General Chemistry Ii Jasperse Acid Base Chemistry Extra Practice Problems

General Chemistry II Jasperse Acid‑Base Chemistry Extra Practice Problems

Acid‑base chemistry is a cornerstone of General Chemistry II, and mastering it requires both conceptual understanding and plenty of problem‑solving practice. The Jasperse approach—known for its clear step‑by‑step methodology and emphasis on equilibrium thinking—provides an effective framework for tackling everything from simple pH calculations to complex polyprotic titrations. Below you will find a concise review of the essential concepts, a set of extra practice problems modeled after Jasperse‑style questions, detailed solutions, and strategic tips to help you build confidence and speed.

Key Concepts in Acid‑Base Chemistry (Jasperse Style)

Before diving into the practice set, refresh the core ideas that repeatedly appear in Jasperse’s problem sets:

• Brønsted‑Lowry definition: Acids donate protons (H⁺); bases accept them. Every acid‑base reaction involves a conjugate acid‑base pair.
• Strength classification: Strong acids/bases dissociate completely in water (e.g., HCl, NaOH). Weak acids/bases only partially dissociate; their equilibrium is described by Kₐ or K_b.
• pH and pOH: pH = –log[H⁺]; pOH = –log[OH⁻]; at 25 °C, pH + pOH = 14.
• Acid dissociation constant (Kₐ): For HA ⇌ H⁺ + A⁻, Kₐ = [H⁺][A⁻]/[HA]. Smaller Kₐ → weaker acid.
• Base dissociation constant (K_b): Related by Kₐ·K_b = K_w (1.0×10⁻¹⁴ at 25 °C).
• Polyprotic acids: Dissociate in steps (H₂A ⇌ H⁺ + HA⁻, HA⁻ ⇌ H⁺ + A²⁻). Each step has its own Kₐ₁, Kₐ₂, etc.
• Buffer solutions: Resist pH change; composed of a weak acid and its conjugate base (or weak base/conjugate acid). Henderson‑Hasselbalch equation: pH = pKₐ + log([A⁻]/[HA]).
• Titration curves: Plot pH vs. volume of titrant. Key points: initial pH, buffer region, half‑equivalence point (pH = pKₐ), equivalence point, and post‑equivalence region.
• Salt hydrolysis: Anions of weak acids (e.g., acetate) and cations of weak bases (e.g., NH₄⁺) react with water to affect pH.

These concepts form the backbone of the Jasperse extra practice problems that follow.

Extra Practice Problems

Problem 1 – Simple Strong Acid/Base pH

Calculate the pH of a 0.025 M solution of nitric acid (HNO₃), assuming complete dissociation.

Solution
HNO₃ is a strong acid → [H⁺] = 0.025 M.
pH = –log(0.025) = –log(2.5×10⁻²) = 1.60 (rounded to two decimal places).

Problem 2 – Weak Acid pH Using Approximation

Find the pH of a 0.10 M solution of acetic acid (CH₃COOH, Kₐ = 1.8×10⁻⁵).

Solution
Set up ICE table:

Kₐ = x²/(0.10 – x) ≈ x²/0.10 (since x ≪ 0.10).
x² = Kₐ·0.10 = (1.8×10⁻⁵)(0.10) = 1.8×10⁻⁶ → x = √(1.8×10⁻⁶) = 1.34×10⁻³ M.

Check approximation: x/0.10 = 0.013 → 1.3 % (<5 %), acceptable. pH = –log(1.34×10⁻³) = 2.87.

Problem 3 – Buffer Preparation (Henderson‑Hasselbalch)

You need 100 mL of a phosphate buffer at pH 7.20 using Na₂HPO₄ (base) and NaH₂PO₄ (acid). The pKₐ of H₂PO₄⁻/HPO₄²⁻ is 7.20. How many grams of each component are required? (Molar masses: Na₂HPO₄ = 141.96 g/mol, NaH₂PO₄ = 119.98 g/mol)

Solution
When pH = pKₐ, the Henderson‑Hasselbalch equation gives log([base]/[acid]) = 0 → [base] = [acid].
Thus, equal molar amounts are needed.

Let the concentration of each be C (M). Total volume = 0.100 L → moles of each = C·0.100 L.

Choose a convenient total buffer concentration, e.g., 0.10 M (common for lab buffers). Then each component is 0.05 M.

Moles of each = 0.05 mol/L × 0.100 L = 0.0050 mol.

Mass Na₂HPO₄ = 0.0050 mol × 141.96 g/mol = 0.71 g.
Mass NaH₂PO₄ = 0.0050 mol × 119.98 g/mol = 0.60 g.

Dissolve both in water and dilute to 100 mL.

Problem 4 – Polyprotic Acid Titration

A 2

0.00 mL sample of 0.10 M H₃PO₄ is titrated with 0.10 M NaOH. Calculate the pH at the first equivalence point (where all H₃PO₄ is converted to H₂PO₄⁻).

Solution
H₃PO₄ is triprotic; the first equivalence point occurs after adding 20.00 mL of 0.10 M NaOH (equal moles of acid and base). At this point, the solution contains only H₂PO₄⁻, the amphoteric intermediate.

H₂PO₄⁻ can act as both an acid and a base. Its Kₐ₁ = 7.5×10⁻³ and Kₐ₂ = 6.2×10⁻⁸. For the amphiprotic ion, the pH at the first equivalence point can be approximated by:

pH ≈ (pKₐ₁ + pKₐ₂)/2

pKₐ₁ = –log(7.5×10⁻³) = 2.12
pKₐ₂ = –log(6.2×10⁻⁸) = 7.21

pH ≈ (2.12 + 7.21)/2 = 4.67

Problem 5 – Salt Hydrolysis and pH

Calculate the pH of a 0.20 M solution of ammonium acetate (NH₄CH₃COO). Given: Kₐ (NH₄⁺) = 5.6×10⁻¹⁰, K_b (CH₃COO⁻) = 5.6×10⁻¹⁰.

Solution
Both ions hydrolyze:

NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺ (acidic)
CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ (basic)

Since Kₐ(NH₄⁺) = K_b(CH₃COO⁻), the solution is neutral at 25°C. The hydrolysis constants are equal, so the contributions to [H⁺] and [OH⁻] are equal, yielding pH = 7.00.

Problem 6 – Titration Curve Analysis

A 25.00 mL sample of 0.100 M formic acid (pKₐ = 3.75) is titrated with 0.100 M NaOH. Determine:

a) The pH at the half-equivalence point.
b) The pH at the equivalence point.
c) The volume of NaOH required to reach the equivalence point.

Solution
a) At half-equivalence, [HA] = [A⁻], so pH = pKₐ = 3.75.
b) At equivalence, all formic acid is converted to formate ion (HCOO⁻). The solution contains 0.050 mol formate in 50.00 mL (0.100 M). Formate is a weak base:

K_b = K_w / K_a = 1.0×10⁻¹⁴ / 1.78×10⁻⁴ = 5.62×10⁻¹¹

Using ICE table for HCOO⁻ + H₂O ⇌ HCOOH + OH⁻:

K_b = x² / (0.100 – x) ≈ x² / 0.100 → x = √(K_b·0.100) = √(5.62×10⁻¹²) = 7.50×10⁻⁶ M = [OH⁻]

pOH = –log(7.50×10⁻⁶) = 5.12 → pH = 14.00 – 5.12 = 8.88
c) Moles of formic acid = 0.02500 L × 0.100 M = 0.00250 mol. Same moles of NaOH needed. Volume NaOH = 0.00250 mol / 0.100 M = 0.0250 L = 25.00 mL.

Problem 7 – Mixed Solution pH

Calculate the pH of a solution containing 0.050 M acetic acid (Kₐ = 1.8×10⁻⁵) and 0.020 M sodium acetate.

Solution
This is a buffer. Use Henderson-Hasselbalch:

pH = pKₐ + log([A⁻]/[HA])
pKₐ = –log(1.8×10⁻⁵) = 4.74
pH = 4.74 + log(0.020/0.050) = 4.74 + log(0.40) = 4.74 – 0.398 = 4.34

Problem 8 – Dilution Effect on pH

A 0.10 M solution of a weak acid (Kₐ = 1.0×10⁻⁴) has pH = 2.87. What is the pH after diluting the solution 10-fold?

Solution
Initial [H⁺] = 10⁻²·⁸⁷ = 1.35×10⁻³ M.
After 10-fold dilution, [HA] = 0.010 M. Let x = [H⁺] at new equilibrium.

Kₐ = x² / (0.010 – x) ≈ x² / 0.010 (since x ≪ 0.010)
x² = Kₐ·0.010 = (1.0×10⁻⁴)(0.010) = 1.0×10⁻⁶ → x = 1.0×10⁻³ M

pH = –log(1.0×10⁻³) = 3.00

Problem 9 – Common Ion Effect

A 0.10 M solution of acetic acid (Kₐ = 1.8×10⁻⁵) has pH = 2.