General Chemistry Ii Jasperse Acid Base Chemistry Extra Practice Problems

Mastering Acid-Base Chemistry: Extra Practice Problems for General Chemistry II

Acid-base chemistry forms the cornerstone of General Chemistry II, moving far beyond simple pH calculations into the intricate dance of complex equilibria, buffer systems, and titration curves. For many students, this is the first topic where conceptual understanding and mathematical precision must meet seamlessly. This collection of extra practice problems is designed to bridge that gap, targeting the challenging concepts that often appear on exams and in advanced coursework. By working through these varied scenarios—from weak acid-strong base titrations to polyprotic acid systems—you will build the analytical toolkit needed to confidently tackle any acid-base question. The goal is not just to find an answer, but to understand the why behind every calculation, transforming uncertainty into mastery.

Core Concepts Revisited: The Foundation for Problem-Solving

Before diving into complex problems, a solid grasp of the fundamental principles is non-negotiable. Acid-base chemistry in this context is governed by equilibrium constants. For any weak acid (HA), the dissociation is defined by its acid dissociation constant, Ka: [ K_a = \frac{[H_3O^+][A^-]}{[HA]} ] Its conjugate base, A⁻, has a corresponding base dissociation constant, Kb, related by the ion product of water: [ K_w = K_a \times K_b \quad (\text{where } K_w = 1.0 \times 10^{-14} \text{ at 25°C}) ] The pH scale ((pH = -\log[H_3O^+])) and pOH ((pOH = -\log[OH^-])) provide a logarithmic framework, with (pH + pOH = 14). The Henderson-Hasselbalch equation is your primary tool for buffer calculations: [ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) ] For bases, a similar form exists using pOH and pKb. Percent ionization and percent hydrolysis quantify the extent of dissociation. Finally, understanding titration curves—the plot of pH vs. volume of titrant—requires identifying the equivalence point, half-equivalence point, and buffer regions for different acid-base combinations (strong-strong, weak-strong, weak-weak).

Problem Set 1: Weak Acid & Weak Base Equilibria (Beyond the ICE Table)

These problems test your ability to make justified approximations and handle systems where both the acid and its conjugate base (or base and conjugate acid) are present in comparable amounts.

Problem 1.1: A 0.150 M solution of a weak monoprotic acid, HA, has a pH of 2.95. Calculate the acid dissociation constant (Ka) for HA and its percent ionization.

• Step 1: Find [H₃O⁺] from pH: ([H_3O^+] = 10^{-2.95} = 1.12 \times 10^{-3} \text{ M}).
• Step 2: Set up the ICE table for (HA \rightleftharpoons H_3O^+ + A^-). Initial [HA] = 0.150 M, change = -x, final [HA] ≈ 0.150 - x, [H₃O⁺] = [A⁻] = x = (1.12 \times 10^{-3}) M.
• Step 3: Check the 5% rule: ((1.12 \times 10^{-3} / 0.150) \times 100% = 0.75% < 5%). The approximation is valid.
• Step 4: Calculate (K_a = \frac{(1.12 \times 10^{-3})^2}{0.150} = 8.37 \times 10^{-6}).
• Step 5: Percent ionization = ((x / \text{initial [HA]}) \times 100% = 0.75%).

Problem 1.2: Calculate the pH of a 0.250 M solution of ammonia (NH₃), given (K_b = 1.8 \times 10^{-5}). Then, calculate the pH of a 0.250 M solution of ammonium chloride (NH₄Cl). Finally, predict the pH of a solution containing both 0.250 M NH₃ and 0.250 M NH₄Cl.

• For NH₃: Use ICE table. (x = \sqrt{K_b \times C} = \sqrt{(1.8 \times 10^{-5})(0.250)} = 2.12 \times 10^{-3} \text{ M}). pH = 14 - pOH = 14 - (-\log(2.12e-3)) = 11.33.
• For NH₄⁺ (from NH₄Cl): NH₄⁺ is a weak acid. Find (K_a) for NH₄⁺: (K_a = K_w / K_b = (1.0 \times 10^{-14}) / (1.8 \times 10^{-5}) = 5.56 \times 10^{-10}). Then, ([H_3O^+] = \sqrt{K_a \times C} = \sqrt{(5.56 \times 10^{-10})(0.250)} = 1.18 \times 10^{-5} \text{ M}). pH = -\log(1.18e-5) = 4.93.
• For the buffer (NH₃/NH₄⁺): Use Henderson-Hasselbalch. First, find pKₐ of NH₄⁺: pKₐ = -\log(5.56e-10) = 9.25. pH = pKₐ + \log([NH₃]/[NH₄⁺]) = 9.25 + \log(0.250/0.250) = 9.25 + 0 = 9.25. Notice the buffer pH is exactly between the pH of the weak base and its conjugate acid salt.

Problem Set 2: The Art of Titration Curves

Sketching and interpreting titration curves is a hallmark of acid-base chemistry. You must know how to calculate pH at every key point.

Problem 2.1: Sketch the titration curve for

Problem 2.1 (continued): Sketch the titration curve for the titration of a 0.100 M solution of acetic acid (CH₃COOH, Kₐ = 1.8 × 10⁻⁵) with 0.100 M NaOH.

1. Initial point (0 mL NaOH added).
   Treat the solution as a weak acid alone.
   [ [\mathrm{H_3O^+}] = \sqrt{K_a C}= \sqrt{(1.8\times10^{-5})(0.100)}=1.34\times10^{-3}\ \text{M} ]
   pH = −log(1.34 × 10⁻³) = 2.87.

2. Buffer region (before equivalence). After adding V mL of NaOH, moles of acid neutralized = 0.100 M × V L.
   Remaining HA = (0.100 M × 0.050 L − 0.100 M × V) and formed A⁻ = 0.100 M × V.
   Use Henderson–Hasselbalch:
   [ \text{pH}=pK_a+\log\frac{[\mathrm{A^-}]}{[\mathrm{HA}]} ]
   where pKₐ = −log(1.8 × 10⁻⁵) = 4.74.
   Plotting pH versus added base gives a gradual rise; at half‑equivalence (V = 25 mL) the ratio [A⁻]/[HA] = 1, so pH = pKₐ = 4.74.

3. Equivalence point (V = 50 mL NaOH).
   All HA converted to acetate (A⁻). Total volume ≈ 100 mL, so [A⁻] ≈ 0.050 M.
   Acetate hydrolyzes:
   [ K_b=\frac{K_w}{K_a}= \frac{1.0\times10^{-14}}{1.8\times10^{-5}}=5.56\times10^{-10} ]
   [ [\mathrm{OH^-}]=\sqrt{K_b C}= \sqrt{(5.56\times10^{-10})(0.050)}=5.27\times10^{-6}\ \text{M} ]
   pOH = −log(5.27 × 10⁻⁶) = 5.28 → pH = 14 − 5.28 = 8.72.

4. Beyond equivalence.
   Excess OH⁻ dominates. After adding V > 50 mL,
   [ [\mathrm{OH^-}]=\frac{0.100,(V-0.050)}{0.050+V} ]
   pOH = −log[OH⁻], pH = 14 − pOH. The curve approaches that of a strong base solution.

The resulting sketch shows a low‑pH start, a gentle buffer slope, a sharp rise at equivalence (pH ≈ 8.7), and a steep basic tail.

Problem 2.2: Titration of a 0.100 M solution of ammonia (NH₃, K_b = 1.8 × 10⁻⁵) with 0.100 M HCl.

• Initial pH: Treat as weak base.
  [ [\mathrm{OH^-}]=\sqrt{K_b C}= \sqrt{(1.8\times10^{-5})(0.100)}=1.34\times10^{-3}\ \text{M}

• Buffer region (before equivalence).
  After adding V mL of 0.100 M HCl, the moles of acid introduced are 0.100 M × V L. These protons convert NH₃ to its conjugate acid NH₄⁺.
  [ \begin{aligned} n_{\mathrm{NH_3,,remaining}} &= 0.100;\text{M}\times0.050;\text{L} - 0.100;\text{M}\times V \ n_{\mathrm{NH_4^+,,formed}} &= 0.100;\text{M}\times V \end{aligned} ] The total volume is (V_{\text{tot}} = 0.050;\text{L}+V).
  Using the Henderson–Hasselbalch expression for the conjugate‑acid pair (pKₐ of NH₄⁺ = 9.25): [ \text{pH}=pK_a+\log\frac{[\mathrm{NH_3}]}{[\mathrm{NH_4^+}]} =9.25+\log\frac{0.100(0.050-V)}{0.100V} =9.25+\log\frac{0.050-V}{V} ] As base is consumed, the ratio ([\mathrm{NH_3}]/[\mathrm{NH_4^+}]) falls, and the pH drops gradually. At the half‑equivalence point ((V=25;\text{mL})), the ratio equals 1, giving pH = pKₐ = 9.25 – the mirror image of the acetic‑acid/acetate buffer centered at pKₐ = 4.74.

• Equivalence point (V = 50 mL HCl). All NH₃ has been protonated to NH₄⁺. The total volume is now ≈ 100 mL, so
  [ [\mathrm{NH_4^+}] \approx \frac{0.100;\text{M}\times0.050;\text{L}}{0.100;\text{L}} = 0.050;\text{M} ] NH₄⁺ is a weak acid (Kₐ = K_w/K_b = 5.56 × 10⁻¹⁰). Its hydrolysis gives: [ [\mathrm{H_3O^+}] = \sqrt{K_a C}= \sqrt{(5.56\times10^{-10})(0.050)} = 5.27\times10^{-6};\text{M} ] [ \text{pH}= -\log(5.27\times10^{-6}) = 5.28 ] Thus the equivalence point lies on the acidic side of the scale (pH ≈ 5.3), in stark contrast to the basic equivalence point (pH ≈ 8.7) observed for the acetic‑acid/NaOH titration.

• Beyond equivalence (excess HCl).
  For V > 50 mL, the added strong acid dominates: [ [\mathrm{H_3O^+}] = \frac{0.100,(V-0.050)}{0.050+V} ] pH = −log[H₃O⁺]. As the volume of titrant increases, the pH approaches that of the 0.100 M HCl solution (≈ 1.0) asymptotically.

• Sketch of the curve.
  Starting at pH ≈ 2.87 (weak base), the curve rises slowly through the buffer region, passes through pH = 9.25 at half‑equivalence, then drops sharply at the equivalence point to pH ≈ 5.3, after which it descends steeply with excess strong acid. The shape is the inverse of the acetic‑acid/NaOH curve: a low‑pH start, a gradual increase, a steep acidic drop at equivalence, and a continued decline toward the pH of the titrant.

Conclusion

Mastering titration curves hinges on recognizing four regimes: the initial pure acid/base solution, the buffer region governed by the Henderson–Hasselbalch equation, the equivalence point where the salt of the weak acid or base

Conclusion

Masteringtitration curves hinges on recognizing four distinct regimes: the initial solution, the buffer region governed by the Henderson-Hasselbalch equation, the equivalence point where the salt of the weak acid or base hydrolyzes, and the excess titrant phase. For ammonia, this translates to a low-pH start (pH ~2.87), a gradual rise through the buffer zone, a sharp acidic drop at equivalence (pH ~5.3), and a steep descent toward the titrant's pH (~1.0) with excess acid. This inverse profile contrasts sharply with acetic acid's titration, which begins basic, rises to a buffer peak, drops to a basic equivalence point (~8.7), and then declines. Understanding these regimes allows chemists to predict pH changes, identify equivalence points, and interpret analytical data across diverse systems, from environmental monitoring to biochemical assays. The ammonia/HCl curve thus serves as a powerful pedagogical tool, illustrating fundamental principles of acid-base chemistry and titration kinetics.