This question tests the understanding of how pH affects the solubility of salts with anions that are conjugate bases of weak acids. In Beaker 2 with 0.10 M HCl, the high [H+] from the acid reacts with $CO3^2$- ions produced by the dissolution of CaCO3 to form HCO3- and H2CO3. This reaction lowers the $[CO3^2$-] in solution, causing the ion product Q to be less than Ksp and shifting the equilibrium toward more dissolution according to Le Chatelier's principle. Therefore, the solubility of CaCO3 increases in the acidic solution compared to pure water. A tempting distractor is choice B, which misapplies the common-ion effect by claiming acid increases $[CO3^2$-], but this ignores the acid-base reaction that actually decreases $[CO3^2$-]. To predict pH effects on solubility, determine if the anion can be protonated by checking if it is the conjugate base of a weak acid, and apply Le Chatelier's principle to see how it shifts the equilibrium.

This question tests understanding of how the strength of the conjugate acid affects pH-dependent solubility. BaSO₄ dissolves to produce Ba²⁺ and SO₄²⁻, where SO₄²⁻ is the conjugate base of HSO₄⁻, which is a strong acid (Ka ≈ 10⁻²). Because HSO₄⁻ is a strong acid, SO₄²⁻ is an extremely weak base that is only slightly protonated even in strongly acidic solution, meaning the concentration of SO₄²⁻ remains essentially unchanged and the solubility is not significantly affected by pH. Choice A incorrectly suggests complete conversion to H₂SO₄, but the first protonation to HSO₄⁻ is already minimal. When the anion is the conjugate base of a strong or very strong acid, pH changes have little effect on solubility.

This question tests understanding of pH effects on carbonate salt solubility. Barium carbonate dissolves according to BaCO₃(s) ⇌ Ba²⁺(aq) + CO₃²⁻(aq), where CO₃²⁻ is the conjugate base of the weak acid HCO₃⁻. In Solution II (0.10 M HCl), H⁺ ions react with CO₃²⁻ to form HCO₃⁻ and H₂CO₃, removing CO₃²⁻ from the equilibrium: CO₃²⁻ + H⁺ ⇌ HCO₃⁻ and HCO₃⁻ + H⁺ ⇌ H₂CO₃. By Le Chatelier's principle, reducing [CO₃²⁻] shifts the dissolution equilibrium to the right, increasing BaCO₃ solubility. Choice A incorrectly claims Cl⁻ is a common ion with BaCO₃, but Cl⁻ does not appear in the BaCO₃ dissolution equilibrium. For carbonate salts, acidic conditions always increase solubility by converting CO₃²⁻ to HCO₃⁻ and H₂CO₃.

This question tests understanding of how pH affects the solubility of metal sulfides. Iron(II) sulfide dissolves according to FeS(s) ⇌ Fe²⁺(aq) + S²⁻(aq), where S²⁻ is a very basic anion (conjugate base of the weak acid HS⁻). In acidic solution (Solution 2 with HCl), H⁺ ions react with S²⁻ in two steps: S²⁻ + H⁺ ⇌ HS⁻ and HS⁻ + H⁺ ⇌ H₂S, effectively removing S²⁻ from the equilibrium. By Le Chatelier's principle, reducing [S²⁻] shifts the dissolution equilibrium to the right, causing more FeS to dissolve. Choice A incorrectly identifies Cl⁻ as a common ion with FeS, but Cl⁻ does not appear in the FeS dissolution equilibrium. When dealing with metal sulfides, remember that acidic conditions dramatically increase solubility by converting S²⁻ to HS⁻ and H₂S.

This question tests understanding of how pH affects metal hydroxide solubility. Zinc hydroxide dissolves according to Zn(OH)₂(s) ⇌ Zn²⁺(aq) + 2OH⁻(aq). In Beaker 2 (0.10 M HNO₃, pH ≈ 1), the high concentration of H⁺ ions reacts with OH⁻ to form water: H⁺ + OH⁻ → H₂O. This neutralization reaction removes OH⁻ from the dissolution equilibrium, and by Le Chatelier's principle, the equilibrium shifts to the right to produce more ions, increasing Zn(OH)₂ solubility. Choice D incorrectly states that lowering pH decreases metal hydroxide solubility by forming more OH⁻, which is backwards—acidic conditions consume OH⁻, not produce it. For metal hydroxides, acidic conditions increase solubility by neutralizing OH⁻ ions to water.

This question tests understanding of the common ion effect on metal hydroxide solubility. Aluminum hydroxide dissolves according to Al(OH)₃(s) ⇌ Al³⁺(aq) + 3OH⁻(aq). Beaker Y contains 0.10 M NaOH, which provides a high concentration of OH⁻ ions (common ion). According to Le Chatelier's principle, adding OH⁻ to the equilibrium shifts the reaction to the left, favoring solid Al(OH)₃ formation and decreasing solubility. The solubility product expression Ksp = [Al³⁺][OH⁻]³ shows that when [OH⁻] increases, [Al³⁺] must decrease dramatically (cubic relationship) to maintain constant Ksp. Choice D incorrectly claims higher pH always increases metal hydroxide solubility, but the common ion effect causes the opposite. When a solution already contains hydroxide ions, the solubility of metal hydroxides decreases due to the common ion effect.

This question tests the effect of low pH on the solubility of fluoride salts through formation of weak acid HF. At pH 2, the high [H+] protonates F- ions from CaF2 dissolution to form HF, reducing [F-] in solution. This decrease in [F-] makes Q < Ksp, driving the equilibrium toward greater dissolution per Le Chatelier's principle. As a result, the solubility of CaF2 increases in the acidic solution compared to pure water. A tempting distractor is choice B, which asserts fluoride cannot react with H+ because it's from a strong acid, but this misconceives HF as weak, allowing protonation. To determine pH effects on solubility, identify if the anion is a conjugate base of a weak acid and evaluate how protonation shifts the dissolution equilibrium.

This question tests understanding of how pH affects the solubility of salts containing basic anions. When CaCO₃ dissolves, it produces Ca²⁺ and CO₃²⁻ ions, where CO₃²⁻ is a basic anion that can accept protons. In acidic solution (low pH), the high concentration of H⁺ ions reacts with CO₃²⁻ to form HCO₃⁻ and H₂CO₃, effectively removing CO₃²⁻ from the solution. According to Le Chatelier's principle, removing a product (CO₃²⁻) shifts the dissolution equilibrium CaCO₃(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq) to the right, increasing solubility. Choice B incorrectly suggests Cl⁻ is a common ion with CaCO₃, but CaCO₃ contains no chloride ions. To solve pH-solubility problems, identify whether the anion is basic (can accept H⁺) or neutral, then apply Le Chatelier's principle to predict how pH changes affect the dissolution equilibrium.

This question tests understanding of the common ion effect with metal hydroxides at high pH. Cu(OH)₂ dissolves according to: Cu(OH)₂(s) ⇌ Cu²⁺(aq) + 2OH⁻(aq). At pH 10, the solution has a relatively high [OH⁻] concentration (10⁻⁴ M) compared to pure water. These OH⁻ ions are a common ion with the dissolution products, so according to Le Chatelier's principle, the increased [OH⁻] shifts the equilibrium to the left, decreasing the solubility of Cu(OH)₂. Choice C incorrectly suggests that buffers prevent equilibrium shifts, but buffers only maintain constant pH—they don't prevent the common ion effect. For metal hydroxides, high pH (high [OH⁻]) always decreases solubility through the common ion effect, while low pH increases solubility by removing OH⁻ through neutralization.

This question tests understanding of how weak acids affect the solubility of salts containing their conjugate bases. Silver acetate dissolves as: AgC₂H₃O₂(s) ⇌ Ag⁺(aq) + C₂H₃O₂⁻(aq). Acetic acid (HC₂H₃O₂) is a weak acid that partially dissociates, providing H⁺ ions that can protonate the acetate ion (C₂H₃O₂⁻) to form more HC₂H₃O₂. This removes C₂H₃O₂⁻ from the solution, and by Le Chatelier's principle, the dissolution equilibrium shifts right to produce more ions, increasing solubility. Choice D incorrectly claims that acetic acid supplies acetate ions, but weak acids actually consume their conjugate bases through protonation rather than supplying them. When a weak acid is added to a solution containing its conjugate base as part of a sparingly soluble salt, the acid will increase the salt's solubility by removing the anion through protonation.