1 question, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 1
A student measures the molar solubility of four sparingly soluble salts, W, X, Y and Z, in solutions buffered at different pH values at 25°C. The graph summarizes the student's results. Based on the results, which salt is most likely a metal hydroxide such as Mg(OH)₂?
Answer and reasoning
ASalt W A student who thinks adding acid shifts a dissolution equilibrium toward the solid picks the line that rises with pH. H₃O⁺ removes OH⁻, a product of the dissolution, so a hydroxide is more soluble at low pH, not less.
BSalt XCorrect Lowering the pH removes OH⁻ and raises the solubility of such a hydroxide, and raising the pH adds OH⁻, a common ion that lowers it. Only a line that falls steadily across the whole pH range, in basic as well as acidic solutions, shows this behavior.
CSalt Y A student who thinks pH affects a hydroxide's solubility only in acidic solution picks the line that is level above pH 7. Above pH 7 the added OH⁻ is a common ion, so a hydroxide's solubility keeps falling as the pH rises.
DSalt Z A student who thinks a constant Ksp means a constant solubility picks the level line. Ksp is fixed at 25°C, but H₃O⁺ and OH⁻ change [OH⁻], so a hydroxide's solubility changes with pH.
Working No calculation. For a hydroxide such as Mg(OH)₂, M(OH)₂(s) ⇌ M²⁺ + 2OH⁻, H₃O⁺ removes OH⁻ at low pH (more dissolves), and added OH⁻ at high pH is a common ion (less dissolves), so the solubility falls steadily as the pH rises across the whole range. The salt whose line falls steadily from pH 0 to 14 fits.
In preparation: 0 of 1 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
8.11.A.1 pH-dependent solubility Fix
pH-dependent solubility
The solubility of a salt depends on pH when one of its ions is a weak acid, a weak base, or the hydroxide ion, because H₃O⁺ or OH⁻ can react with that ion and change its concentration in the dissolution equilibrium.
Salt with a weakly basic anion
For a salt such as CaF₂, added H₃O⁺ converts the anion to its weak conjugate acid (F⁻ + H₃O⁺ → HF + H₂O). Removing F⁻ shifts CaF₂(s) ⇌ Ca²⁺ + 2F⁻ toward dissolution, so the salt is more soluble at lower pH.
Metal hydroxide solubility and pH
For a hydroxide such as Mg(OH)₂, added H₃O⁺ removes OH⁻ (H₃O⁺ + OH⁻ → 2H₂O), so more solid dissolves at lower pH; added OH⁻ is a common ion that shifts the equilibrium toward the solid, so the hydroxide is less soluble at higher pH.
Salt of a strong acid's anion
The anion of a strong acid, such as Cl⁻ or NO₃⁻, has a negligible tendency to accept a proton, so the solubility of a salt such as AgCl is essentially unaffected by adding a strong acid such as HNO₃.
Le Châtelier's principle applied to solubility
Removing one of the ions of a dissolution equilibrium (for example, by reaction with H₃O⁺) shifts the equilibrium toward dissolution; adding one of its ions shifts it toward the solid. Ksp itself does not change at constant temperature.
Students often think Because Ksp is constant at a given temperature, the solubility of a salt and the concentrations of its ions in a saturated solution cannot change when the pH changes. In fact No. Ksp is constant at a given temperature, but when H₃O⁺ or OH⁻ reacts with one of the ions, that ion's concentration changes and more or less solid dissolves until the ion product again equals Ksp. The solubility changes even though Ksp does not.
Students often think Adding acid to a saturated solution of a hydroxide or of a salt with a weakly basic anion shifts the dissolution equilibrium toward the solid, so less of the salt dissolves. In fact No. The added H₃O⁺ removes OH⁻ (or converts the basic anion to its weak acid), lowering the concentration of a product of the dissolution, so by Le Châtelier's principle the equilibrium shifts toward dissolution.
4 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 4
Saturated solutions of silver acetate, AgCH₃COO, and silver chloride, AgCl, are each in contact with excess solid at 25°C. A small amount of HNO₃(aq) is added to each mixture at constant temperature. For which of the two salts, if either, does the solubility increase, and why?
Answer and reasoning
AAgCl only, because Cl⁻ is the conjugate base of a strong acid and CH₃COO⁻ is not A student who thinks the anion of a strong acid is itself a strong base, reacting readily with H₃O⁺, picks this. The stronger the acid, the weaker its conjugate base: Cl⁻ does not accept a proton in water, so the solubility of AgCl is essentially unchanged, while CH₃COO⁻ is converted to CH₃COOH.
BBoth salts, because H₃O⁺ reacts with the anion of any sparingly soluble salt A student who thinks acid makes every sparingly soluble salt dissolve picks this. H₃O⁺ reacts with an anion that is a weak base, such as CH₃COO⁻, or with OH⁻; Cl⁻ does not react with it.
CNeither salt, because the Ksp of each salt is a constant at this temperature A student who thinks a constant Ksp means a constant solubility picks this. Ksp is unchanged, but H₃O⁺ lowers [CH₃COO⁻], so more AgCH₃COO dissolves until the ion product again equals Ksp.
DAgCH₃COO only, because CH₃COO⁻ is the conjugate base of a weak acid and Cl⁻ is notCorrect CH₃COO⁻ is a weak base: H₃O⁺ converts it to CH₃COOH, which lowers [CH₃COO⁻] and shifts AgCH₃COO(s) ⇌ Ag⁺ + CH₃COO⁻ toward dissolution. Cl⁻, the anion of the strong acid HCl, has a negligible tendency to accept a proton, so the solubility of AgCl is essentially unchanged.
Working No calculation. AgCH₃COO(s) ⇌ Ag⁺(aq) + CH₃COO⁻(aq): CH₃COO⁻ is a weak base, so CH₃COO⁻ + H₃O⁺ → CH₃COOH + H₂O lowers [CH₃COO⁻] and the equilibrium shifts toward dissolution (Le Châtelier's principle). AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq): Cl⁻, the conjugate base of the strong acid HCl, does not react with H₃O⁺, and NO₃⁻ is not an ion of either salt, so the equilibrium does not shift. Ksp of each salt is unchanged.
Solid CaF₂ is in equilibrium with its saturated solution: CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq). HF is a weak acid. A small amount of HNO₃(aq) is added at constant temperature, and solid CaF₂ remains. Which statement correctly describes [Ca²⁺] after equilibrium is re-established, and why?
Answer and reasoning
AIt is higher, because adding acid increases the value of Ksp for CaF₂ A student who thinks adding acid changes Ksp picks this. [Ca²⁺] does increase, but Ksp depends only on temperature; the solubility rises because H₃O⁺ removes F⁻.
BIt is unchanged, because Ksp for CaF₂ is a constant at this temperature A student who thinks a constant Ksp means fixed ion concentrations picks this. Ksp is constant, but when [F⁻] falls the ion product drops below Ksp, so more CaF₂ dissolves and [Ca²⁺] rises.
CIt is higher, because H₃O⁺ converts F⁻ to HF, so that more CaF₂ dissolvesCorrect F⁻ is the conjugate base of the weak acid HF, so added H₃O⁺ converts F⁻ to HF. Removing F⁻ shifts CaF₂(s) ⇌ Ca²⁺ + 2F⁻ toward dissolution, so [Ca²⁺] increases while Ksp stays the same.
DIt is lower, because the added H₃O⁺ shifts the equilibrium toward solid CaF₂ A student who thinks adding acid pushes a dissolution equilibrium toward the solid picks this. H₃O⁺ removes F⁻, a product, so the equilibrium shifts toward dissolution, not toward the solid.
Working No calculation. F⁻ is a weak base: F⁻ + H₃O⁺ → HF + H₂O lowers [F⁻]. Q = [Ca²⁺][F⁻]² < Ksp, so more CaF₂ dissolves (Le Châtelier's principle), raising [Ca²⁺]. Ksp is unchanged at constant temperature.
The box labeled Before represents a small volume of a saturated solution of a metal hydroxide, M(OH)₂, in contact with some of the solid. A small amount of HNO₃(aq) is added, and equilibrium is re-established at the same temperature with solid still present. Which numbered diagram best represents the same volume afterward? Water molecules, H₃O⁺ ions and NO₃⁻ ions are not shown, and the diagrams show the direction of each change, not exact amounts.
Answer and reasoning
ADiagram 1Correct H₃O⁺ removes OH⁻, so the equilibrium shifts toward dissolution: some solid dissolves (fewer blocks), the number of M²⁺ ions rises, and OH⁻ stays below its original level. The box with four M²⁺, two OH⁻ and four blocks of solid shows all three changes.
BDiagram 2 A student who thinks adding acid shifts a dissolution equilibrium toward the solid picks this box, with fewer M²⁺ ions and more solid. Removing OH⁻, a product, shifts the equilibrium toward dissolution.
CDiagram 3 A student who thinks a constant Ksp means that nothing in a saturated solution can change picks this box, identical to Before. Ksp is unchanged, but the added H₃O⁺ removes OH⁻, so more solid dissolves.
DDiagram 4 A student who thinks the solid takes no part in the equilibrium picks this box, in which OH⁻ has been removed but the solid and the M²⁺ ions are unchanged. The lower [OH⁻] makes more solid dissolve, increasing [M²⁺].
Working No calculation. H₃O⁺ + OH⁻ → 2H₂O lowers [OH⁻]; the equilibrium M(OH)₂(s) ⇌ M²⁺ + 2OH⁻ shifts toward dissolution, so some solid dissolves: less solid, more M²⁺, and [OH⁻] lower than before (the ion product returns to Ksp with more M²⁺ and less OH⁻).
Solid Mg(OH)₂ is in equilibrium with its saturated solution at 25°C. A small amount of solid NaOH is dissolved in the solution, and the temperature is kept at 25°C. How does the solubility of Mg(OH)₂ change, and why?
Answer and reasoning
AIt increases, because a basic solution dissolves a basic solid more readily A student who applies 'like dissolves like' to bases picks this. Added OH⁻ is a product ion of the dissolution, so it shifts the equilibrium toward the solid and lowers the solubility.
BIt is unchanged, because Ksp for Mg(OH)₂ is constant at this temperature A student who thinks a constant Ksp means a constant solubility picks this. Ksp is unchanged, but the larger [OH⁻] makes the ion product exceed Ksp, so Mg(OH)₂ precipitates until it again equals Ksp.
CIt decreases, because added OH⁻ shifts the equilibrium toward the solidCorrect OH⁻ is one of the ions of Mg(OH)₂(s) ⇌ Mg²⁺ + 2OH⁻. Adding it is a common-ion stress that shifts the equilibrium toward the solid, so less Mg(OH)₂ dissolves at the higher pH.
DIt is unchanged, because no H₃O⁺ is added to react with the OH⁻ ions A student who thinks pH affects a hydroxide's solubility only when H₃O⁺ is present picks this. Raising the pH with OH⁻ also changes the solubility, through the common-ion effect.
Working No calculation. Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq). Added OH⁻ is a common ion; Q > Ksp, so the equilibrium shifts toward the solid (Le Châtelier's principle): the solubility decreases. Ksp is unchanged.
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
