pH and Solubility - AP Chemistry

Solid copper(II) hydroxide, $\text{Cu(OH)}_2(s)$ , 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 $[\text{H}^+]$ is kept very low and $[\text{OH}^-]$ is relatively high). Compared with Beaker 1, what happens to the solubility of $\text{Cu(OH)}_2(s)$ in the pH 10 buffer, and why?
Answer: 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. 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.
Solid iron(II) sulfide, $\text{FeS}(s)$ , is placed into two solutions at the same temperature. Solution 1 is pure water. Solution 2 is $0.10\ \text{M}$ HCl (pH $\approx 1$ ). Compared with Solution 1, what happens to the solubility of $\text{FeS}(s)$ in Solution 2, and why?
Answer: 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. 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.
A student tests the solubility of solid barium sulfate, $\text{BaSO}_4(s)$ , in pure water and in 0.10 M HCl at the same temperature. Which statement best describes the effect of lowering pH on the solubility of $\text{BaSO}_4(s)$ and why?
Answer: Solubility is essentially unchanged because $\text{SO}_4^{2-}$ is the conjugate base of a strong acid, so added $\text{H}^+$ does not significantly remove $\text{SO}_4^{2-}$ from solution. This question tests understanding of pH effects on salts containing anions of strong acids. BaSO₄ dissolves according to: BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq). Unlike previous examples, SO₄²⁻ is the conjugate base of HSO₄⁻, which is a strong acid in its second ionization. This means SO₄²⁻ is an extremely weak base that does not significantly protonate even in acidic solutions. Therefore, adding HCl does not remove SO₄²⁻ from solution through protonation, and the solubility remains essentially unchanged. Choice A incorrectly claims that H⁺ completely protonates SO₄²⁻ to H₂SO₄, but this reaction has negligible extent because SO₄²⁻ is such a weak base. When evaluating pH effects on solubility, consider the basicity of the anion—anions of strong acids (like SO₄²⁻, Cl⁻, NO₃⁻) show little to no pH-dependent solubility changes.
Solid magnesium hydroxide, $\text{Mg(OH)}_2(s)$ , is added to two beakers at the same temperature: Beaker 1 contains pure water and Beaker 2 contains a solution adjusted to pH 12 with strong base. Compared with Beaker 1, the solubility of $\text{Mg(OH)}_2$ in Beaker 2 will be
Answer: smaller because added $\text{OH}^-$ is a common ion that shifts dissolution to the left. This question tests understanding of the common ion effect on solubility equilibria. When Mg(OH)₂ dissolves, it produces Mg²⁺ and OH⁻ ions according to the equilibrium: Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq). In Beaker 2 with pH 12, the solution already contains a high concentration of OH⁻ ions (10⁻² M) from the added strong base. This added OH⁻ acts as a common ion, shifting the dissolution equilibrium to the left and decreasing the solubility of Mg(OH)₂. Choice E incorrectly claims OH⁻ is not included in the dissolution equilibrium, when it clearly appears as a product. When a solution already contains one of the ions produced by dissolution, apply the common ion effect - the solubility will decrease.
Solid calcium phosphate, $\text{Ca}_3(\text{PO}_4)_2(s)$ , is added to two beakers at the same temperature. Beaker 1 contains pure water. Beaker 2 contains a solution at pH $3$ (adjusted with a strong acid). Which statement best describes the solubility in Beaker 2 and why?
Answer: 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. 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.
A student tests the solubility of solid lead(II) sulfide, $\text{PbS}(s)$ , in two solutions at the same temperature: Solution 1 is pure water (about pH 7) and Solution 2 is 0.10 M HCl (about pH 1). How does lowering the pH affect the solubility of $\text{PbS}(s)$ and why?
Answer: The solubility increases in 0.10 M HCl because $\text{H}^+$ protonates $\text{S}^{2-}$ to form $\text{HS}^-$ and $\text{H}_2\text{S}$ , lowering $[\text{S}^{2-}]$ and shifting dissolution forward. This question tests understanding of how pH affects salts containing the conjugate base of a very weak acid. Lead(II) sulfide dissolves as PbS(s) ⇌ Pb²⁺(aq) + S²⁻(aq), where S²⁻ is the conjugate base of the weak acid HS⁻. In acidic solution (0.10 M HCl), H⁺ ions protonate S²⁻ stepwise to form HS⁻ and H₂S, with H₂S being a weak acid that can even escape as a gas. This effectively removes S²⁻ from the dissolution equilibrium, and by Le Chatelier's principle, removing a product shifts the equilibrium to the right, causing more PbS to dissolve. Choice C incorrectly claims PbS is insoluble at all pH values, failing to recognize that while PbS has a very low Ksp, its solubility can still be increased by removing sulfide ions through protonation. For metal sulfides, acidic conditions significantly increase solubility by converting S²⁻ to H₂S.
Solid calcium oxalate, $\text{CaC}_2\text{O}_4(s)$ , is added to two solutions: one is pure water, the other is 0.10 M HCl(aq). Oxalate, $\text{C}_2\text{O}_4^{2-}$ , is a base (conjugate base of a weak acid). How does the solubility change in acid and why?
Answer: It increases in acid because $\text{H}^+$ protonates $\text{C}_2\text{O}_4^{2-}$ to $\text{HC}_2\text{O}_4^-$ , reducing free oxalate and shifting dissolution forward. The skill being tested is evaluating acidity's influence on oxalate solubility. The correct answer is that solubility increases in acid because H+ protonates $C2O4^2$ - to HC2O4-, reducing free oxalate and shifting dissolution forward via Le Châtelier's principle. In pure water, no protonation limits solubility. Oxalate is basic as conjugate of weak acid. A tempting distractor is C, unchanged as only hydroxides are pH-dependent, but this misconceives that any salt with basic anions shows pH sensitivity. Identify conjugate acid strength to determine if acidity enhances solubility.
A student compares the solubility of $\text{CaF}_2(s)$ in pure water versus a solution at pH $2$ (adjusted with a strong acid). Which choice best describes how the solubility changes at pH $2$ and why?
Answer: The solubility increases because $\text{H}^+$ reacts with $\text{F}^-$ to form HF, lowering $[\text{F}^-]$ and shifting dissolution toward products. 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 HF is a weak acid, so F⁻ is a weak base that H⁺ protonates. 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.
A student investigates the solubility of $\text{Ag}_2\text{CO}_3(s)$ in two beakers at the same temperature. Beaker 1 contains pure water. Beaker 2 contains a solution adjusted to pH $2$ using a strong acid. Which statement best describes the solubility in Beaker 2 and why?
Answer: 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. 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.
A student compares the solubility of $\text{Al(OH)}_3(s)$ in two solutions at the same temperature: Solution 1 is pH $7$ water, and Solution 2 is pH $2$ (adjusted with a strong acid). Which statement best describes the solubility in Solution 2 and why?
Answer: The solubility increases because $\text{H}^+$ consumes $\text{OH}^-$ to form water, decreasing $[\text{OH}^-]$ and pulling the dissolution equilibrium forward. This question tests how low pH increases the solubility of amphoteric hydroxides like Al(OH)3 by removing hydroxide ions. In Solution 2 at pH 2, H+ consumes OH- from Al(OH)3 dissolution to form water, lowering [OH-]. This makes the ion product Q < Ksp, pulling the equilibrium toward more dissolution. Thus, solubility increases compared to pH 7 water. A tempting distractor is choice E, which incorrectly identifies H+ as a common ion suppressing dissolution, but this misconceives H+ as part of the Ksp instead of its role in neutralizing OH-. When examining pH impacts on hydroxide solubility, consider neutralization effects and apply Le Chatelier's principle to ion concentration changes.
Solid magnesium hydroxide, $\text{Mg(OH)}_2(s)$ , is added to two solutions at the same temperature until equilibrium is reached. Solution X is pure water (pH $\approx 7$ ). Solution Y is $0.10\ \text{M}$ NaOH (pH $\approx 13$ ). Compared with Solution X, how does the solubility of $\text{Mg(OH)}_2(s)$ change in Solution Y, and why?
Answer: The solubility decreases because added $\text{OH}^-$ is a common ion that shifts the dissolution equilibrium toward solid. This question tests understanding of the common ion effect on solubility equilibria. Magnesium hydroxide dissolves according to Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq). Solution 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 the formation of solid Mg(OH)₂ and decreasing its solubility. The equilibrium expression Ksp = [Mg²⁺][OH⁻]² shows that when [OH⁻] increases, [Mg²⁺] must decrease to maintain constant Ksp. Choice D incorrectly claims that higher pH always increases solubility, but for metal hydroxides, the opposite is true due to the common ion effect. When a solution already contains one of the ions from a sparingly soluble salt, the solubility of that salt decreases.
A student compares the solubility of calcium carbonate, $\text{CaCO}_3(s)$ , in three beakers: (1) pure water, (2) 0.10 M HCl(aq), and (3) 0.10 M NaOH(aq). Which statement best predicts how changing pH affects the solubility of $\text{CaCO}_3$ and why?
Answer: 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. 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.
Solid iron(II) hydroxide, $\text{Fe(OH)}_2(s)$ , is added to two beakers at the same temperature until undissolved solid remains. Beaker 1 contains pure water. Beaker 2 contains $0.10\,\text{M}$ HCl. Compared with Beaker 1, what happens to the solubility in Beaker 2 and why?
Answer: The solubility increases because $\text{H}^+$ reacts with $\text{OH}^-$ to form water, decreasing $[\text{OH}^-]$ and shifting dissolution toward ions. This question tests how acidic conditions enhance the solubility of metal hydroxides by neutralizing hydroxide ions. In Beaker 2 with 0.10 M HCl, H+ reacts with OH- from Fe(OH)2 dissolution to form water, decreasing [OH-] in solution. This lowers the ion product Q below Ksp, shifting the equilibrium toward more dissolution according to Le Chatelier's principle. Therefore, the solubility increases in the acidic beaker relative to pure water. A tempting distractor is choice B, which incorrectly claims Cl- causes a common-ion effect leading to FeCl2 precipitation, but this ignores that Cl- is not part of the Fe(OH)2 equilibrium and misapplies precipitation concepts. When assessing solubility in acids, consider if the anion can be neutralized or protonated, and apply equilibrium principles to predict changes in ion concentrations.
A saturated solution of $\text{Al(OH)}_3(s)$ is at equilibrium. The pH is then lowered by adding a small amount of strong acid. Ignoring complex-ion formation, which change is expected and why?
Answer: More $\text{Al(OH)}_3$ dissolves because $\text{H}^+$ consumes $\text{OH}^-$ , decreasing $[\text{OH}^-]$ and shifting dissolution to the right. The skill being tested is predicting dissolution changes in saturated hydroxide solutions upon acidification. The correct answer is that more Al(OH)3 dissolves because H+ consumes OH-, decreasing [OH-] and shifting equilibrium right by Le Châtelier's principle. In the saturated system, this increases dissolved $Al^3$ +. Ignoring complexes focuses on basic anion consumption. A tempting distractor is B, less dissolves due to H+ as common ion, but this misconceives H+'s role in neutralizing OH-. Integrate Le Châtelier's principle across linked equilibria for pH adjustments in saturated solutions.
A student investigates the solubility of solid iron(II) sulfide, $\text{FeS}(s)$ , in two solutions at the same temperature: one is neutral water and the other is strongly acidic (pH 1). Which statement best describes the solubility of $\text{FeS}$ in the acidic solution and why?
Answer: The solubility increases because $\text{H}^+$ reacts with $\text{S}^{2-}$ to form $\text{HS}^-$ and $\text{H}_2\text{S}$ , reducing $[\text{S}^{2-}]$ . This question tests understanding of how pH affects the solubility of metal sulfides. When FeS dissolves, it produces Fe²⁺ and S²⁻ ions, where S²⁻ is the conjugate base of the very weak acid HS⁻. In strongly acidic solution (pH 1), the high concentration of H⁺ ions readily reacts with S²⁻ to form HS⁻ and H₂S, removing S²⁻ from solution and shifting the dissolution equilibrium to the right according to Le Chatelier's principle. Choice C incorrectly claims sulfide is not part of any acid-base equilibrium, when S²⁻ is actually a strong base that readily accepts protons. For metal sulfides, remember that S²⁻ is highly basic and will be extensively protonated in acid, dramatically increasing solubility.
A sparingly soluble salt, $\text{ZnS}(s)$ , is placed into two solutions at the same temperature. Solution 1 is pure water. Solution 2 is $0.10\,\text{M}$ HNO $_3$ (pH $\approx 1$ ). Which choice best describes the solubility of $\text{ZnS}(s)$ in Solution 2 compared with Solution 1 and gives the correct reason?
Answer: The solubility increases because $\text{H}^+$ reacts with $\text{S}^{2-}$ to form $\text{HS}^-$ and $\text{H}_2\text{S}$ , reducing $[\text{S}^{2-}]$ and pulling dissolution forward. This question tests how acidic pH influences the solubility of metal sulfides through protonation of the sulfide ion. In Solution 2 with 0.10 M HNO3, the high [H+] reacts with $S^2$ - from dissolving ZnS to form HS- and H2S, effectively reducing [ $S^2$ -] in solution. This decrease in [ $S^2$ -] makes Q < Ksp, driving more ZnS to dissolve to reestablish equilibrium per Le Chatelier's principle. Thus, the solubility increases in the acidic solution compared to pure water. A tempting distractor is choice A, which wrongly identifies NO3- as a common ion, but this misconstrues the common-ion effect since NO3- is not part of the ZnS equilibrium. For solubility in varying pH, assess if the anion is basic and can form protonated species, then predict the equilibrium shift using Le Chatelier's principle.
Solid silver acetate, $\text{AgC}_2\text{H}_3\text{O}_2(s)$ , is added to two beakers at the same temperature: Beaker A contains pure water, and Beaker B contains 0.10 M HC $_2$ H $_3$ O $_2$ (acetic acid). Compared with Beaker A, what happens to the solubility of $\text{AgC}_2\text{H}_3\text{O}_2(s)$ in Beaker B, and why?
Answer: 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. 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.
Solid zinc hydroxide, $\text{Zn(OH)}_2(s)$ , is added to two solutions at the same temperature: Solution M is pure water, and Solution N is 0.10 M HNO $_3$ . Compared with Solution M, what happens to the solubility of $\text{Zn(OH)}_2(s)$ in Solution N, and why?
Answer: Solubility increases because $\text{H}^+$ neutralizes $\text{OH}^-$ to form water, reducing $[\text{OH}^-]$ and shifting dissolution forward. 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.
Solid copper(II) sulfide, $\text{CuS}(s)$ , is placed into two solutions: Solution A is pH 1 and Solution B is pH 7. Which statement best predicts the relative solubility and the reason?
Answer: More soluble at pH 1 because $\text{H}^+$ converts $\text{S}^{2-}$ to $\text{HS}^-/$ \text{H}_2\text{S}$, lowering free $\text{S}^{2-}$ and shifting dissolution. The skill being tested is comparing sulfide solubility across pH values. The correct answer is that CuS is more soluble at pH 1 because H+ converts $S^2$ - to HS-/H2S, lowering free [ $S^2$ -] and shifting dissolution forward per Le Châtelier's principle. At pH 7, minimal protonation occurs, resulting in lower solubility. $S^2$ - is a strong base, highly sensitive to acidity. A tempting distractor is C, same solubility as pH affects only acids/bases not salts, but this misconceives that salts with basic anions are pH-dependent. Assess the basicity of anions to predict acidity's enhancement of solubility.
A student investigates the solubility of solid barium carbonate, $\text{BaCO}_3(s)$ , in two solutions at the same temperature. Solution I is $0.10\ \text{M}$ NaCl (approximately neutral). Solution II is $0.10\ \text{M}$ HCl (pH $\approx 1$ ). Compared with Solution I, how does the solubility of $\text{BaCO}_3(s)$ change in Solution II, and why?
Answer: 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. 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.

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