pH and SolubilitypH 与溶解度

Big Idea · Scale, Proportion, and Quantity (SPQ)AP Science Practice 6 · ArgumentationAP link · Unit 7~40 min

Learning goals

  • Identify which slightly soluble salts become more soluble as pH decreases (those whose anion is OH⁻ or the conjugate base of a weak acid) and which are essentially unaffected by pH
  • Use Le Châtelier's principle and the comparison of Q with Ksp to explain why an acid dissolves such salts
  • Calculate the solubility of a metal hydroxide in a buffered solution, and combine K values to find K for a “salt + acid” reaction

Before you start

  • pH and pOH (2.12.3)
  • Properties of the equilibrium constant (U7.6)
  • Common-ion effect (U7.12)

Which slightly soluble salts does acid dissolve?

If the anion of a slightly soluble salt can combine with H⁺, the salt becomes more soluble as pH decreases:

  • OH⁻: H⁺ + OH⁻ → H₂O. Hydroxides such as Mg(OH)₂ and Fe(OH)₃ dissolve in acid.
  • Conjugate bases of weak acids, such as F⁻, CO₃²⁻, and PO₄³⁻: H⁺ turns them into HF, HCO₃⁻ (or H₂CO₃ and CO₂), HPO₄²⁻, and so on, so CaF₂, CaCO₃, and Ca₃(PO₄)₂ dissolve in acid.
  • Conjugate bases of strong acids: Cl⁻, Br⁻, and I⁻ barely combine with H⁺, so the solubility of AgCl, AgBr, and PbI₂ hardly changes with pH.

The reasoning is the reverse of the common-ion effect: H⁺ uses up the anion produced by dissolving, making Q < Ksp, so the equilibrium shifts toward dissolving and more solid dissolves. Tooth decay (acid dissolving the hydroxyapatite in enamel) and acid rain eating away limestone both work this way.

See how the solubility of several salts changes with pH (the pH is held fixed by a buffer):

Try it: how does pH affect solubility?

Example 1: Mg(OH)₂ in buffered solutions

At 25 °C, Ksp for Mg(OH)₂ is 8.9 × 10⁻¹². Find its molar solubility in buffers at pH 9.00 and at pH 11.00.

Step-by-step solution

  1. Write the equilibrium and Ksp.

    Mg(OH)X2(s)⇌MgX2++2 OHX−Ksp=[MgX2+][OHX−]2\begin{gathered} \ce{Mg(OH)2(s) <=> Mg^2+ + 2OH-} \\ K_{sp} = [\ce{Mg^2+}][\ce{OH-}]^2 \end{gathered}

  2. Approach: the buffer fixes [OH⁻]. Any OH⁻ released by dissolving is absorbed by the buffer, so the pH, and therefore [OH⁻], stays at a known value. Every Mg(OH)₂ that dissolves becomes one Mg²⁺, so s = [Mg²⁺]:

    s=[MgX2+]=Ksp[OHX−]2s = [\ce{Mg^2+}] = \frac{K_{sp}}{[\ce{OH-}]^2}

  3. [OH⁻] at pH 9.00 (in M).

    pOH=14.00−9.00=5.00[OHX−]=10−5.00=1.0×10−5\begin{aligned} \text{pOH} &= 14.00 - 9.00 = 5.00 \\ [\ce{OH-}] &= 10^{-5.00} = 1.0 \times 10^{-5} \end{aligned}

  4. Solubility at pH 9.00.

    s=8.9×10−12(1.0×10−5)2=8.9×10−121.0×10−10=0.089 M\begin{aligned} s &= \frac{8.9 \times 10^{-12}}{(1.0 \times 10^{-5})^2} = \frac{8.9 \times 10^{-12}}{1.0 \times 10^{-10}} \\ &= \boxed{0.089\ \text{M}} \end{aligned}

  5. At pH 11.00. pOH = 3.00, so [OH⁻] = 1.0 × 10⁻³ M:

    s=8.9×10−12(1.0×10−3)2=8.9×10−121.0×10−6=8.9×10−6 M\begin{aligned} s &= \frac{8.9 \times 10^{-12}}{(1.0 \times 10^{-3})^2} = \frac{8.9 \times 10^{-12}}{1.0 \times 10^{-6}} \\ &= \boxed{8.9 \times 10^{-6}\ \text{M}} \end{aligned}

  6. Compare.

    0.0898.9×10−6=1.0×104\frac{0.089}{8.9 \times 10^{-6}} = 1.0 \times 10^{4}

    Lowering the pH by just 2 raises the solubility 10⁴-fold: [OH⁻] becomes 1/100 as large, and because [OH⁻] is squared in Ksp, (1/100)² = 1/10⁴.

Example 2: K for CaF₂ dissolving in acid

At 25 °C, Ksp for CaF₂ is 4.0 × 10⁻¹¹ and Ka for HF is 6.4 × 10⁻⁴ (OpenStax). Find K for CaF₂(s) + 2H⁺(aq) ⇌ Ca²⁺(aq) + 2HF(aq).

Step-by-step solution

  1. Split the target into two steps. First the solid dissolves, then F⁻ picks up H⁺. Adding the steps cancels the intermediate F⁻.

  2. Step 1: dissolving.

    CaFX2(s)⇌CaX2++2 FX−K1=Ksp=4.0×10−11\begin{gathered} \ce{CaF2(s) <=> Ca^2+ + 2F-} \\ K_1 = K_{sp} = 4.0 \times 10^{-11} \end{gathered}

  3. Step 2: F⁻ picks up H⁺. This is the ionization of HF (HF ⇌ H⁺ + F⁻) reversed and multiplied by 2. Reversing takes the reciprocal of K; multiplying by 2 squares it:

    2 HX++2 FX−⇌2 HFK2=(1Ka)2=1(6.4×10−4)2\begin{gathered} \ce{2H+ + 2F- <=> 2HF} \\ K_2 = \left(\frac{1}{K_a}\right)^2 = \frac{1}{(6.4 \times 10^{-4})^2} \end{gathered}

  4. Calculate K₂.

    K2=14.10×10−7=2.44×106K_2 = \frac{1}{4.10 \times 10^{-7}} = 2.44 \times 10^{6}

  5. Add the steps and multiply the K values. 2F⁻ cancels, leaving exactly the target (U7.6):

    K=K1×K2=(4.0×10−11)(2.44×106)=9.8×10−5\begin{aligned} K &= K_1 \times K_2 \\ &= (4.0 \times 10^{-11})(2.44 \times 10^{6}) \\ &= \boxed{9.8 \times 10^{-5}} \end{aligned}

  6. Meaning. K is about 10⁶ times larger than Ksp: in acid, H⁺ keeps “taking away” F⁻ as HF, so the dissolving equilibrium keeps shifting forward and much more CaF₂ dissolves.

Practice: each slightly soluble salt is placed in dilute nitric acid. Compared with pure water, its solubility:
  1. Mg(OH)₂
  2. AgCl
  3. CaCO₃
  4. AgBr
  5. CaF₂
  6. Fe(OH)₃
  7. PbI₂
  8. Ag₃PO₄

Common mistakes

Mistake 1: “Every slightly soluble salt dissolves in acid.” Only those whose anion can combine with H⁺. AgCl, AgBr, and AgI are about as soluble in dilute acid as in water.

Mistake 2: “Adding acid increases Ksp.” Ksp changes only with temperature. The acid uses up the anion in solution, making Q smaller, so more solid dissolves; Ksp itself is unchanged.

Mistake 3: “Only hydroxides have pH-dependent solubility.” Whenever the anion is the conjugate base of any weak acid (F⁻, CO₃²⁻, PO₄³⁻, and so on), the solubility increases as pH decreases.

Quick check

Which of these slightly soluble salts is much more soluble in 0.1 M HNO₃ than in pure water?

How many times as soluble is Mg(OH)₂ in a buffer at pH 10.00 as in a buffer at pH 12.00?

What is the net ionic equation for marble (CaCO₃) reacting with hydrochloric acid?

The solubility of AgCl hardly changes with pH. What is the best explanation?

Mg(OH)₂ has Ksp = 8.9 × 10⁻¹² (OpenStax), and Kw = 1.0 × 10⁻¹⁴. Which expression gives K for Mg(OH)₂(s) + 2H⁺(aq) ⇌ Mg²⁺(aq) + 2H₂O(l)?

Tooth enamel is mostly hydroxyapatite, Ca₅(PO₄)₃OH. Why can frequent acidic foods damage teeth?

AP-style free response

AP-style free response

At 25 °C, Ksp for Cu(OH)₂ is 2.2 × 10⁻²⁰ (OpenStax). Wastewater containing Cu²⁺ is often treated by adding base to raise the pH so that the Cu²⁺ precipitates as Cu(OH)₂.

(a) Write the Ksp expression for Cu(OH)₂.

(b) Calculate the molar solubility of Cu(OH)₂ in buffers at pH 6.00 and at pH 8.00.

(c) Use the idea of a shifting equilibrium to explain why more Cu(OH)₂ dissolves at lower pH.

(d) To bring [Cu²⁺] in the wastewater below 1.0 × 10⁻⁶ M, what is the minimum pH?

(e) Could the same approach (adjusting the pH) remove Ag⁺ from wastewater as AgCl? Justify your answer.

Model answer and scoring

(a) Ksp = [Cu²⁺][OH⁻]².

(b) pH 6.00: [OH⁻] = 1.0 × 10⁻⁸ M, so s = 2.2 × 10⁻²⁰ ÷ (1.0 × 10⁻⁸)² = 2.2 × 10⁻⁴ M. pH 8.00: [OH⁻] = 1.0 × 10⁻⁶ M, so s = 2.2 × 10⁻²⁰ ÷ (1.0 × 10⁻⁶)² = 2.2 × 10⁻⁸ M.

(c) Cu(OH)₂(s) ⇌ Cu²⁺ + 2OH⁻. At low pH, [OH⁻] is small (H⁺ combines with OH⁻), so the equilibrium shifts toward dissolving and more Cu(OH)₂ dissolves.

(d) [OH⁻] ≥ √(2.2 × 10⁻²⁰ ÷ 1.0 × 10⁻⁶) = 1.5 × 10⁻⁷ M, so pOH ≤ 6.83 and pH ≥ 7.17: above about 7.2.

(e) No. Cl⁻ is the conjugate base of a strong acid, so the solubility of AgCl doesn’t depend on pH; to precipitate Ag⁺, add Cl⁻ (the common-ion effect).

Scoring: (a) the expression; (b) [OH⁻] from pH, then s; (c) the equilibrium shift; (d) working back from Ksp to [OH⁻] and pH; (e) conjugate bases of strong acids are unaffected by pH.

Summary

  • Slightly soluble salts whose anion is OH⁻ or the conjugate base of a weak acid become more soluble as pH decreases; those whose anion is the conjugate base of a strong acid (Cl⁻, Br⁻, I⁻) are essentially unaffected.
  • The reason: H⁺ uses up the anion, making Q < Ksp, so the equilibrium shifts toward dissolving; Ksp is unchanged.
  • For a hydroxide in a buffer, s = Ksp ÷ [OH⁻]ⁿ; for “salt + acid,” K = Ksp ÷ Kaⁿ.

AP tip: To decide whether pH affects a salt’s solubility, look at the anion: is its conjugate acid weak? If so, pH matters. In explanations, write the equation for H⁺ combining with the anion, then say which way the equilibrium shifts.

Practice this chapter: Unit 7 · Equilibrium · 12 questionsPractice this topic: U7.13 · pH and Solubility · 6 questions

Found a mistake in this lesson, something unclear, or have a suggestion?

Feedback / Report a problem

Type

We only receive what you write, the page you’re on and any contact you choose to leave. No names or other personal data (to stop spam we keep a one-way hash of your IP, never the IP itself).