Module 5 Acids, Bases, and Solubility Equilibrium
In this lesson, we explore how chemical systems maintain stability through buffer mechanisms and how we can analyze these systems using titration techniques. You'll learn to predict and control pH changes in biological and analytical systems—skills essential for healthcare, environmental science, and laboratory work.
In this lesson, you'll master two critical learning objectives:
We begin with the common ion effect—a foundational concept that explains how adding salts affects equilibrium systems. This leads naturally to buffer theory, where you'll learn why some solutions resist pH changes and how to calculate their effectiveness. Finally, we'll explore titration analysis, learning to interpret the characteristic curves and select proper indicators for different acid-base combinations.
Why This Matters: Buffer systems are critical in biological processes (blood pH regulation, enzyme function), pharmaceutical formulation (IV fluids, drug stability), environmental monitoring (natural water systems, pollution control), and analytical chemistry (quality control, research applications). Titration skills are fundamental to quantitative analysis in every chemistry laboratory.
How to Succeed: Watch all video segments carefully, paying special attention to the real-world examples in blood chemistry and toxicity. Practice the Henderson-Hasselbalch calculations immediately after learning the theory. Work through titration problems systematically, focusing on identifying the type of titration and selecting appropriate calculation methods.
Overby/Chang: Chemistry, 14th Ed. - Chapter 16: Sections 16.2-16.5
Common Ion Effect and Buffer Foundation
Titration Theory and Practice
The tabs to the left indicate you have two videos to watch for fundamental concepts.
Apply LO5.1.3 Foundation: Use your understanding of equilibrium shifts to predict and calculate the effect of adding salts to weak acid solutions. This prepares you for buffer calculations in the next section.
PROBLEM 1 - BASIC Calculate the pH of a 0.10 M acetic acid solution before and after adding 0.050 M sodium acetate. Ka for acetic acid = 1.8 × 10⁻⁵
Answer & Step-by-Step Solution
Answer: pH without salt = 2.87, pH with salt = 4.44
Not what you got? Study this walk-through to understand the common ion effect.
Without sodium acetate:
Ka = [H⁺][CH₃COO⁻]/[CH₃COOH] = x²/0.10 = 1.8 × 10⁻⁵
x = [H⁺] = 1.34 × 10⁻³ M, pH = 2.87
With sodium acetate (common ion effect):
Initial [CH₃COO⁻] = 0.050 M from complete dissociation of salt
Ka = x(0.050 + x)/0.10 ≈ x(0.050)/0.10 = 1.8 × 10⁻⁵
x = [H⁺] = 3.6 × 10⁻⁵ M, pH = 4.44
Key Insight: Adding the common ion (acetate) suppresses acid ionization, increasing pH significantly!
The tabs to the left indicate you have two videos covering critical buffer concepts and real-world applications.
Understanding Buffer Resistance: Buffers work by having both a weak acid (to neutralize added base) and its conjugate base (to neutralize added acid) present in significant concentrations.
The Problem:
The Solution:
When antifreeze is consumed, glycolic acid overwhelms the buffer capacity, causing fatal acidosis as shown in the video.
Apply LO5.1.3: Master buffer calculations using the Henderson-Hasselbalch equation. Problems progress from basic buffer pH to buffer preparation and capacity.
PROBLEM 2 - BASIC Calculate the pH of a buffer containing 0.15 M acetic acid and 0.25 M sodium acetate. pKa of acetic acid = 4.75
Answer & Step-by-Step Solution
Answer: pH = 4.97
Step 1: Identify buffer components: CH₃COOH (weak acid) and CH₃COO⁻ (conjugate base from salt)
Step 2: Apply Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA])
Step 3: Substitute values: pH = 4.75 + log(0.25/0.15)
Step 4: Calculate: pH = 4.75 + log(1.67) = 4.75 + 0.22 = 4.97
Key Point: Since [base] > [acid], pH > pKa as expected!
PROBLEM 3 - INTERMEDIATE How many grams of sodium acetate must be added to 500 mL of 0.20 M acetic acid to create a buffer with pH = 5.00? (MW of sodium acetate = 82.03 g/mol)
Answer & Step-by-Step Solution
Answer: 14.6 g sodium acetate needed
Step 1: Use Henderson-Hasselbalch: 5.00 = 4.75 + log([CH₃COO⁻]/[CH₃COOH])
Step 2: Solve for ratio: 0.25 = log([CH₃COO⁻]/0.20), so [CH₃COO⁻]/0.20 = 10^0.25 = 1.78
Step 3: Calculate acetate concentration needed: [CH₃COO⁻] = 1.78 × 0.20 = 0.356 M
Step 4: Calculate moles needed: 0.356 mol/L × 0.500 L = 0.178 mol
Step 5: Convert to mass: 0.178 mol × 82.03 g/mol = 14.6 g
PROBLEM 4 - ADVANCED A buffer contains 0.10 M NH₃ and 0.12 M NH₄Cl. Calculate the pH change when 0.010 mol HCl is added to 1.0 L of this buffer. Kb for NH₃ = 1.8 × 10⁻⁵
Answer & Step-by-Step Solution
Answer: pH decreases by 0.11 units (9.17 → 9.06)
Step 1: Calculate initial pH using Henderson-Hasselbalch for base buffer:
pOH = pKb + log([BH⁺]/[B]) = 4.75 + log(0.12/0.10) = 4.83
Initial pH = 14.00 - 4.83 = 9.17
Step 2: HCl reaction: NH₃ + HCl → NH₄Cl (converts base to conjugate acid)
Step 3: After reaction concentrations:
[NH₃] = 0.10 - 0.010 = 0.09 M
[NH₄⁺] = 0.12 + 0.010 = 0.13 M
Step 4: Calculate new pH:
pOH = 4.75 + log(0.13/0.09) = 4.75 + 0.16 = 4.91
Final pH = 14.00 - 4.91 = 9.09
Step 5: pH change = 9.09 - 9.17 = -0.08 (minimal change demonstrates buffer effectiveness!)
Henderson-Hasselbalch Success:
Buffer Capacity Insights:
Master acid-base titrations through comprehensive video content and practice. The tabs organize content by titration type and complexity.
Understanding Titration Regions: Each type of titration has characteristic curve shapes that reveal the chemistry occurring at each stage.
Click to match each titration feature with its correct characteristic or cause.
Apply LO5.1.4: Master titration calculations for different regions and learn to select appropriate indicators. Work through problems systematically by identifying the titration type and region.
PROBLEM 5 - BASIC Calculate the pH when 30.0 mL of 0.150 M NaOH is added to 25.0 mL of 0.200 M HCl.
Answer & Step-by-Step Solution
Answer: pH = 12.10
Step 1: Calculate moles of each reactant:
mol HCl = 0.0250 L × 0.200 M = 0.00500 mol
mol NaOH = 0.0300 L × 0.150 M = 0.00450 mol
Step 2: Determine limiting reactant and excess:
HCl + NaOH → NaCl + H₂O (1:1 stoichiometry)
NaOH is limiting; excess HCl = 0.00500 - 0.00450 = 0.00050 mol
Wait! Actually NaOH neutralizes all HCl with 0.00050 mol NaOH excess!
Step 3: Calculate [OH⁻] from excess NaOH:
Total volume = 25.0 + 30.0 = 55.0 mL = 0.0550 L
[OH⁻] = 0.00050 mol ÷ 0.0550 L = 0.00909 M
Step 4: Calculate pH: pOH = -log(0.00909) = 2.04, pH = 14.00 - 2.04 = 11.96
PROBLEM 6 - INTERMEDIATE Calculate the pH at the equivalence point when 25.0 mL of 0.100 M acetic acid is titrated with 0.100 M NaOH. Ka = 1.8 × 10⁻⁵
Answer & Step-by-Step Solution
Answer: pH = 8.72
Step 1: At equivalence point, all acetic acid converts to acetate:
CH₃COOH + NaOH → CH₃COONa + H₂O
mol CH₃COOH = 0.0250 L × 0.100 M = 0.00250 mol
Volume NaOH needed = 0.00250 mol ÷ 0.100 M = 0.0250 L
Step 2: Calculate acetate concentration at equivalence:
Total volume = 25.0 + 25.0 = 50.0 mL = 0.0500 L
[CH₃COO⁻] = 0.00250 mol ÷ 0.0500 L = 0.0500 M
Step 3: Acetate hydrolyzes (acts as base):
CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻
Kb = Kw/Ka = 1.0×10⁻¹⁴/1.8×10⁻⁵ = 5.56×10⁻¹⁰
Step 4: Solve for [OH⁻]: Kb = x²/0.0500, x = 5.27×10⁻⁶ M
pOH = 5.28, pH = 14.00 - 5.28 = 8.72
Key Point: Equivalence point pH > 7 due to base hydrolysis - use phenolphthalein!
Problem-Solving Strategy:
Indicator Selection Guide: