 stoichiometry for lead(II) sulfate dissolution.

Flashcard 4: What is the $K_{sp}$ expression for $Fe(OH)_3$?

Answer: $K_{sp} = [Fe^{3+}][OH^-]^3$. Stoichiometry: 1 $Fe^{3+}$ and 3 $OH^-$ ions per formula unit.

Flashcard 5: How does $pH$ affect solubility of compounds containing basic anions?

Answer: Increases solubility at lower $pH$. Acidic conditions protonate basic anions, reducing their concentration.

Flashcard 6: What is the unit of $K_{sp}$ for $CaF_2$?

Answer: mol^3/L^3. Units from $[Ca^{2+}][F^-]^2$ concentration terms.

Flashcard 7: How does the presence of a common ion affect solubility?

Answer: Decreases solubility. Le Chatelier's principle: excess ion shifts equilibrium toward solid.

Flashcard 8: Identify the $K_{sp}$ expression for $CdS$.

Answer: $K_{sp} = [Cd^{2+}][S^{2-}]$. 1:1 stoichiometry for cadmium sulfide dissolution.

Flashcard 9: Which compound is more soluble: $CaF_2$ or $SrF_2$?

Answer: $SrF_2$ (higher $K_{sp}$ value). Compare $K_{sp}$ values directly for same formula type.

Flashcard 10: Calculate $K_{sp}$ for $Ag_2S$ with $[Ag^+] = 2x$ and $[S^{2-}] = x$.

Answer: $K_{sp} = 4x^3$. For $Ag_2S$: $K_{sp} = [Ag^+]^2[S^{2-}] = (2x)^2(x) = 4x^3$.

Flashcard 11: Identify the $K_{sp}$ expression for $Ca_3(PO_4)_2$.

Answer: $K_{sp} = [Ca^{2+}]^3[PO_4^{3-}]^2$. Stoichiometry: 3 $Ca^{2+}$ and 2 $PO_4^{3-}$ per formula unit.

Flashcard 12: How does complex ion formation affect solubility?

Answer: Increases solubility. Complex formation removes free metal ions from solution.

Flashcard 13: Which has higher solubility: $BaSO_4$ or $SrSO_4$?

Answer: $SrSO_4$ (higher $K_{sp}$ value). Compare $K_{sp}$ values; larger value indicates higher solubility.

Flashcard 14: State the formula to calculate $Q_{sp}$.

Answer: $Q_{sp} = [cation]^m[anion]^n$. Same form as $K_{sp}$ but uses current concentrations, not equilibrium.

Flashcard 15: Determine $K_{sp}$ for $Ag_2S$ if $[Ag^+] = 2x$ and $[S^{2-}] = x$ at equilibrium.

Answer: $K_{sp} = 4x^3$. Substitute equilibrium concentrations: $(2x)^2(x) = 4x^3$.

Flashcard 16: What is the relationship between $K_{sp}$ and $Q_{sp}$?

Answer: Precipitation occurs if $Q_{sp} > K_{sp}$. $Q_{sp}$ compares current ion concentrations to equilibrium values.

Flashcard 17: What is the effect of adding $NH_3$ to $AgCl$ solubility?

Answer: Increases due to complexation. $NH_3$ forms $[Ag(NH_3)_2]^+$ complex, removing $Ag^+$ ions.

Flashcard 18: Explain the role of $pH$ in the solubility of $CaCO_3$.

Answer: Solubility increases in acidic conditions. Acid protonates $CO_3^{2-}$ to form $HCO_3^-$ and $H_2CO_3$.

Flashcard 19: What does a large $K_{sp}$ value indicate about solubility?

Answer: Compound is more soluble. Higher $K_{sp}$ means more ions can dissolve at equilibrium.

Flashcard 20: Which compound will precipitate first: $AgCl$ or $AgBr$?

Answer: $AgBr$ (lower $K_{sp}$ value). Lower $K_{sp}$ means less soluble, so precipitates first.

Flashcard 21: What is the effect of $pH$ on the solubility of $Mg(OH)_2$?

Answer: Solubility decreases at higher $pH$. Higher $pH$ increases $[OH^-]$, shifting equilibrium toward solid.

Flashcard 22: What is the $K_{sp}$ expression for $Al(OH)_3$?

Answer: $K_{sp} = [Al^{3+}][OH^-]^3$. Aluminum hydroxide: 1 $Al^{3+}$ and 3 $OH^-$ ions.

Flashcard 23: Identify the effect of temperature on $K_{sp}$ for endothermic dissolution.

Answer: $K_{sp}$ increases with temperature. Higher temperature favors endothermic dissolution process.

Flashcard 24: How does $K_{sp}$ change with temperature for exothermic reactions?

Answer: $K_{sp}$ decreases with temperature. Lower temperature favors exothermic dissolution process.

Flashcard 25: Identify the $K_{sp}$ expression for $Hg_2Cl_2$.

Answer: $K_{sp} = [Hg_2^{2+}][Cl^-]^2$. Dimeric mercury(I) ion has +2 charge with two chlorides.

Flashcard 26: What is $K_{sp}$ for $PbCl_2$ given solubility $s$?

Answer: $K_{sp} = 4s^3$. For $PbCl_2$: $K_{sp} = s(2s)^2 = 4s^3$.

Flashcard 27: Identify the common ion effect in solubility equilibria.

Answer: Reduction in solubility due to a common ion. Adding an ion already present shifts equilibrium toward solid formation.

Flashcard 28: What is the solubility product constant ($K_{sp}$) expression for $AgCl$?

Answer: $K_{sp} = [Ag^+][Cl^-]$. Product of ion concentrations at equilibrium for the saturated solution.