Module 5 Acids, Bases, and Solubility Equilibrium
Welcome to Module 5, where we explore one of the most fundamental and widely applicable areas of chemistry: acid-base chemistry. This lesson establishes the theoretical foundation and computational skills you'll need to understand acid-base behavior in both laboratory and biological systems.
In this lesson, you'll master two essential learning objectives:
We begin with Brønsted-Lowry theory—the proton transfer model that explains most aqueous acid-base reactions—before expanding to Lewis theory, which provides a broader electron-pair framework. You'll then develop quantitative skills through pH calculations, starting with straightforward strong acid/base problems and progressing to more complex weak acid equilibrium systems.
Why This Matters: Acid-base chemistry is critical for understanding biochemical processes (blood pH regulation, enzyme function), environmental science (ocean acidification, acid rain), analytical chemistry (titrations, buffer preparation), and pharmaceutical applications (drug solubility, delivery mechanisms). The concepts and calculations you master here form the foundation for advanced topics in buffers, titrations, and solubility equilibria.
How to Succeed: Watch all 11 video segments carefully, practice the interactive activities immediately after each section, and work through the calculation problems step-by-step. Don't skip the guided solutions—understanding the problem-solving process is as important as getting the right answer.
Foundation Theory
pH and Equilibrium Concepts
Applications and Advanced Topics
Overby/Chang: Chemistry, 14th Ed. - Chapter 15: Complete Chapter (15.1-15.12)
The tabs to the left indicate you have two videos to watch.
Apply LO5.1.1: Use your understanding of Brønsted-Lowry acid-base theory from the videos above to identify conjugate acid-base pairs. Remember: a conjugate acid-base pair differs by exactly one proton (H⁺). Note: You'll practice Lewis acid-base pairs later in this lesson after learning Lewis theory.
Click to match each acid with its conjugate base. Each pair differs by exactly one H⁺ ion.
The tabs to the left indicate you have four videos to watch.
Apply LO5.1.2: Use the calculation methods from the videos above to solve pH and pOH problems. Problems progress from basic to advanced difficulty. Try each problem before checking the answer!
PROBLEM 1 - BASIC Calculate the pH of a 0.010 M HCl solution.
Answer & Step-by-Step Solution
Answer: pH = 2.00
Not what you got? Study this walk-through to understand where you went wrong.
Step 1: Recognize that HCl is a strong acid that completely ionizes: HCl → H⁺ + Cl⁻
Step 2: For strong acids, [H⁺] = initial acid concentration = 0.010 M
Step 3: Calculate pH using pH = -log[H⁺] = -log(0.010) = -log(1.0 × 10⁻²) = 2.00
PROBLEM 2 - BASIC Calculate the pH and pOH of a 0.0050 M NaOH solution at 25°C.
Answer & Step-by-Step Solution
Answer: pOH = 2.30, pH = 11.70
Not what you got? Study this walk-through to understand where you went wrong.
Step 1: Recognize that NaOH is a strong base: NaOH → Na⁺ + OH⁻
Step 2: For strong bases, [OH⁻] = initial base concentration = 0.0050 M
Step 3: Calculate pOH: pOH = -log[OH⁻] = -log(0.0050) = -log(5.0 × 10⁻³) = 2.30
Step 4: Use pH + pOH = 14.00 at 25°C: pH = 14.00 - 2.30 = 11.70
PROBLEM 3 - INTERMEDIATE If a strong acid solution has a pH of 1.85, what is the concentration of H⁺ ions?
Answer & Step-by-Step Solution
Answer: [H⁺] = 0.014 M or 1.4 × 10⁻² M
Not what you got? Study this walk-through to understand where you went wrong.
Step 1: Start with the pH equation: pH = -log[H⁺]
Step 2: Substitute the given pH: 1.85 = -log[H⁺]
Step 3: Solve for [H⁺] using the antilog: [H⁺] = 10⁻ᵖᴴ = 10⁻¹·⁸⁵
Step 4: Calculate: [H⁺] = 0.014 M (or 1.4 × 10⁻² M in scientific notation)
PROBLEM 4 - INTERMEDIATE Acetic acid (CH₃COOH) has a pKₐ of 4.75. Calculate the Kₐ value for acetic acid.
Answer & Step-by-Step Solution
Answer: Kₐ = 1.8 × 10⁻⁵
Not what you got? Study this walk-through to understand where you went wrong.
Step 1: Recall the relationship between pKₐ and Kₐ: pKₐ = -log(Kₐ)
Step 2: Substitute the given pKₐ: 4.75 = -log(Kₐ)
Step 3: Solve for Kₐ using the antilog: Kₐ = 10⁻ᵖᴷₐ = 10⁻⁴·⁷⁵
Step 4: Calculate: Kₐ = 1.8 × 10⁻⁵
PROBLEM 5 - ADVANCED Calculate the pH of a 0.10 M solution of formic acid (HCOOH) given that Kₐ = 1.8 × 10⁻⁴.
Answer & Step-by-Step Solution
Answer: pH = 2.37
Not what you got? Study this walk-through to understand where you went wrong.
Step 1: Set up the ICE table for HCOOH ⇌ H⁺ + HCOO⁻:
Initial: [HCOOH] = 0.10 M, [H⁺] = 0, [HCOO⁻] = 0
Change: [HCOOH] = -x, [H⁺] = +x, [HCOO⁻] = +x
Equilibrium: [HCOOH] = 0.10-x, [H⁺] = x, [HCOO⁻] = x
Step 2: Write the Kₐ expression: Kₐ = [H⁺][HCOO⁻]/[HCOOH] = x²/(0.10-x)
Step 3: Substitute Kₐ: 1.8 × 10⁻⁴ = x²/(0.10-x)
Step 4: Check if x << 0.10: If so, approximate as x²/0.10 = 1.8 × 10⁻⁴
Step 5: Solve: x² = 1.8 × 10⁻⁵, so x = 4.24 × 10⁻³ M = [H⁺]
Step 6: Calculate pH: pH = -log(4.24 × 10⁻³) = 2.37
PROBLEM 6 - ADVANCED At 25°C, if [OH⁻] = 2.5 × 10⁻³ M, calculate [H⁺], pH, and pOH. Use Kw = 1.0 × 10⁻¹⁴.
Answer & Step-by-Step Solution
Answer: [H⁺] = 4.0 × 10⁻¹² M, pH = 11.40, pOH = 2.60
Not what you got? Study this walk-through to understand where you went wrong.
Step 1: Use the water autoionization constant: Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴
Step 2: Solve for [H⁺]: [H⁺] = Kw/[OH⁻] = (1.0 × 10⁻¹⁴)/(2.5 × 10⁻³)
Step 3: Calculate [H⁺]: [H⁺] = 4.0 × 10⁻¹² M
Step 4: Calculate pOH: pOH = -log[OH⁻] = -log(2.5 × 10⁻³) = 2.60
Step 5: Calculate pH: pH = -log[H⁺] = -log(4.0 × 10⁻¹²) = 11.40
Step 6: Verify: pH + pOH = 11.40 + 2.60 = 14.00 ✓
Problems 1-2: Master basic strong acid/base pH and pOH calculations
Problems 3-4: Practice converting between pH/pOH and concentrations, pK relationships
Problems 5-6: Apply equilibrium calculations to weak acids and Kw relationships
Do not let the title fool you. These videos are required and do contain content, not just additional examples. The tabs to the left indicate you have six videos to watch.
Apply LO5.1.1: Master advanced acid-base concepts including Lewis theory, polyprotic systems, and structural effects. These activities reinforce concepts from videos 007-011 and prepare you for Lewis theory applications.
Apply Lewis theory to identify electron acceptors and electron donors. Remember: Lewis acids accept electron pairs, Lewis bases donate electron pairs.
Click to match each Lewis acid with its corresponding Lewis base in these reaction examples.
Click each card to explore how different acid-base theories apply to various chemical systems.
Which acid-base theory explains this reaction?
No proton transfer occurs. NH₃ donates electron pair to BF₃. Brønsted-Lowry cannot explain this reaction.
Which theories explain this reaction?
Brønsted-Lowry: H⁺ transfer from HCl to H₂O. Lewis: H⁺ accepts electron pair from H₂O.
Complex ion formation - which theory applies?
Cu²⁺ accepts electron pairs from H₂O ligands. No protons involved - pure Lewis acid-base chemistry.
Sort these chemical species and reactions based on their acid-base behavior. Use concepts from videos 007-011.
Drag each item to the correct category based on its acid-base behavior.
Apply concepts from videos 007-010 to solve complex acid-base problems.
ADVANCED Calculate the pH of a 0.10 M Al(NO₃)₃ solution. Ka for [Al(H₂O)₆]³⁺ = 1.4 × 10⁻⁵
Answer & Step-by-Step Solution
Answer: pH = 2.93
This applies concepts from Video 007: Cations as Weak Acids
Step 1: Recognize Al³⁺ forms hydrated complex: [Al(H₂O)₆]³⁺
Step 2: The complex acts as weak acid: [Al(H₂O)₆]³⁺ ⇌ [Al(H₂O)₅OH]²⁺ + H⁺
Step 3: Set up ICE table with [Al³⁺] = 0.10 M initially
Step 4: Ka = 1.4 × 10⁻⁵ = [H⁺]²/(0.10 - [H⁺])
Step 5: Solve: [H⁺] = 1.18 × 10⁻³ M
Step 6: pH = -log(1.18 × 10⁻³) = 2.93
EXPERT For 0.10 M H₂CO₃: Ka1 = 4.3 × 10⁻⁷, Ka2 = 5.6 × 10⁻¹¹. Calculate pH and [CO₃²⁻].
Answer & Step-by-Step Solution
Answer: pH = 3.67, [CO₃²⁻] = 5.6 × 10⁻¹¹ M
This applies concepts from Video 010: Diprotic and Triprotic Acids
Step 1: For polyprotic acids, first ionization dominates pH
Step 2: H₂CO₃ ⇌ H⁺ + HCO₃⁻ (use Ka1 only for pH)
Step 3: Ka1 = [H⁺][HCO₃⁻]/[H₂CO₃] = x²/(0.10-x)
Step 4: 4.3 × 10⁻⁷ = x²/0.10, so x = 2.1 × 10⁻⁴ M
Step 5: pH = -log(2.1 × 10⁻⁴) = 3.67
Step 6: For diprotic acids: [CO₃²⁻] = Ka2 = 5.6 × 10⁻¹¹ M
Based on Video 010 content, select ALL statements that correctly describe polyprotic acids.
Select all characteristics that apply to polyprotic acids like H₂SO₄, H₃PO₄, and H₂CO₃.
Lewis Theory Applications:
Advanced Equilibrium Skills: