Unit 8 · Acids and BasesPractice · 酸与碱

pH = 14.00 − 3.40 = 10.60, so [H₃O⁺] = 10^−10.60 = 2.5 × 10⁻¹¹ M. Or find [OH⁻] = 10^−3.40 first and use Kw ÷ [OH⁻]. Two decimal places in pH means two significant figures in the concentration.

n(H⁺) = 30.0 × 0.100 = 3.00 mmol. Each Sr(OH)₂ gives 2 OH⁻, so n(OH⁻) = 2 × 20.0 × 0.100 = 4.00 mmol. OH⁻ is in excess by 1.00 mmol in 50.0 mL, so [OH⁻] = 0.0200 M, pOH = 1.70, and pH = 12.30. Forgetting the factor of 2 wrongly leaves the acid in excess.

HNO₂ ⇌ H⁺ + NO₂⁻, x² ÷ (0.15 − x) = 4.6 × 10⁻⁴. Shortcut: x ≈ √(4.6 × 10⁻⁴ × 0.15) = 8.31 × 10⁻³, which is 5.5%, just over 5%, so solve the quadratic: x = 8.08 × 10⁻³ and pH = 2.09. The shortcut gives 2.08, only 0.01 off, which is within tolerance.

Both are 5.0 mmol, so HNO₂ + OH⁻ → NO₂⁻ + H₂O goes to completion, leaving 0.10 M KNO₂. NO₂⁻ is the conjugate base of a weak acid, with K_b = K_w ÷ K_a = 2.2 × 10⁻¹¹, so the solution is weakly basic (pH ≈ 8.17). A weak acid exactly neutralized by a strong base is not neutral, and with no HNO₂ left it isn’t a buffer.

At the equivalence point all the HCN has become CN⁻, which reacts with water to make OH⁻, so pH > 7 (HCN is very weak, so CN⁻ is a fairly strong base and the pH is quite high). pH = pKa holds at the half-equivalence point, where HCN and CN⁻ are equal.

Acid strength depends on how stable the conjugate base is. In C₆H₅O⁻ the negative charge spreads over several ring carbons by resonance, making it more stable than C₂H₅O⁻, where the charge sits on one O atom, so phenol gives up H⁺ more easily. A stronger O–H bond would make an acid weaker, and the ring H atoms don’t ionize.

pH = pKa + log([A⁻]/[HA]), so log of the ratio = 5.74 − 4.74 = 1.00 and the ratio = 10^1.00 = 10. One pH unit above the pKa, the base form (CH₃COO⁻) is 10 times the acid form.

A buffer needs a weak conjugate acid–base pair: HCO₃⁻ (a weak acid) and CO₃²⁻ (its conjugate base) can neutralize both added OH⁻ and added H⁺. HCl and HNO₃ are strong acids, and Cl⁻ and NO₃⁻ are essentially not basic, so they can’t neutralize added acid; NaOH with NaCl has no weak acid to neutralize added base.

pH = pKa + log([NO₂⁻]/[HNO₂]) = 3.34 + log(0.40/0.25) = 3.34 + 0.20 = 3.54. There is more base form than acid form, so the pH is a little above the pKa.

Added H⁺ is neutralized by the base form: NH₃ + H⁺ → NH₄⁺. How much HCl can be handled depends on the moles of NH₃: 0.50 mol in the first, and 0.050 mol in each of the others (and the fourth starts with far less NH₃ than NH₄⁺). Buffer capacity depends on moles, not just concentration.

HCl ionizes completely, so [H⁺] = 0.10 M and pH = 1.00. HCN is very weak: [H⁺] ≈ √(4.9 × 10⁻¹⁰ × 0.10) = 7.0 × 10⁻⁶ M, pH ≈ 5.15. Same concentration, hugely different ionization. Forgetting the square root gives 10.31.

NH₄⁺ is the conjugate acid of NH₃: K_a = K_w ÷ K_b = 1.0 × 10⁻¹⁴ ÷ 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰. NH₄⁺ ⇌ NH₃ + H⁺, so [H⁺] ≈ √(5.6 × 10⁻¹⁰ × 0.20) = 1.1 × 10⁻⁵ M and pH = 4.98. Cl⁻ is the conjugate base of a strong acid and doesn’t affect the pH.