U7.13 · pH and SolubilityPractice · pH 与溶解度
Lessons covered:U7.6 Properties of the Equilibrium ConstantU7.13 pH and Solubility
- 1Select all
Which of these slightly soluble salts become noticeably more soluble when a strong acid is added? Select all that apply.
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ExplanationWhen the anion is a base (OH⁻) or the conjugate base of a weak acid (CO₃²⁻, PO₄³⁻), H⁺ combines with it, lowering its concentration so Q < Ksp and more salt dissolves. I⁻ and Br⁻ are conjugate bases of the strong acids HI and HBr, so they barely combine with H⁺, and those salts’ solubility hardly depends on pH.
Review: U7.13 pH and Solubility
- 2Numeric
What is the molar solubility of Fe(OH)₃ (Ksp = 4 × 10−38) in a buffer at pH = 3.00?
× 10MExplanationpH = 3.00, so pOH = 11.00 and [OH⁻] = 1 × 10−11 M, held fixed by the buffer. [Fe³⁺] = s = Ksp ÷ [OH⁻]³ = 4 × 10−38 ÷ (1 × 10−11)³ = 4 × 10−5 M. It is almost insoluble in neutral water, but the more acidic the solution, the more dissolves, which is why acid removes rust.
Review: U7.13 pH and Solubility
- 3Multiple choice
A marble countertop (mostly CaCO₃) is etched where vinegar is spilled. Which is the best explanation?
ExplanationCaCO₃(s) ⇌ Ca²⁺ + CO₃²⁻. CO₃²⁻ is the conjugate base of a weak acid, so the H⁺ from acetic acid uses it up and the equilibrium keeps shifting toward dissolving, while CO₂ escapes. Acetic acid is weak, calcium acetate is soluble, and Ksp changes only with temperature.
Review: U7.13 pH and Solubility
- 4Multiple choice
Tooth enamel is mostly hydroxyapatite, Ca₅(PO₄)₃OH. Fluoride toothpaste converts some of it to fluorapatite, Ca₅(PO₄)₃F, making teeth more acid-resistant. Which is the best explanation?
ExplanationAcid in the mouth (H⁺) combines with the OH⁻ and PO₄³⁻ in hydroxyapatite, dissolving enamel. F⁻ is the conjugate base of HF and far less basic than OH⁻, so H⁺ hardly combines with it, and fluorapatite dissolves less in acid. F⁻ doesn’t neutralize acid, and fluorapatite still contains phosphate.
Review: U7.13 pH and Solubility
- 5Numeric
At 25 °C, Ksp = 8.9 × 10−12 for Mg(OH)₂ and Kw = 1.0 × 10−14. Find K for Mg(OH)₂(s) + 2H⁺(aq) ⇌ Mg²⁺(aq) + 2H₂O(l).
× 10ExplanationAdd Mg(OH)₂(s) ⇌ Mg²⁺ + 2OH⁻ (Ksp) and 2 × [H⁺ + OH⁻ ⇌ H₂O] (each with K = 1/Kw); the 2OH⁻ cancel. K = Ksp × (1/Kw)² = 8.9 × 10−12 ÷ (1.0 × 10−14)² = 8.9 × 1016. K is huge, so Mg(OH)₂ dissolves readily in acid, which is how it works as an antacid.
Review: U7.13 pH and SolubilityU7.6 Properties of the Equilibrium Constant
- 6Multiple choice
The graph shows how the solubilities of two slightly soluble salts, X (CaF₂) and Y (AgCl), change with pH (each relative to its solubility at pH 7). Below about pH 4, X becomes much more soluble but Y does not. Why?
ExplanationHF is a weak acid (Ka ≈ 6.4 × 10⁻⁴), so when the pH approaches or drops below its pKa (about 3.2), most F⁻ is captured by H⁺ as HF. Lower [F⁻] shifts CaF₂(s) ⇌ Ca²⁺ + 2F⁻ to the right, so the solubility rises: about 6.5 times its pH 7 value at pH 2. HCl is a strong acid, so Cl⁻ hardly combines with H⁺ and AgCl’s solubility doesn’t change with pH. Ca²⁺ doesn’t react with H⁺, and AgCl’s Ksp is actually much smaller than CaF₂’s.
Review: U7.13 pH and Solubility
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