pH and Solubility - AP Chemistry

SECTION 1

Historical Context & Motivation

The relationship between pH and solubility has been central to chemistry since the earliest attempts to understand why certain minerals dissolve in acidic waters but remain stubbornly insoluble in neutral streams. Miners in the sixteenth century observed that acidic mine drainage could dissolve metal ores that resisted dissolution in clean water, though they lacked a theoretical framework to explain the phenomenon. It was not until the development of ionic theory and the quantitative pH scale that chemists could rigorously connect hydrogen ion concentration to the dissolution behavior of sparingly soluble salts. This connection lies at the intersection of two pillars of general chemistry: acid–base equilibrium and solubility equilibrium, and understanding their interplay is essential for predicting precipitation, designing buffer systems, and interpreting real-world phenomena such as ocean acidification and kidney stone formation.

1887 • Arrhenius Ion Theory

Svante Arrhenius proposes that salts dissociate into ions in solution, establishing the ionic framework required to understand dissolution and precipitation equilibria.

1909 • The pH Scale

Søren Sørensen introduces the pH scale at the Carlsberg Laboratory, providing a quantitative logarithmic measure of hydrogen ion activity that enables precise studies of solution acidity.

1923 • Brønsted–Lowry Theory

Johannes Brønsted and Thomas Lowry independently define acids as proton donors and bases as proton acceptors, broadening acid–base chemistry to encompass conjugate pairs relevant to solubility.

1938 • Systematic Ksp Measurements

Systematic tabulation of solubility product constants (K_sp) becomes standard in analytical chemistry, enabling quantitative prediction of precipitation as a function of ion concentrations.

1960s • Environmental Applications

Environmental chemists apply pH-dependent solubility models to acid rain research, demonstrating how lowered pH mobilizes toxic metal ions from soil and rock into aquatic ecosystems.

The central question this lesson addresses is deceptively simple: Why does changing the pH of a solution alter the solubility of certain ionic compounds but not others? Answering this question requires combining Le Châtelier's principle with K_sp expressions and acid–base chemistry—skills that are heavily tested on the AP Chemistry exam and foundational for future work in analytical chemistry, geochemistry, and biochemistry.

SECTION 2

Core Principles & Definitions

Before examining the pH–solubility relationship in detail, it is necessary to establish several foundational concepts. The solubility product constant (K_sp) is the equilibrium constant for the dissolution of a sparingly soluble ionic compound in water. Unlike a simple solubility value expressed in g/L, K_sp is expressed in terms of ion concentrations raised to stoichiometric powers. The key insight is that when the anion of a sparingly soluble salt is the conjugate base of a weak acid, added H⁺ ions can react with that anion and shift the dissolution equilibrium to the right, thereby increasing solubility. Conversely, anions derived from strong acids (such as Cl⁻ and NO₃⁻) do not react appreciably with H⁺, so the solubility of those salts is essentially pH-independent.

Solubility Product (Ksp)

The equilibrium constant for the dissolution of a sparingly soluble salt. For A_mB_n(s) ⇌ mAⁿ⁺ + nB^m−, K_sp = [Aⁿ⁺]^m[B^m−]ⁿ.

Common Ion Effect

Adding a common ion to a solution shifts the dissolution equilibrium to the left, reducing solubility. This is a direct application of Le Châtelier's principle to the K_sp expression.

pH-Dependent Solubility

When the anion of a sparingly soluble salt is the conjugate base of a weak acid, lowering pH (adding H⁺) removes that anion via protonation, shifting the dissolution equilibrium rightward and increasing solubility.

pH-Independent Solubility

Salts whose anions are conjugate bases of strong acids (e.g., AgCl, BaSO₄ to a rough approximation, PbI₂) show minimal solubility change with pH because their anions do not react with H⁺ to any significant extent.

Le Châtelier's Principle

When a system at equilibrium is disturbed, it shifts to counteract the disturbance. Removing anion from solution (by protonation) constitutes a stress that drives the dissolution forward.

✦

KEY TAKEAWAY

Think of a sparingly soluble salt as a partially opened faucet dripping ions into solution. Lowering the pH is like opening a drain on the anion side—if H⁺ reacts with and removes the anion (protonation), the solution's ion product drops below K_sp, so more solid dissolves to re-establish equilibrium. If the anion is from a strong acid, it ignores H⁺ entirely—the drain is plugged, and pH has no leverage.

SECTION 3

Visual Explanation

How pH Shifts the Dissolution Equilibrium

The diagram contrasts two categories of sparingly soluble salts. On the left, CaF₂ contains F⁻ (conjugate base of the weak acid HF), so adding H⁺ removes F⁻ and shifts dissolution rightward. On the right, AgCl contains Cl⁻ (conjugate base of the strong acid HCl), so H⁺ has no effect. The decision rule at the bottom summarizes the critical test: identify the parent acid of the anion.

The diagram above illustrates the conceptual core of pH-dependent solubility. For CaF₂, the fluoride ion is the conjugate base of HF (K_a = 6.6 × 10⁻⁴), so in acidic solution F⁻ is consumed by protonation. Le Châtelier's principle then demands that additional CaF₂ dissolves to replenish the fluoride. By contrast, for AgCl the chloride ion derives from HCl, a strong acid with no meaningful tendency to accept a proton, so pH has no thermodynamic leverage on the dissolution equilibrium. When analyzing any salt on the AP exam, the first question to ask is: What acid does the anion come from, and is that acid weak or strong?

Fluorite at a solution interface Fluorite, calcium fluoride, contacts a solution in which dissolved fluoride can participate in acid–base equilibria.

SECTION 4

Mathematical Framework

To treat pH and solubility quantitatively, we combine the K_sp expression for dissolution with the K_a expression for the anion's protonation. When both equilibria operate simultaneously, the overall process has an effective equilibrium constant that is the product of K_sp and 1/K_a (or K_b of the anion), as governed by Hess's law applied to equilibrium constants.

SOLUBILITY PRODUCT EXPRESSION

Ksp = [Ca²⁺][F⁻]²

For CaF₂(s) ⇌ Ca²⁺(aq) + 2 F⁻(aq). K_sp = 3.45 × 10⁻¹¹ at 25 °C. In pure water, molar solubility s yields [Ca²⁺] = s and [F⁻] = 2s.

ANION PROTONATION EQUILIBRIUM

F⁻(aq) + H⁺(aq) ⇌ HF(aq) K = 1/Ka = 1/(6.6 × 10⁻⁴) ≈ 1.5 × 10³

Since HF is a weak acid, its conjugate base F⁻ is a reasonably effective proton acceptor. The large value of 1/K_a means this reaction lies far to the right, ensuring significant anion removal in acidic solution.

OVERALL DISSOLUTION IN ACID

CaF₂(s) + 2 H⁺(aq) ⇌ Ca²⁺(aq) + 2 HF(aq) K = Ksp × (1/Ka)²

The overall equilibrium constant K = K_sp × (1/K_a)² = (3.45 × 10⁻¹¹)(1.5 × 10³)² ≈ 7.8 × 10⁻⁵. Because K > K_sp by many orders of magnitude, solubility is dramatically enhanced in acid.

💡

AP Exam Tip

On the AP Chemistry exam, you are rarely asked to compute the exact solubility in a buffered solution. Instead, the exam tests whether you can predict the direction of the solubility change (increase or no change) when pH is lowered, and justify your reasoning using Le Châtelier's principle and the identity of the anion. Always write the dissolution equilibrium and the protonation reaction explicitly in free-response answers.

SECTION 5

Classifying Salts by pH Sensitivity

A systematic approach to predicting pH effects on solubility begins with classifying the anion. The table below organizes common sparingly soluble salts by whether their solubility increases, decreases, or remains unchanged as pH decreases. Note that hydroxide salts represent a special case: the anion is OH⁻, so adding H⁺ directly neutralizes it, and solubility increases dramatically in acidic solution. Similarly, salts of polyprotic acid anions (CO₃²⁻, PO₄³⁻, S²⁻) are especially pH-sensitive because multiple protonation steps can occur.

Anion	Parent Acid	Acid Strength	pH Effect on Solubility	Example Salt
OH⁻	H₂O	Amphiprotic	Increases	Fe(OH)₃
F⁻	HF	Weak	Increases	CaF₂
CO₃²⁻	H₂CO₃	Weak (diprotic)	Increases strongly	CaCO₃
PO₄³⁻	H₃PO₄	Weak (triprotic)	Increases strongly	Ca₃(PO₄)₂
S²⁻	H₂S	Weak (diprotic)	Increases	CuS
Cl⁻	HCl	Strong	No change	AgCl
Br⁻	HBr	Strong	No change	AgBr
I⁻	HI	Strong	No change	PbI₂

Classification of common anions by pH sensitivity of their sparingly soluble salts

Qualitative graph showing how molar solubility varies with pH for three representative salts. CaCO₃ (pink) shows the greatest pH sensitivity because CO₃²⁻ can accept two protons. CaF₂ (violet) also increases in solubility as pH drops, but less steeply. AgCl (amber) remains flat, confirming pH independence.

The graph reinforces a critical pattern: the weaker the parent acid, the more basic the anion, and the more dramatically pH affects solubility. Carbonate (from H₂CO₃, K_a1 = 4.3 × 10⁻⁷, K_a2 = 4.7 × 10⁻¹¹) is an exceptionally basic anion, so CaCO₃ dissolves readily even in mildly acidic solution—a fact that explains why limestone caves form in regions where slightly acidic groundwater contacts calcium carbonate bedrock over geological timescales.

SECTION 6

Worked Example

Will CaF₂ Dissolve More in a pH 3.00 Buffer Than in Pure Water?

Predicting the Effect of Acid on CaF₂ Solubility

1
Step 1 — Write the Dissolution Equilibrium

CaF₂(s) ⇌ Ca²⁺(aq) + 2 F⁻(aq), with K_sp = 3.45 × 10⁻¹¹.

Ksp = 3.45 × 10⁻¹¹
2
Step 2 — Identify the Anion and Its Parent Acid

The anion F⁻ is the conjugate base of HF, a weak acid with K_a = 6.6 × 10⁻⁴. Because HF is weak, F⁻ will react with H⁺ in the buffered solution.

F⁻ is a basic anion → pH-dependent solubility
3
Step 3 — Apply Le Châtelier's Principle

At pH 3.00, [H⁺] = 1.0 × 10⁻³ M. The protonation reaction F⁻ + H⁺ → HF removes F⁻ from solution. This lowers the ion product Q relative to K_sp, so additional CaF₂ dissolves to restore equilibrium.

Equilibrium shifts RIGHT → more CaF₂ dissolves
4
Step 4 — Calculate Solubility in Pure Water (Baseline)

Let s = molar solubility. Then [Ca²⁺] = s and [F⁻] = 2s. K_sp = (s)(2s)² = 4s³. Solving: s = (K_sp/4)^1/3 = (3.45 × 10⁻¹¹/4)^1/3 = (8.63 × 10⁻¹²)^1/3 ≈ 2.05 × 10⁻⁴ M.

s (pure water) ≈ 2.05 × 10⁻⁴ M
5
Step 5 — Conclude

Because F⁻ is protonated at pH 3.00, the effective solubility of CaF₂ in the buffer is significantly greater than 2.05 × 10⁻⁴ M. On an AP free-response question, a full-credit answer would state: "CaF₂ is more soluble at pH 3.00 because F⁻, the conjugate base of the weak acid HF, reacts with H⁺ to form HF. This removes F⁻ from the dissolution equilibrium, shifting it to the right and increasing solubility."

CaF₂ is MORE soluble in the pH 3.00 buffer than in pure water.

SECTION 7

Strengths & Limitations of the pH–Solubility Model

The qualitative model presented above—asking whether the anion is the conjugate base of a weak acid—is powerful and sufficient for the AP Chemistry exam. However, like all models in chemistry, it has boundaries. The table below compares the strengths and limitations of this approach, and identifies situations where more advanced treatments are required.

Strengths	Limitations
Simple decision rule: identify the anion's parent acid and check if it is weak or strong.	Does not account for ion pairing or activity coefficients at high ionic strength.
Correctly predicts direction of solubility change for the vast majority of common salts.	Cannot quantify exact solubility in buffered solutions without simultaneous equilibrium calculations.
Integrates Le Châtelier's principle with Ksp and Ka, reinforcing conceptual unity.	Metal hydroxide amphoterism (e.g., Al(OH)₃ dissolving in base) requires additional consideration of complex-ion formation.
Applicable across environmental, biological, and industrial contexts.	Temperature dependence of Ksp is neglected in most AP-level analyses.

Strengths and limitations of the qualitative pH–solubility model

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KEY TAKEAWAY

The qualitative pH–solubility model is analogous to using a free-body diagram in physics: it captures the essential forces (equilibria) at play and reliably predicts the direction of the outcome, even though it does not account for every real-world complication such as friction (activity effects) or air resistance (complex-ion formation). For the AP exam, the qualitative model is the expected level of analysis; quantitative extensions are reserved for college-level analytical chemistry.

Three securely supported glass test tubes contrast a white gelatinous precipitate in the center with clear colorless liquid in the two outer tubes; separate pipettes are poised above those outer tubes. Aluminum hydroxide is amphoteric: acid dissolves it by forming water and hydrated dissolved aluminum, while excess strong base forms soluble Al(OH)₄⁻.

One solid, two dissolution routes. Aluminum hydroxide is amphoteric: acid dissolves it by forming water and hydrated dissolved aluminum, while excess strong base forms soluble Al(OH)₄⁻. Aluminum hydroxide is amphoteric: acid dissolves it by forming water and hydrated dissolved aluminum, while excess strong base forms soluble Al(OH)₄⁻.

SECTION 8

Connections to Advanced Theory

The pH–solubility relationship studied in AP Chemistry is a gateway to several more sophisticated topics encountered in upper-division and graduate courses. Understanding how equilibrium constants combine when multiple reactions occur simultaneously prepares you for simultaneous equilibrium calculations, complex-ion formation, and speciation diagrams (alpha plots) used in analytical and environmental chemistry.

AP Chemistry Level	Advanced / Graduate Level
Qualitative prediction: solubility increases or stays the same at lower pH.	Quantitative speciation: alpha (α) fraction plots showing distribution of all protonated forms as a function of pH.
Combine Ksp with 1/Ka to get an overall K.	Solve systems of mass balance, charge balance, and multiple equilibrium expressions simultaneously (e.g., using logarithmic concentration diagrams).
Assume ideal behavior: concentrations ≈ activities.	Apply Debye–Hückel or Davies equations to correct for non-ideal activity coefficients at high ionic strength.
pH affects dissolution only via anion protonation.	pH also affects complex-ion formation (e.g., Al(OH)₄⁻ in base), redox equilibria, and surface adsorption in environmental contexts.

AP-level treatment versus advanced analytical chemistry

One particularly elegant extension is the concept of conditional solubility product, K_sp′, which incorporates the fraction of the anion that exists in its fully deprotonated form at a given pH. At low pH, only a small fraction of total dissolved fluoride exists as F⁻ (the rest is HF), so the apparent solubility product is much larger than the thermodynamic K_sp. This formalism is central to gravimetric analysis, where analysts choose pH conditions to selectively precipitate one ion while keeping others in solution. Mastering the AP-level treatment now provides the conceptual scaffolding for these powerful techniques.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Which of the following sparingly soluble salts will have its molar solubility most significantly increased when dissolved in a solution of pH 2.0 compared to pure water?

A. AgCl
B. PbI₂
C. Ca₃(PO₄)₂
D. BaSO₄

Show answer (C)

C is correct. Ca₃(PO₄)₂ contains PO₄³⁻, the conjugate base of the weak triprotic acid H₃PO₄. In acidic solution, PO₄³⁻ undergoes multiple protonation steps (PO₄³⁻ → HPO₄²⁻ → H₂PO₄⁻ → H₃PO₄), consuming H⁺ and dramatically shifting the dissolution equilibrium to the right. A is incorrect because Cl⁻ is the conjugate base of HCl (a strong acid) and does not react with H⁺. B is incorrect because I⁻ is the conjugate base of HI (a strong acid) and is similarly unaffected by pH. D is incorrect because SO₄²⁻ derives from H₂SO₄, a strong acid in its first dissociation (K_a1 is very large), so SO₄²⁻ shows only negligible protonation at pH 2 (K_a2 for HSO₄⁻ is 0.012, so there is a minor effect, but far less than the triprotic phosphate system).

Question 2: BASIC CALCULATION

The K_sp of Mg(OH)₂ is 5.6 × 10⁻¹². What is the molar solubility of Mg(OH)₂ in a solution buffered at pH 9.00?

A. 5.6 × 10⁻² M
B. 1.1 × 10⁻⁴ M
C. 5.6 × 10⁻¹² M
D. 2.4 × 10⁻⁶ M

Show answer (A)

A is correct. Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2 OH⁻(aq). At pH 9.00, pOH = 14.00 − 9.00 = 5.00, so [OH⁻] = 1.0 × 10⁻⁵ M. Because the solution is buffered, [OH⁻] is fixed and does not change as the solid dissolves. K_sp = [Mg²⁺][OH⁻]², so [Mg²⁺] = K_sp / [OH⁻]² = (5.6 × 10⁻¹²) / (1.0 × 10⁻⁵)² = (5.6 × 10⁻¹²) / (1.0 × 10⁻¹⁰) = 5.6 × 10⁻² M. The molar solubility equals [Mg²⁺] = 0.056 M. B is incorrect; 1.1 × 10⁻⁴ M is the solubility in pure water, found by solving 4s³ = K_sp, i.e., s = (K_sp / 4)^(1/3) =

(1.4 × 10⁻¹²)^{1/3}

≈ 1.1 × 10⁻⁴ M. This approach ignores the buffered pH constraint. C is incorrect; 5.6 × 10⁻¹² M confuses K_sp itself with the molar solubility. D is incorrect; 2.4 × 10⁻⁶ M likely results from dividing K_sp by [OH⁻] instead of [OH⁻]², an exponent error in the calculation.

Question 3: INTERMEDIATE

Iron(III) hydroxide, Fe(OH)₃, has K_sp = 2.8 × 10⁻³⁹. A water treatment engineer wants to ensure that [Fe³⁺] remains below 1.0 × 10⁻⁶ M. What is the minimum pH that must be maintained?

A. 3.0
B. 5.0
C. 7.0
D. 3.2

Show answer (D)

D is correct. Fe(OH)₃(s) ⇌ Fe³⁺(aq) + 3 OH⁻(aq). K_sp = [Fe³⁺][OH⁻]³. Setting [Fe³⁺] = 1.0 × 10⁻⁶ M: [OH⁻]³ = K_sp/[Fe³⁺] = (2.8 × 10⁻³⁹)/(1.0 × 10⁻⁶) = 2.8 × 10⁻³³. [OH⁻] =

(2.8 × 10⁻³³)^{1/3}

≈ 1.41 × 10⁻¹¹ M. pOH = −log(1.41 × 10⁻¹¹) ≈ 10.85. pH = 14.00 − 10.85 ≈ 3.15. The minimum pH must be at or above 3.15 to keep [Fe³⁺] ≤ 1.0 × 10⁻⁶ M. Rounding up to one decimal place gives a minimum pH of 3.2. Verification: at pH 3.2, pOH = 10.8, [OH⁻] = 1.58 × 10⁻¹¹ M, [OH⁻]³ = 3.95 × 10⁻³³, and [Fe³⁺] = 2.8 × 10⁻³⁹/3.95 × 10⁻³³ = 7.1 × 10⁻⁷ M, which is below the 1.0 × 10⁻⁶ M limit. A is incorrect because pH 3.0 is below the threshold, giving [Fe³⁺] > 1.0 × 10⁻⁶ M. B and C are incorrect because at higher pH values, [Fe³⁺] would be far below the limit—those are sufficient but not the minimum pH required.

Question 4: APPLIED

10 pts

Calcium carbonate (CaCO₃) is the primary mineral in limestone. Its K_sp = 3.4 × 10⁻⁹. A geologist collects a groundwater sample buffered at pH 5.60 and finds it is saturated with CaCO₃. (a) Write the dissolution equilibrium for CaCO₃ and the relevant protonation reactions for CO₃²⁻ in acidic solution. (b) Explain, using Le Châtelier's principle, why CaCO₃ is more soluble in this mildly acidic groundwater than in neutral water. (c) The geologist observes that a nearby stream with pH 8.30 contains much less dissolved Ca²⁺. Is this observation consistent with the pH–solubility relationship? Justify your answer. (d) Propose one environmental consequence of increased CaCO₃ solubility due to acid rain (pH ≈ 4.2) on limestone formations.

Show answer

Part (a) — 3 points: • (1 pt) CaCO₃(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq) • (1 pt) CO₃²⁻(aq) + H⁺(aq) ⇌ HCO₃⁻(aq) • (1 pt) HCO₃⁻(aq) + H⁺(aq) ⇌ H₂CO₃(aq) [or CO₂(aq) + H₂O(l)] Part (b) — 3 points: • (1 pt) CO₃²⁻ is the conjugate base of the weak acid HCO₃⁻ (K_a2 of H₂CO₃ = 4.7 × 10⁻¹¹), so it reacts with H⁺. • (1 pt) Protonation of CO₃²⁻ removes it from the dissolution equilibrium. • (1 pt) By Le Châtelier's principle, the equilibrium shifts to the right (toward products), dissolving additional CaCO₃ and increasing [Ca²⁺]. Part (c) — 2 points: • (1 pt) Yes, this is consistent. At pH 8.30, [H⁺] is much lower than at pH 5.60, so less protonation of CO₃²⁻ occurs. • (1 pt) With less anion removal, the dissolution equilibrium is not shifted as far to the right, so less CaCO₃ dissolves and [Ca²⁺] is lower. Part (d) — 2 points: • (1 pt) Increased dissolution of CaCO₃ from limestone by acid rain leads to physical erosion, formation of sinkholes, and degradation of limestone monuments and buildings. • (1 pt) The dissolved Ca²⁺ and HCO₃⁻ enter waterways, increasing water hardness and potentially altering aquatic ecosystems by changing the buffering capacity and mineral content of the water.

Question 5: CRITICAL THINKING

10 pts

A student dissolves samples of three sparingly soluble salts—AgCl, PbF₂, and CaCO₃—in buffer solutions at five different pH values and measures the molar solubility. The data are shown in the table below. Table: Measured Molar Solubility (× 10⁻⁴ M) pH 2.0: AgCl = 1.33, PbF₂ = 8.75, CaCO₃ = 95.0 pH 4.0: AgCl = 1.33, PbF₂ = 4.20, CaCO₃ = 42.0 pH 7.0: AgCl = 1.33, PbF₂ = 2.60, CaCO₃ = 0.58 pH 10.0: AgCl = 1.33, PbF₂ = 2.58, CaCO₃ = 0.56 pH 12.0: AgCl = 1.33, PbF₂ = 2.58, CaCO₃ = 0.55 (a) For each salt, state whether its solubility is pH-dependent or pH-independent. Justify your answer by referencing specific data trends. (b) For the salts that show pH-dependent solubility, write the relevant protonation reaction(s) and explain why solubility increases at lower pH. (c) Explain why the solubility of CaCO₃ shows a much steeper increase from pH 7 to pH 2 than PbF₂ does. (d) A classmate claims that at very high pH (e.g., pH 14), the solubility of PbF₂ should decrease below its value at pH 7 due to the common ion effect from OH⁻. Evaluate this claim.

Show answer

Part (a) — 3 points: • (1 pt) AgCl: pH-independent. The data show a constant solubility of 1.33 × 10⁻⁴ M across all pH values. Cl⁻ is the conjugate base of HCl (a strong acid), so it does not react with H⁺. • (1 pt) PbF₂: pH-dependent. Solubility increases from 2.58 × 10⁻⁴ M at pH 10–12 to 8.75 × 10⁻⁴ M at pH 2. F⁻ is the conjugate base of the weak acid HF. • (1 pt) CaCO₃: pH-dependent. Solubility increases dramatically from ~0.55 × 10⁻⁴ M at pH 10–12 to 95.0 × 10⁻⁴ M at pH 2. CO₃²⁻ is the conjugate base of the weak diprotic acid H₂CO₃. Part (b) — 3 points: • (1 pt) For PbF₂: F⁻(aq) + H⁺(aq) → HF(aq). This removes F⁻ from solution, shifting PbF₂(s) ⇌ Pb²⁺(aq) + 2 F⁻(aq) to the right by Le Châtelier's principle. • (1 pt) For CaCO₃: CO₃²⁻(aq) + H⁺(aq) → HCO₃⁻(aq), and at lower pH, HCO₃⁻(aq) + H⁺(aq) → H₂CO₃(aq). • (1 pt) Each protonation step removes CO₃²⁻ from the dissolution equilibrium, shifting CaCO₃(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq) to the right and increasing [Ca²⁺]. Part (c) — 2 points: • (1 pt) CO₃²⁻ is a much stronger base than F⁻ (K_a2 for H₂CO₃ ≈ 4.7 × 10⁻¹¹ vs.

K_a

for HF ≈ 6.6 × 10⁻⁴), so CO₃²⁻ has a greater affinity for protons and is more completely removed from solution at any given acidic pH. • (1 pt) Additionally, CO₃²⁻ undergoes two protonation steps (CO₃²⁻ → HCO₃⁻ → H₂CO₃), consuming 2 mol H⁺ per mol CO₃²⁻, which provides a double driving force for dissolution compared to the single protonation of F⁻. Part (d) — 2 points: • (1 pt) The claim is incorrect as stated. OH⁻ is not a common ion with PbF₂, which dissociates into Pb²⁺ and F⁻, not OH⁻. The common ion effect requires the addition of an ion already present in the K_sp expression. • (1 pt) However, at very high pH, Pb²⁺ could form Pb(OH)₂ or Pb(OH)₃⁻ complexes (amphoteric behavior), which is a separate phenomenon from the common ion effect. The data show essentially no change from pH 10 to pH 12 (2.58 × 10⁻⁴ M), confirming that neither common-ion nor complex-ion effects are significant in this range for PbF₂. The classmate's reasoning (common ion from OH⁻) is fundamentally flawed.

SUMMARY

pH and Solubility — Summary

Summary

The solubility of a sparingly soluble salt depends on pH when its anion is the conjugate base of a weak acid. In acidic solution, H⁺ protonates the anion, removing it from the dissolution equilibrium. By Le Châtelier's principle, the equilibrium shifts to the right, and additional solid dissolves. Anions such as F⁻, OH⁻, CO₃²⁻, PO₄³⁻, and S²⁻ exhibit this behavior. By contrast, anions from strong acids (Cl⁻, Br⁻, I⁻, NO₃⁻) show pH-independent solubility because they do not react with H⁺.

Quantitatively, the overall equilibrium constant for dissolution in acid is K = Ksp × (1/Ka)ⁿ, where n is the stoichiometric coefficient of the anion. The weaker the parent acid (smaller K_a), the larger 1/K_a and the more dramatically solubility increases at low pH. On the AP exam, focus on identifying the anion's parent acid, writing the dissolution and protonation equilibria explicitly, and applying Le Châtelier's principle to justify the direction of the solubility change.