Define solubility-product constant ( K* sp ) .

The equilibrium constant for the dissolution of a sparingly soluble ionic compound. It equals the product of the ion concentrations in a saturated solution, each raised to the power of its stoichiometric coefficient in the dissolution equation.

Why is K sp only useful for sparingly soluble (slightly soluble) salts, and not for highly soluble salts such as NaCl?

For highly soluble salts, ion–ion interactions and activity effects become significant at the high concentrations involved, so the simple product of concentrations no longer accurately represents the equilibrium. K sp is derived assuming ideal dilute solution behaviour, which only holds at the very low ion concentrations found in slightly soluble salt solutions.

True or False? For ionic compounds that dissolve to give the same number of ions, a smaller K sp value indicates a lower molar solubility.

True. A smaller K sp means fewer ions are present in a saturated solution, so less of the compound dissolves. This comparison only holds when the ion stoichiometry is the same — for example, when comparing two 1:1 salts like AgCl and BaSO 4 .

How do you use an ICE table to calculate K sp from the molar solubility of a salt?

Set up an ICE table with 0 M for each ion initially and let the molar solubility = s ; use stoichiometric coefficients to express equilibrium concentrations in terms of s (e.g. for MX 2 : [M 2+ ] = s , [X - ] = 2 s ). Substitute into the K sp expression and calculate.

Solubility rules indicate that all Group 1 salts are soluble. Why does this mean K sp cannot be applied to sodium chloride (NaCl)?

K sp applies only to sparingly soluble salts where a genuine dynamic equilibrium exists between undissolved solid and a very dilute solution of its ions. NaCl is highly soluble — it dissolves completely under normal conditions and no such solid–ion equilibrium is established.

True or False? If Q > K sp for a dissolved salt, a precipitate will form.

True. When Q exceeds K sp , the ion product is greater than the solubility product, so the solution is supersaturated. The system shifts toward the reactants (the solid), and the excess salt precipitates until Q = K sp .

For the dissolution equilibrium MgCl 2 (s) ⇌ Mg 2+ (aq) + 2Cl - (aq), the solubility-product expression is K sp = .......... .

For the dissolution equilibrium MgCl 2 (s) ⇌ Mg 2+ (aq) + 2Cl - (aq), the solubility-product expression is K sp = [Mg 2+ (aq)][Cl - (aq)] 2 .

Define common-ion effect .

The decrease in solubility of a sparingly soluble ionic compound that occurs when a soluble salt sharing one of its ions is added to the solution. The added ion shifts the dissolution equilibrium toward the solid (reactants), causing precipitation.

Why does dissolving KCl in a saturated solution of AgCl cause AgCl to precipitate?

KCl dissociates to give additional Cl - ions, raising the ion product [Ag + ][Cl - ] above K sp for AgCl. To re-establish equilibrium, the system shifts in the reverse direction — AgCl precipitates — until the ion product equals K sp again.

True or False? When two solutions are mixed in equal volumes, the concentration of every ion in the mixture is halved compared to its original solution.

False. Each ion is diluted according to C 1 V 1 = C 2 V 2 . When equal volumes are mixed, each individual ion concentration is halved. However, a common ion present in both solutions is not simply halved — its total concentration in the mixture is the sum of its diluted contributions from each source.

How do you calculate the concentration of a common ion after mixing two solutions of equal volume, one of which already contains that ion?

Apply C 1 V 1 = C 2 V 2 separately to each source of the ion; each solution is diluted into the combined volume. The total concentration of the common ion in the mixture equals the sum of its diluted contributions from each solution.

True or False? Adding a common ion to a saturated solution increases the molar solubility of the sparingly soluble salt.

False. Adding a common ion increases the ion product Q above K sp , shifting the equilibrium toward the solid. The salt precipitates, meaning its molar solubility decreases — less of the salt can remain dissolved in the presence of the common ion.

Adding a soluble salt that shares an ion with a sparingly soluble compound .......... the solubility of that compound, because the added ion causes the ion product Q to exceed .......... and the equilibrium shifts toward the solid.

Adding a soluble salt that shares an ion with a sparingly soluble compound decreases the solubility of that compound, because the added ion causes the ion product Q to exceed K sp and the equilibrium shifts toward the solid.

AP®ChemistryCollege BoardFlashcardsUnit 7: EquilibriumSolubility Equilibria

Solubility Equilibria (College Board AP® Chemistry): Flashcards

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Define solubility-product constant (K*_sp).

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Define solubility-product constant (K*_sp).
The equilibrium constant for the dissolution of a sparingly soluble ionic compound. It equals the product of the ion concentrations in a saturated solution, each raised to the power of its stoichiometric coefficient in the dissolution equation.
Why is K_sp only useful for sparingly soluble (slightly soluble) salts, and not for highly soluble salts such as NaCl?
For highly soluble salts, ion–ion interactions and activity effects become significant at the high concentrations involved, so the simple product of concentrations no longer accurately represents the equilibrium. K_sp is derived assuming ideal dilute solution behaviour, which only holds at the very low ion concentrations found in slightly soluble salt solutions.
True or False?
For ionic compounds that dissolve to give the same number of ions, a smaller K_sp value indicates a lower molar solubility.
True.
A smaller K_sp means fewer ions are present in a saturated solution, so less of the compound dissolves. This comparison only holds when the ion stoichiometry is the same — for example, when comparing two 1:1 salts like AgCl and BaSO₄.
How do you use an ICE table to calculate K_sp from the molar solubility of a salt?
Set up an ICE table with 0 M for each ion initially and let the molar solubility = s; use stoichiometric coefficients to express equilibrium concentrations in terms of s (e.g. for MX₂: [M²⁺] = s, [X^-] = 2s). Substitute into the K_sp expression and calculate.
Solubility rules indicate that all Group 1 salts are soluble. Why does this mean K_sp cannot be applied to sodium chloride (NaCl)?
K_sp applies only to sparingly soluble salts where a genuine dynamic equilibrium exists between undissolved solid and a very dilute solution of its ions. NaCl is highly soluble — it dissolves completely under normal conditions and no such solid–ion equilibrium is established.
True or False?
If Q > K_sp for a dissolved salt, a precipitate will form.
True.
When Q exceeds K_sp, the ion product is greater than the solubility product, so the solution is supersaturated. The system shifts toward the reactants (the solid), and the excess salt precipitates until Q = K_sp.
For the dissolution equilibrium MgCl₂ (s) ⇌ Mg²⁺ (aq) + 2Cl^- (aq), the solubility-product expression is K_sp = .......... .
For the dissolution equilibrium MgCl₂ (s) ⇌ Mg²⁺ (aq) + 2Cl^- (aq), the solubility-product expression is K_sp = [Mg²⁺ (aq)][Cl^- (aq)]².
Define common-ion effect.
The decrease in solubility of a sparingly soluble ionic compound that occurs when a soluble salt sharing one of its ions is added to the solution. The added ion shifts the dissolution equilibrium toward the solid (reactants), causing precipitation.
Why does dissolving KCl in a saturated solution of AgCl cause AgCl to precipitate?
KCl dissociates to give additional Cl^- ions, raising the ion product [Ag⁺][Cl^-] above K_sp for AgCl. To re-establish equilibrium, the system shifts in the reverse direction — AgCl precipitates — until the ion product equals K_sp again.
True or False?
When two solutions are mixed in equal volumes, the concentration of every ion in the mixture is halved compared to its original solution.
False.
Each ion is diluted according to C₁V₁ = C₂V₂. When equal volumes are mixed, each individual ion concentration is halved. However, a common ion present in both solutions is not simply halved — its total concentration in the mixture is the sum of its diluted contributions from each source.
How do you calculate the concentration of a common ion after mixing two solutions of equal volume, one of which already contains that ion?
Apply C₁V₁ = C₂V₂ separately to each source of the ion; each solution is diluted into the combined volume. The total concentration of the common ion in the mixture equals the sum of its diluted contributions from each solution.
True or False?
Adding a common ion to a saturated solution increases the molar solubility of the sparingly soluble salt.
False.
Adding a common ion increases the ion product Q above K_sp, shifting the equilibrium toward the solid. The salt precipitates, meaning its molar solubility decreases — less of the salt can remain dissolved in the presence of the common ion.
Adding a soluble salt that shares an ion with a sparingly soluble compound .......... the solubility of that compound, because the added ion causes the ion product Q to exceed .......... and the equilibrium shifts toward the solid.
Adding a soluble salt that shares an ion with a sparingly soluble compound decreases the solubility of that compound, because the added ion causes the ion product Q to exceed K_sp and the equilibrium shifts toward the solid.

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