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

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

Flashcard 8: How does the presence of a common ion affect solubility?

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

Flashcard 9: Identify the $K_{sp}$ expression for $CdS$.

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

Flashcard 10: 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 11: 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 12: 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 13: How does complex ion formation affect solubility?

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

Flashcard 14: 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 15: How does $K_{sp}$ change with temperature for exothermic reactions?

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

Flashcard 16: 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 17: 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 18: 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 19: 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 20: 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 21: 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 22: 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 23: 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 24: Which compound will precipitate first: $AgCl$ or $AgBr$?

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

Flashcard 25: 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 26: 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 27: Identify the effect of temperature on $K_{sp}$ for endothermic dissolution.

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

Flashcard 28: How does $K_{sp}$ change with temperature for exothermic reactions?

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

Flashcard 29: 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 30: 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 31: 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 32: 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.