AP Chemistry Quiz: Ph And Solubility

What this quiz covers

This quiz focuses on Ph And Solubility, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The solubility stays the same because HCl acts as a buffer that keeps $[\text{CO}_3^{2-}]$ constant in solution.
2. The solubility decreases because adding acid increases $[\text{CO}_3^{2-}]$ through the common-ion effect, shifting dissolution backward.
3. The solubility stays the same because $K_{sp}$ 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 $\text{H}^+$ reacts with $\text{CO}_3^{2-}$ to form $\text{HCO}_3^-$ and $\text{H}_2\text{CO}_3$, lowering $[\text{CO}_3^{2-}]$ and shifting dissolution forward. (correct answer)

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

1. The solubility increases greatly because $\text{H}^+$ converts $\text{SO}_4^{2-}$ completely into $\text{H}_2\text{SO}_4$, removing sulfate.
2. The solubility decreases because $\text{H}^+$ is a common ion with $\text{Ba}^{2+}$ and shifts dissolution left.
3. The solubility is approximately unchanged because $\text{SO}_4^{2-}$ is the conjugate base of a strong acid and is only weakly protonated. (correct answer)
4. The solubility decreases because acids always reduce solubility by lowering the value of $K_{sp}$.
5. The solubility increases because the acid buffers the solution and therefore prevents precipitation of $\text{BaSO}_4$.

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

1. The solubility decreases because $\text{Cl}^-$ is a common ion with $\text{BaCO}_3$ and shifts equilibrium toward the solid.
2. The solubility increases because $\text{H}^+$ consumes $\text{CO}_3^{2-}$ to form $\text{HCO}_3^-$/$\text{H}_2\text{CO}_3$, lowering$\text{CO}_3^{2-}$ and shifting dissolution toward products. (correct answer)
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 $K_{sp}$ 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

1. The solubility decreases because $\text{Cl}^-$ is a common ion that suppresses dissolution of $\text{FeS}(s)$.
2. The solubility increases because $\text{H}^+$ reacts with $\text{S}^{2-}$ to form HS$^-$ and H$_2$S, reducing $\text{S}^{2-}$ and pulling dissolution forward. (correct answer)
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 $\text{S}^{2-}$ 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

1. The solubility stays the same because nitrate does not appear in the $K_{sp}$ expression for $\text{Zn(OH)}_2$.
2. The solubility decreases because $\text{NO}_3^-$ is a common ion that shifts dissolution toward the solid.
3. The solubility increases because $\text{H}^+$ neutralizes $\text{OH}^-$ to form water, reducing $\text{OH}^-$ and shifting dissolution toward ions. (correct answer)
4. The solubility decreases because lowering pH always decreases the solubility of metal hydroxides by forming more $\text{OH}^-$.
5. The solubility stays the same because strong acids act as buffers that keep $\text{OH}^-$ 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

1. The solubility decreases because added $\text{OH}^-$ is a common ion that shifts the dissolution equilibrium toward the solid. (correct answer)
2. The solubility increases because added $\text{Na}^+$ reacts with $\text{OH}^-$ to form NaOH(aq), removing $\text{OH}^-$ 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 $\text{Al}^{3+}$ 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

1. The solubility decreases because $\text{H}^+$ is a common ion with $\text{CaF}_2$ 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 $\text{H}^+$.
3. The solubility increases because $\text{H}^+$ reacts with $\text{F}^-$ to form HF, lowering $[\text{F}^-]$ and shifting dissolution toward products. (correct answer)
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

1. Solubility increases because $\text{H}^+$ reacts with $\text{CO}_3^{2-}$ to form $\text{HCO}_3^-$ and $\text{H}_2\text{CO}_3$, reducing $[\text{CO}_3^{2-}]$ and shifting dissolution forward. (correct answer)
2. Solubility decreases because added $\text{Cl}^-$ is a common ion with $\text{CaCO}_3$ and shifts the equilibrium toward the solid.
3. Solubility is unchanged because $K_{sp}$ depends only on temperature and pH cannot affect any equilibrium concentrations.
4. Solubility decreases because a lower pH forces carbonate to remain as $\text{CO}_3^{2-}$, increasing $[\text{CO}_3^{2-}]$ 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

1. Solubility increases because buffers consume $\text{Cu}^{2+}$, removing it from solution and pulling dissolution forward.
2. Solubility decreases because the higher $[\text{OH}^-]$ acts as a common ion and shifts $\text{Cu(OH)}_2(s) \rightleftharpoons \text{Cu}^{2+}+2\text{OH}^-$ toward the solid. (correct answer)
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 $\text{OH}^-$ 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

1. Solubility decreases because acetic acid supplies acetate ions directly, increasing $[\text{C}_2\text{H}_3\text{O}_2^-]$ and shifting toward the solid.
2. Solubility increases because $\text{H}^+$ protonates $\text{C}_2\text{H}_3\text{O}_2^-$ to form HC$_2$H$_3$O$_2$, reducing $[\text{C}_2\text{H}_3\text{O}_2^-]$ and shifting dissolution forward. (correct answer)
3. Solubility decreases because adding any acid adds a common ion $\text{H}^+$ that appears in the solubility product expression for acetate salts.
4. Solubility is unchanged because $K_{sp}$ 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.

Question 11

1. Solubility increases because $\text{H}^+$ protonates $\text{PO}_4^{3-}$ to form $\text{HPO}_4^{2-}$ and $\text{H}_2\text{PO}_4^-$, lowering $[\text{PO}_4^{3-}]$ and shifting dissolution forward. (correct answer)
2. Solubility decreases because added acid increases $[\text{PO}_4^{3-}]$ by converting $\text{H}_2\text{PO}_4^-$ into $\text{PO}_4^{3-}$, driving precipitation.
3. Solubility is unchanged because HCl is a strong acid and strong acids do not participate in equilibrium shifts.
4. Solubility is unchanged because $K_{sp}$ is constant and therefore the amount that dissolves cannot depend on pH.
5. Solubility decreases because $\text{Cl}^-$ is a common ion with phosphate salts and shifts the equilibrium toward the solid.

Explanation: This question tests understanding of how pH affects the solubility of salts containing polyprotic basic anions. Ca₃(PO₄)₂ dissolves to produce Ca²⁺ and PO₄³⁻ ions, where PO₄³⁻ is a strongly basic anion that readily accepts protons. In acidic solution (HCl), H⁺ ions protonate PO₄³⁻ stepwise to form HPO₄²⁻, H₂PO₄⁻, and even H₃PO₄, effectively removing PO₄³⁻ from the solution. By Le Chatelier's principle, removing the product PO₄³⁻ shifts the dissolution equilibrium to the right, significantly increasing the solubility of Ca₃(PO₄)₂. Choice D incorrectly reverses the acid-base chemistry, claiming acid increases [PO₄³⁻] when it actually decreases it through protonation. For salts containing highly basic anions (especially polyprotic ones like PO₄³⁻, CO₃²⁻), acidic conditions dramatically increase solubility by removing the anion through multiple protonation steps.

Question 12

1. Solubility decreases because nitrate is a common ion with $\text{Zn(OH)}_2$ and shifts the equilibrium toward the solid.
2. Solubility is unchanged because strong acids prevent any equilibrium shift by fully dissociating.
3. Solubility increases because $\text{H}^+$ neutralizes $\text{OH}^-$ to form water, reducing $[\text{OH}^-]$ and shifting dissolution forward. (correct answer)
4. Solubility decreases because adding acid increases $[\text{OH}^-]$ through water autoionization, shifting toward precipitation.
5. Solubility is unchanged because the solid controls the ion concentrations, so adding $\text{H}^+$ cannot change solubility.

Explanation: This question tests understanding of how pH affects the solubility of metal hydroxides. Zn(OH)₂ dissolves according to: Zn(OH)₂(s) ⇌ Zn²⁺(aq) + 2OH⁻(aq). In acidic solution (HNO₃), H⁺ ions react with OH⁻ ions to form water through the neutralization reaction: H⁺ + OH⁻ → H₂O. This effectively removes OH⁻ from the solution, and by Le Chatelier's principle, the dissolution equilibrium shifts to the right to produce more ions, increasing solubility. Choice D incorrectly claims that adding acid increases [OH⁻], when acids actually decrease hydroxide concentration by neutralization. For metal hydroxides, acidic conditions always increase solubility because H⁺ removes OH⁻ through neutralization, while basic conditions decrease solubility through the common ion effect.

Question 13

1. The solubility decreases because $\text{H}^+$ is a common ion for $\text{Ag}_2\text{CO}_3$ and shifts dissolution toward the solid.
2. The solubility stays the same because $K_{sp}$ depends only on temperature and pH cannot change concentrations at equilibrium.
3. The solubility increases because $\text{H}^+$ converts $\text{CO}_3^{2-}$ to $\text{HCO}_3^-$ and $\text{H}_2\text{CO}_3$, lowering $[\text{CO}_3^{2-}]$ and favoring dissolution. (correct answer)
4. The solubility decreases because the strong acid increases ionic strength, which always decreases solubility for ionic solids.
5. The solubility stays the same because the acid acts as a buffer that keeps the carbonate concentration fixed.

Explanation: This question tests the impact of low pH on the solubility of carbonate salts via acid-base reactions with the carbonate ion. In Beaker 2 at pH 2, the added H+ from the strong acid converts CO3^2- from Ag2CO3 dissolution into HCO3- and H2CO3, lowering [CO3^2-]. This reduction in [CO3^2-] causes Q to fall below Ksp, promoting more dissolution to restore equilibrium. Consequently, the solubility of Ag2CO3 increases in the acidic beaker compared to pure water. A tempting distractor is choice A, which mistakenly applies the common-ion effect to H+, but H+ is not an ion in the Ksp expression and instead participates in an acid-base reaction. To evaluate pH-dependent solubility, check if the anion can accept protons and use Le Chatelier's principle to analyze how removing the anion affects the equilibrium.

Question 14

1. More soluble at pH 2 because $\text{H}^+$ converts $\text{CO}_3^{2-}$ to $\text{HCO}_3^-$/$\text{H}_2\text{CO}_3$, reducing free$\text{CO}_3^{2-}$. (correct answer)
2. Less soluble at pH 2 because adding acid introduces a common ion that shifts equilibrium left.
3. Same solubility because $\text{Ag}^+$ is not acidic or basic, so pH has no effect.
4. More soluble at pH 12 because $\text{OH}^-$ removes $\text{CO}_3^{2-}$ by forming $\text{HCO}_3^-$.
5. Same solubility because pH affects only reaction rates, not equilibrium solubility.

Explanation: The skill being tested is comparing solubility at different pH levels for carbonates. The correct answer is that Ag2CO3 is more soluble at pH 2 because H+ converts CO3^2- to HCO3-/H2CO3, reducing free [CO3^2-] and promoting dissolution per Le Châtelier's principle. At pH 12, the lack of H+ prevents this reaction, resulting in lower solubility similar to neutral conditions. CO3^2- acts as a base, making acidity enhance solubility. A tempting distractor is A, that it's more soluble at pH 12 because OH- removes CO3^2-, but this misconceives the acid-base behavior, as OH- does not effectively protonate or remove carbonate. Identify the acid-base nature of ions and use equilibrium shifts to assess pH impacts on solubility.

Question 15

1. Solubility is highest in pure water because adding acid or base always adds a common ion that suppresses dissolution.
2. Solubility is highest in 0.10 M NaOH because $\text{OH}^-$ neutralizes $\text{Ca}^{2+}$, removing it and shifting dissolution forward.
3. Solubility is the same in all three because $K_{sp}$ fixes the solubility regardless of pH.
4. Solubility is lowest in 0.10 M HCl because added ions make the solution buffered and prevent further dissolving.
5. Solubility is highest in 0.10 M HCl because $\text{H}^+$ consumes $\text{CO}_3^{2-}$ to form $\text{HCO}_3^-$/$\text{H}_2\text{CO}_3$, shifting dissolution forward. (correct answer)

Explanation: The skill being tested is understanding how pH affects the solubility of salts with basic anions through acid-base equilibria. The correct answer is that solubility is highest in 0.10 M HCl because H+ consumes CO3^2- to form HCO3-/H2CO3, shifting the dissolution equilibrium forward by Le Châtelier's principle. In pure water, there is no excess H+ to react with CO3^2-, resulting in lower solubility limited by Ksp. In 0.10 M NaOH, the high [OH-] does not remove CO3^2- and may slightly suppress solubility due to indirect effects, but the key is the acid's enhancement. A tempting distractor is C, that solubility is the same because Ksp fixes it regardless of pH, but this misconceives that while Ksp is constant, coupled acid-base reactions alter effective ion concentrations. Always consider if dissolved ions can participate in pH-dependent equilibria when predicting solubility changes.

Question 16

1. It decreases because acid always decreases the solubility of ionic solids by neutralizing ions.
2. It decreases because $\text{Cl}^-$ is a common ion with $\text{F}^-$ and suppresses dissolution.
3. It stays the same because $K_{sp}$ is constant and does not depend on pH.
4. It increases because $\text{H}^+$ reacts with $\text{F}^-$ to form HF, lowering $[\text{F}^-]$ and shifting dissolution forward. (correct answer)
5. It increases because the solution becomes buffered and buffers dissolve more solids.

Explanation: The skill being tested is analyzing pH-dependent solubility for salts with weakly basic anions. The correct answer is that solubility of CaF2 increases in 0.10 M HCl because H+ reacts with F- to form HF, lowering [F-] and shifting dissolution forward via Le Châtelier's principle. In pure water, no such protonation occurs, limiting solubility to Ksp constraints. This is due to F- being the conjugate base of weak acid HF, making it susceptible to protonation. A tempting distractor is C, that it stays the same because Ksp is constant, but this ignores how acid-base side reactions affect ion concentrations in the Ksp expression. Consider coupled equilibria when predicting solubility changes in acidic or basic environments.

Question 17

1. It decreases because the acid forms a buffer with $\text{OH}^-$ that prevents additional dissolution.
2. It increases because $\text{H}^+$ neutralizes $\text{OH}^-$, lowering $[\text{OH}^-]$ and shifting dissolution to produce more ions. (correct answer)
3. It stays constant because $\text{Fe}^{2+}$ concentration in a saturated solution cannot change.
4. It decreases because added $\text{H}^+$ is a common ion that shifts the equilibrium toward solid.
5. It increases because $\text{H}^+$ directly appears in the $K_{sp}$ expression.

Explanation: The skill being tested is predicting changes in dissolved ion concentration upon pH adjustment. The correct answer is that dissolved Fe^2+ increases because H+ neutralizes OH-, lowering [OH-] and shifting dissolution to produce more ions via Le Châtelier's principle. In the saturated solution, this allows more Fe(OH)2 to dissolve. The effect stems from OH- being consumable by acid. A tempting distractor is A, decreases due to H+ as common ion, but this misconceives that H+ reacts with OH- instead of being a common ion in the solubility equilibrium. Always integrate acid-base and solubility equilibria when analyzing pH perturbations.

Question 18

1. Higher in Beaker B because higher pH increases $K_{sp}$ for hydroxides, allowing more solid to dissolve.
2. Higher in Beaker A because lower pH reduces $[\text{OH}^-]$ by protonation, shifting dissolution toward ions. (correct answer)
3. Equal in both because buffers prevent any shift in the dissolution equilibrium.
4. Higher in Beaker B because the buffer supplies $\text{OH}^-$, which pulls the equilibrium toward dissolved ions.
5. Equal in both because $\text{OH}^-$ is not a product in the dissolution of $\text{Zn(OH)}_2$.

Explanation: This question tests understanding of how pH affects the solubility of metal hydroxides. Zn(OH)₂ dissolves according to: Zn(OH)₂(s) ⇌ Zn²⁺(aq) + 2OH⁻(aq). At pH 5, the concentration of OH⁻ is very low (10⁻⁹ M), while at pH 10, [OH⁻] is much higher (10⁻⁴ M). In the acidic buffer (pH 5), H⁺ ions react with OH⁻ produced by dissolution to form water, effectively removing OH⁻ and shifting the equilibrium to the right, increasing solubility. Choice D incorrectly suggests the buffer supplies OH⁻ at pH 10, when actually the higher pH means more OH⁻ is already present, suppressing dissolution. For metal hydroxides, lower pH always increases solubility by removing OH⁻ through neutralization.

Question 19

1. The solubility stays the same because NaIO$_3$ acts as a buffer that keeps $[\text{IO}_3^-]$ constant at the $K_{sp}$ value.
2. The solubility decreases because $\text{IO}_3^-$ is a common ion that increases $Q$ and shifts the dissolution equilibrium toward the solid. (correct answer)
3. The solubility stays the same because common ions only matter when the added ion comes from a strong acid, not from a salt.
4. The solubility increases because adding any electrolyte increases solubility by separating ions in solution.
5. The solubility increases because adding $\text{IO}_3^-$ neutralizes $\text{Pb}^{2+}$, removing it from solution and driving more dissolution.

Explanation: This question tests the common-ion effect on the solubility of iodate salts when additional iodate is present. In Beaker 2 with 0.10 M NaIO3, the added IO3- increases [IO3-] in solution, elevating the ion product Q above Ksp for Pb(IO3)2. This shift favors the formation of solid Pb(IO3)2, reducing solubility compared to pure water. The effect is purely due to the common ion without pH involvement. A tempting distractor is choice A, which wrongly suggests IO3- neutralizes Pb^2+, but this misinterprets the role of ions in equilibrium rather than recognizing suppression via Le Chatelier's principle. For solubility problems with added salts, check for shared ions and use Q versus Ksp to predict if dissolution is enhanced or suppressed.

Question 20

1. The solubility decreases because $\text{H}^+$ is a product of dissolution of phosphate salts and drives precipitation by the common-ion effect.
2. The solubility increases because $\text{H}^+$ protonates $\text{PO}_4^{3-}$ to form $\text{HPO}_4^{2-}$ and $\text{H}_2\text{PO}_4^-$, lowering $[\text{PO}_4^{3-}]$ and shifting dissolution forward. (correct answer)
3. The solubility stays the same because phosphate is a spectator ion and does not participate in acid-base reactions.
4. The solubility decreases because adding acid increases $[\text{Ca}^{2+}]$ directly, causing a common-ion effect that precipitates the salt.
5. The solubility stays the same because the strong acid acts as a buffer and holds the phosphate concentration constant.

Explanation: This question tests how acidic pH enhances the solubility of phosphate salts by protonating the phosphate ion. In Beaker 2 at pH 3, H+ protonates PO4^3- to form HPO4^2- and H2PO4-, substantially lowering [PO4^3-]. This reduction causes Q < Ksp, driving more Ca3(PO4)2 to dissolve per Le Chatelier's principle. The effect is pronounced because phosphate has high pKa values, making it sensitive to low pH. A tempting distractor is choice A, which misidentifies H+ as a product causing common-ion suppression, but this ignores that H+ removes PO4^3- via protonation rather than being part of Ksp. For solubility of salts with multi-protic anions, evaluate pKa steps relative to solution pH and use equilibrium shifts to predict changes.