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AP Chemistry

AP Chemistry Help: Ph And Solubility

Review real example questions for Ph And Solubility in AP Chemistry.

Question 1

A student compares the solubility of solid calcium carbonate, CaCO3(s)\text{CaCO}_3(s)CaCO3​(s), in two beakers at the same temperature: Beaker 1 contains pure water, and Beaker 2 contains 0.10 M0.10\,\text{M}0.10M HCl (pH ≈1\approx 1≈1). Which statement best describes how the solubility changes in Beaker 2 and why?

  1. The solubility stays the same because HCl acts as a buffer that keeps [CO32−][\text{CO}_3^{2-}][CO32−​] constant in solution.
  2. The solubility decreases because adding acid increases [CO32−][\text{CO}_3^{2-}][CO32−​] through the common-ion effect, shifting dissolution backward.
  3. The solubility stays the same because KspK_{sp}Ksp​ is constant and pH cannot affect equilibrium positions for ionic solids.
  4. The solubility decreases because HCl is a strong electrolyte that screens charges and forces ions to recombine into the solid.
  5. The solubility increases because H+\text{H}^+H+ reacts with CO32−\text{CO}_3^{2-}CO32−​ to form HCO3−\text{HCO}_3^-HCO3−​ and H2CO3\text{H}_2\text{CO}_3H2​CO3​, lowering [CO32−][\text{CO}_3^{2-}][CO32−​] and shifting dissolution forward.
Explanation: 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.

Question 2

Solid barium sulfate, BaSO4(s)\text{BaSO}_4(s)BaSO4​(s), is added to two solutions at the same temperature: one is pure water and the other is adjusted to pH 2 with strong acid. Which statement best describes the solubility change in the acidic solution and why?

  1. The solubility increases greatly because H+\text{H}^+H+ converts SO42−\text{SO}_4^{2-}SO42−​ completely into H2SO4\text{H}_2\text{SO}_4H2​SO4​, removing sulfate.
  2. The solubility decreases because H+\text{H}^+H+ is a common ion with Ba2+\text{Ba}^{2+}Ba2+ and shifts dissolution left.
  3. The solubility is approximately unchanged because SO42−\text{SO}_4^{2-}SO42−​ is the conjugate base of a strong acid and is only weakly protonated.
  4. The solubility decreases because acids always reduce solubility by lowering the value of KspK_{sp}Ksp​.
  5. The solubility increases because the acid buffers the solution and therefore prevents precipitation of BaSO4\text{BaSO}_4BaSO4​.
Explanation: 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.

Question 3

A student investigates the solubility of solid barium carbonate, BaCO3(s)\text{BaCO}_3(s)BaCO3​(s), in two solutions at the same temperature. Solution I is 0.10 M0.10\ \text{M}0.10 M NaCl (approximately neutral). Solution II is 0.10 M0.10\ \text{M}0.10 M HCl (pH ≈1\approx 1≈1). Compared with Solution I, how does the solubility of BaCO3(s)\text{BaCO}_3(s)BaCO3​(s) change in Solution II, and why?

  1. The solubility decreases because Cl−\text{Cl}^-Cl− is a common ion with BaCO3\text{BaCO}_3BaCO3​ and shifts equilibrium toward the solid.
  2. The solubility increases because H+\text{H}^+H+ consumes CO32−\text{CO}_3^{2-}CO32−​ to form HCO3−\text{HCO}_3^-HCO3−​/H2CO3\text{H}_2\text{CO}_3H2​CO3​, lowering CO32−\text{CO}_3^{2-}CO32−​ and shifting dissolution toward products.
  3. The solubility stays the same because both solutions contain chloride, so the effect of pH cancels out.
  4. The solubility decreases because adding acid adds more ions, and more ions always means less dissolves.
  5. The solubility stays the same because KspK_{sp}Ksp​ depends only on temperature, so changing pH cannot change solubility.
Explanation: 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.

Question 4

Solid iron(II) sulfide, FeS(s)\text{FeS}(s)FeS(s), is placed into two solutions at the same temperature. Solution 1 is pure water. Solution 2 is 0.10 M0.10\ \text{M}0.10 M HCl (pH ≈1\approx 1≈1). Compared with Solution 1, what happens to the solubility of FeS(s)\text{FeS}(s)FeS(s) in Solution 2, and why?

  1. The solubility decreases because Cl−\text{Cl}^-Cl− is a common ion that suppresses dissolution of FeS(s)\text{FeS}(s)FeS(s).
  2. The solubility increases because H+\text{H}^+H+ reacts with S2−\text{S}^{2-}S2− to form HS−^-− and H2_22​S, reducing S2−\text{S}^{2-}S2− and pulling dissolution forward.
  3. The solubility stays the same because acids only change reaction rates, not equilibrium solubility.
  4. The solubility decreases because lower pH increases ionic strength, which always decreases solubility.
  5. The solubility stays the same because HCl is a buffer that maintains constant S2−\text{S}^{2-}S2− concentration.
Explanation: 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.

Question 5

Solid zinc hydroxide, Zn(OH)2(s)\text{Zn(OH)}_2(s)Zn(OH)2​(s), is added to two beakers at the same temperature. Beaker 1 contains pure water. Beaker 2 contains 0.10 M0.10\ \text{M}0.10 M HNO3_33​ (pH ≈1\approx 1≈1). Compared with Beaker 1, what happens to the solubility of Zn(OH)2(s)\text{Zn(OH)}_2(s)Zn(OH)2​(s) in Beaker 2, and why?

  1. The solubility stays the same because nitrate does not appear in the KspK_{sp}Ksp​ expression for Zn(OH)2\text{Zn(OH)}_2Zn(OH)2​.
  2. The solubility decreases because NO3−\text{NO}_3^-NO3−​ is a common ion that shifts dissolution toward the solid.
  3. The solubility increases because H+\text{H}^+H+ neutralizes OH−\text{OH}^-OH− to form water, reducing OH−\text{OH}^-OH− and shifting dissolution toward ions.
  4. The solubility decreases because lowering pH always decreases the solubility of metal hydroxides by forming more OH−\text{OH}^-OH−.
  5. The solubility stays the same because strong acids act as buffers that keep OH−\text{OH}^-OH− constant.
Explanation: 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.

Question 6

A student adds solid aluminum hydroxide, Al(OH)3(s)\text{Al(OH)}_3(s)Al(OH)3​(s), to two beakers at the same temperature. Beaker X contains pure water. Beaker Y contains 0.10 M0.10\ \text{M}0.10 M NaOH (pH ≈13\approx 13≈13). Ignoring any complex-ion formation, compared with Beaker X, what happens to the solubility of Al(OH)3(s)\text{Al(OH)}_3(s)Al(OH)3​(s) in Beaker Y, and why?

  1. The solubility decreases because added OH−\text{OH}^-OH− is a common ion that shifts the dissolution equilibrium toward the solid.
  2. The solubility increases because added Na+\text{Na}^+Na+ reacts with OH−\text{OH}^-OH− to form NaOH(aq), removing OH−\text{OH}^-OH− from solution.
  3. The solubility stays the same because bases do not affect solubility equilibria, only acids do.
  4. The solubility increases because higher pH always increases the solubility of metal hydroxides.
  5. The solubility stays the same because NaOH is a buffer that holds the Al3+\text{Al}^{3+}Al3+ concentration constant.
Explanation: 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.

Question 7

A student compares the solubility of CaF2(s)\text{CaF}_2(s)CaF2​(s) in pure water versus a solution at pH 222 (adjusted with a strong acid). Which choice best describes how the solubility changes at pH 222 and why?

  1. The solubility decreases because H+\text{H}^+H+ is a common ion with CaF2\text{CaF}_2CaF2​ and suppresses dissolution by the common-ion effect.
  2. The solubility stays the same because fluoride is the conjugate base of a strong acid and therefore cannot react with H+\text{H}^+H+.
  3. The solubility increases because H+\text{H}^+H+ reacts with F−\text{F}^-F− to form HF, lowering [F−][\text{F}^-][F−] and shifting dissolution toward products.
  4. The solubility decreases because strong acids always decrease the solubility of ionic solids by increasing ionic strength.
  5. The solubility stays the same because changing pH affects only the rate of dissolving, not the equilibrium solubility.
Explanation: 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.

Question 8

A student compares the solubility of solid calcium carbonate, CaCO3(s)\text{CaCO}_3(s)CaCO3​(s), in two beakers at the same temperature: Beaker 1 contains pure water (about pH 7), and Beaker 2 contains 0.10 M HCl (pH about 1). Which statement best describes how lowering the pH affects the solubility of CaCO3(s)\text{CaCO}_3(s)CaCO3​(s) and why?

  1. Solubility increases because H+\text{H}^+H+ reacts with CO32−\text{CO}_3^{2-}CO32−​ to form HCO3−\text{HCO}_3^-HCO3−​ and H2CO3\text{H}_2\text{CO}_3H2​CO3​, reducing [CO32−][\text{CO}_3^{2-}][CO32−​] and shifting dissolution forward.
  2. Solubility decreases because added Cl−\text{Cl}^-Cl− is a common ion with CaCO3\text{CaCO}_3CaCO3​ and shifts the equilibrium toward the solid.
  3. Solubility is unchanged because KspK_{sp}Ksp​ depends only on temperature and pH cannot affect any equilibrium concentrations.
  4. Solubility decreases because a lower pH forces carbonate to remain as CO32−\text{CO}_3^{2-}CO32−​, increasing [CO32−][\text{CO}_3^{2-}][CO32−​] and shifting precipitation.
  5. Solubility is unchanged because strong acids act as buffers and keep the carbonate equilibrium from shifting.
Explanation: 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.

Question 9

Solid copper(II) hydroxide, Cu(OH)2(s)\text{Cu(OH)}_2(s)Cu(OH)2​(s), is added to two beakers at the same temperature: Beaker 1 contains pure water, and Beaker 2 contains a solution buffered at pH 10 (so [H+][\text{H}^+][H+] is kept very low and [OH−][\text{OH}^-][OH−] is relatively high). Compared with Beaker 1, what happens to the solubility of Cu(OH)2(s)\text{Cu(OH)}_2(s)Cu(OH)2​(s) in the pH 10 buffer, and why?

  1. Solubility increases because buffers consume Cu2+\text{Cu}^{2+}Cu2+, removing it from solution and pulling dissolution forward.
  2. Solubility decreases because the higher [OH−][\text{OH}^-][OH−] acts as a common ion and shifts Cu(OH)2(s)⇌Cu2++2OH−\text{Cu(OH)}_2(s) \rightleftharpoons \text{Cu}^{2+}+2\text{OH}^-Cu(OH)2​(s)⇌Cu2++2OH− toward the solid.
  3. Solubility is unchanged because a buffer keeps pH constant, so dissolution equilibria cannot shift.
  4. Solubility increases because higher pH always increases solubility of metal hydroxides by neutralizing the solid.
  5. Solubility is unchanged because OH−\text{OH}^-OH− is not included in the equilibrium for dissolving metal hydroxides.
Explanation: 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.

Question 10

Solid silver acetate, AgC2H3O2(s)\text{AgC}_2\text{H}_3\text{O}_2(s)AgC2​H3​O2​(s), is added to two beakers at the same temperature: Beaker A contains pure water, and Beaker B contains 0.10 M HC2_22​H3_33​O2_22​ (acetic acid). Compared with Beaker A, what happens to the solubility of AgC2H3O2(s)\text{AgC}_2\text{H}_3\text{O}_2(s)AgC2​H3​O2​(s) in Beaker B, and why?

  1. Solubility decreases because acetic acid supplies acetate ions directly, increasing [C2H3O2−][\text{C}_2\text{H}_3\text{O}_2^-][C2​H3​O2−​] and shifting toward the solid.
  2. Solubility increases because H+\text{H}^+H+ protonates C2H3O2−\text{C}_2\text{H}_3\text{O}_2^-C2​H3​O2−​ to form HC2_22​H3_33​O2_22​, reducing [C2H3O2−][\text{C}_2\text{H}_3\text{O}_2^-][C2​H3​O2−​] and shifting dissolution forward.
  3. Solubility decreases because adding any acid adds a common ion H+\text{H}^+H+ that appears in the solubility product expression for acetate salts.
  4. Solubility is unchanged because KspK_{sp}Ksp​ fixes the solubility regardless of other equilibria such as acid-base reactions.
  5. Solubility is unchanged because weak acids do not affect equilibrium concentrations and only strong acids can change solubility.
Explanation: 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.