General Chemistry · Aqueous Equilibria · SOLUBILITY AND pH
Salt solubility and acidity — PbF₂ vs PbI₂
Does a slightly soluble salt dissolve better in an acidic solution? Put two lead salts, PbF₂ and PbI₂, side by side.
Change the pH and calculate the molar solubility with an exact mass-balance equation. Predict first.
Learning objectives After this activity you can …
decide, with reasons, which of PbF₂ and PbI₂ is the salt whose solubility changes in acid.identify the salt whose solubility changes in acid by asking whether its anion is the conjugate base of a weak acid or of a strong acid.
calculate the molar solubility at a given pH from Ksp and the fraction α(F⁻) = Ka/(Ka + [H⁺]).
explain which ion of the dissolution equilibrium the H⁺ of an acidic solution reacts with, and how the equilibrium shifts as a result.explain that H⁺ reacts with the anion (F⁻), not with the metal ion, and shifts the dissolution equilibrium to the right.
The solubility-product constant Ksp is fixed at a given temperature, but the molar solubility can change with other equilibria in the solution.
The ionization constant of a weak acid HA is Ka = [H⁺][A⁻]/[HA].An anion that is the conjugate base of a weak acid combines with H⁺ to form a weak-acid molecule, so the [H⁺] of the solution sets the free concentration of that anion.
To judge the molar solubility, first check whether the ions formed when the salt dissolves take part in other equilibria in the solution.Anions that are conjugate bases of strong acids (Cl⁻, Br⁻, I⁻, NO₃⁻) hardly combine with H⁺ in water.
Solution conditions
PbF₂ · pH 7.00
PbF₂(s) ⇌ Pb²⁺(aq) + 2F⁻(aq)
Ksp = 3.6×10⁻⁸ · Ka(HF) = 6.8×10⁻⁴
The molar solubility and species concentrations are shown after you submit your prediction.
Molar solubility s: moles of salt that dissolve in 1 L of saturated solution (mol/L). · Solubility-product constant Ksp: the equilibrium constant between the solid and its saturated solution (the solid is not written in the expression). ·
Conjugate base: the species left after an acid gives away H⁺ (HF → F⁻). · Fraction α(F⁻) = [F⁻] / ([F⁻] + [HF]) = Ka / (Ka + [H⁺]). ·
Buffered (fixed pH): a solution whose pH hardly changes even when acid is used up.
Predict Think first, before you observe
Saturated solutions of PbF₂ and PbI₂ are acidified from pH 7 to pH 1 (held by a buffer). What happens to the molar solubility of each salt?
Both PbF₂ and PbI₂ increase by tens of timesOnly PbF₂ increases by tens of times; PbI₂ stays almost the sameBoth PbF₂ and PbI₂ stay almost the sameOnly PbI₂ increases by tens of times; PbF₂ stays almost the same
ConfidenceMedium
s = —
Change the salt and the pH.
The explanation opens after you lock in a prediction
The molar solubility is found without approximation by putting the mass balance 2[Pb²⁺] = [F⁻] + [HF] and the Ka expression into the Ksp expression. The pH is assumed to be held by a buffer (why this is needed is explained in "Limits" after the prediction).
Why PbF₂ dissolves better in acid
F⁻ is the conjugate base of the weak acid HF, so it combines with H⁺ to form HF. When free [F⁻] drops, Q = [Pb²⁺][F⁻]² becomes smaller than Ksp and more solid dissolves.
H⁺ reacts with F⁻, not with Pb²⁺. Overall reaction: PbF₂(s) + 2H⁺ ⇌ Pb²⁺ + 2HF.
Ksp = s · (2s·α)² = 4s³α² , α = Ka / (Ka + [H⁺])
PbI₂ and AgCl do not depend on pH
I⁻ and Cl⁻ are conjugate bases of strong acids (HI, HCl), so they hardly combine with H⁺ (α = 1). Adding acid leaves the free anion concentration unchanged, and so the molar solubility is unchanged.
The more basic the anion (the smaller the Ka of its conjugate acid), the larger the effect of pH — check this with the Ka slider.
In soil science this is why the solubility of a heavy-metal salt in acidic soil depends on its anion, not only on the metal.
If [H⁺] ≫ Ka, s ∝ [H⁺]^(2/3)
Limits Limits of this model
The model uses 25 °C, activity coefficients of 1 (ideal solution) and a pH held by a buffer. At pH 1, the HF formed as PbF₂ dissolves uses up about 0.12 M of H⁺, so 0.1 M strong acid alone cannot keep the pH at 1 (this is why a buffer must be assumed).
Ion pairs such as PbF⁺, hydrolysis of Pb²⁺ and activity effects in concentrated solutions (s > 0.1 M) are not included — values in that range are model values.
Ksp values are from Martell and Smith, Critical Stability Constants (PbF₂ 3.6×10⁻⁸ = reference value for this topic, CaF₂ 3.9×10⁻¹¹, AgCl 1.8×10⁻¹⁰). Values that differ by more than 5% from the CRC Handbook are listed in the teacher layer. For PbI₂ the two sources differ by 24%, so its value is shown but not graded.
Check Formative check — pass all items to complete
Q1—
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Q2 · Task—
Now: — · Goal: —
Q3—
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Q4 · Fill in the blank—
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Q5 · Apply (advanced, not required for completion)In a buffer held at pH 2.00, what is the molar solubility of CaF₂ (Ksp 3.9×10⁻¹¹) in units of ×10⁻³ M? (Ka(HF) 6.8×10⁻⁴, one decimal place)
Reflect Reflect — metacognition · completion
My prediction result: Not predicted yet
Explain in one sentence why the idea "adding acid makes all slightly soluble salts dissolve better" is wrong. Use a concept word (conjugate base, HF, Ksp, equilibrium) and a reason word ("because", "so"):
Apply Use the idea in a new context (not required for completion)
Tooth enamel is mainly calcium hydroxyapatite, Ca₅(PO₄)₃OH. It is weakened by acidic food, while Ca₅(PO₄)₃F, formed by fluoride treatment, is weakened less. Explain this in one or two sentences in terms of how basic the anions are (the strength of their conjugate acids).
Next up: dissolving AgCl by complex-ion formation (Kf) → predicting precipitation on mixing (Q and Ksp).
COMPLETE · Completed — all checks passed.
Worksheet — Salt solubility and acidity
Name ________ Class ________ Date ________
1. Predict — From pH 7 to pH 1 (buffered), what happens to the molar solubility of PbF₂ and PbI₂? (both increase / only PbF₂ increases / both stay the same / only PbI₂ increases) Reason for your choice:
2. Calculate — Find the molar solubility of PbF₂ (Ksp 3.6×10⁻⁸, Ka(HF) 6.8×10⁻⁴) at pH 1.00 (buffered).
Step
Equation
Value
α(F⁻)
Ka / (Ka + [H⁺])
Ksp expression
Ksp = s · (2s·α)²
s
(Ksp / 4α²)^(1/3)
s(pH 1) = ________ M, s(pH 7) = ________ M, ratio = ______ times
3. Observe — On the axes below (log vertical scale), sketch the molar solubility vs pH curves of PbF₂, CaF₂, PbI₂ and AgCl. Mark which salts change with pH and which do not.
4. Fill in the blank — PbF₂(s) + 2H⁺(aq) ⇌ Pb²⁺(aq) + 2[＿＿＿](aq)
Answer key (teacher only · not in the student printout) — 1. only PbF₂ increases (about 28 times), PbI₂ unchanged ·
2. α = 6.75×10⁻³, s(pH 1) = 5.8×10⁻² M, s(pH 7) = 2.1×10⁻³ M, ratio about 28 ·
3. the fluorides (PbF₂, CaF₂) rise with slope −2/3 below about pH 4; PbI₂ and AgCl are flat · 4. HF
Teacher notes — open the lesson plan
Lesson plan · 5E phase and time
Explain and Elaborate, 15–20 min. After the common-ion effect, before complex-ion formation. Whole-class demonstration (?mode=teacher) or individual work.
Demonstration in 3 steps
① PbF₂ at pH 7 → s 2.1×10⁻³ M. ② pH 1 (buffered) → 5.8×10⁻² M, about 28 times. ③ Switch to PbI₂ and repeat → a flat line.
Three key questions
"Does H⁺ react with Pb²⁺ or with F⁻?"
"Ksp stays the same. How can the molar solubility increase?"
"Why is a buffer needed to keep the pH at 1?"
Common misconceptions
MC-pHsol-01 "acid dissolves every salt better" → ignores how basic the anion is. PbI₂ and AgCl do not depend on pH.
MC-pHsol-02 "H⁺ reacts with the metal ion" → ions of the same charge do not combine. Correct it with F⁻ + H⁺ ⇌ HF.
Calculation method (E2 exact equations)
At fixed pH, combining Ksp = [M²⁺][F⁻]², [HF] = [H⁺][F⁻]/Ka and 2[M²⁺] = [F⁻] + [HF] gives the exact closed form Ksp = 4s³α² (no approximation). For AgCl s = √Ksp; for PbI₂ s = (Ksp/4)^(1/3).
Buffer assumption: at pH 1 the HF formed, 2s(1 − α) ≈ 0.116 M, uses up H⁺, so 0.1 M HCl alone cannot keep the pH. Students see this in the "H⁺ used up" line on screen.
Sources
Ka(HF) 6.8×10⁻⁴: common reference value (Martell and Smith).
CaF₂ Ksp 3.9×10⁻¹¹ and AgCl 1.80×10⁻¹⁰: common reference values. PbF₂ and PbI₂: Martell and Smith.
Literature check (2 sources: Martell and Smith, Critical Stability Constants Vol. 1 and 4 (tables in Harvey, Analytical Chemistry 2.1) · CRC Handbook 84th ed. (LibreTexts Tables E1 and E3), checked 2026-10-03)
PbF₂: 3.6×10⁻⁸ (Martell–Smith) / 3.3×10⁻⁸ (CRC), 8% difference, both listed — with the CRC value, s at pH 7 is 2.0×10⁻³ M.
HF Ka: 6.8×10⁻⁴ (Martell–Smith) / 6.3×10⁻⁴ (CRC), 7% difference, both listed — with the CRC value the ratio becomes 29 instead of 28.