Module 5 Acids, Bases, and Solubility Equilibrium
Building on your understanding of Brønsted-Lowry acid-base theory, this lesson introduces the quantitative tools that allow chemists to measure and calculate acidity: the pH and pOH scales. You'll master the logarithmic relationships that connect hydrogen ion concentration to pH values, memorize the seven strong acids that form the foundation of all acid-base problem solving, and develop the decision-making skills needed to choose the correct calculation approach for any acid-base problem you encounter.
In this lesson, you'll master the following learning objective:
You'll master the logarithmic relationships that connect hydrogen ion concentrations to pH values, memorize the seven strong acids that every chemist must know instantly, and develop a systematic decision-making framework for solving any acid-base calculation problem. Starting with pH and pOH scales, you'll progress through strong acid/base calculations to weak acid equilibrium problems involving Ka values and ICE tables.
Why This Matters: pH calculations are essential for biochemistry (blood pH regulation, enzyme activity), environmental chemistry (acid rain, ocean acidification), pharmaceutical science (drug formulation), and analytical chemistry (titration analysis). The calculation skills you develop here—especially the ability to distinguish between strong and weak acid problems—are critical for success in all subsequent acid-base topics.
How to Succeed: Master the "Strong Seven" acids first—this single decision determines your entire calculation approach. Watch the four video segments on pH concepts, work through the logarithmic relationships section carefully (many students struggle here), and practice the decision tree until choosing the correct method becomes automatic. Don't rush the calculator mechanics—incorrect button sequences cause more errors than conceptual misunderstandings.
Overby/Chang: Chemistry, 14th Ed. - Chapter 15: Complete Chapter (15.1-15.12)
pH and Equilibrium Concepts
The tabs to the left indicate you have four videos to watch.
⚠️ PREREQUISITE ALERT: Many pH calculation errors stem from logarithmic relationship confusion. Master these concepts before attempting any calculations!
Watch for these labels to understand what type of content you're reading:
pH = -log[H⁺]
The NEGATIVE sign is why the relationship is inverse!
More H⁺ ions = more acidic = LOWER pH number
Fewer H⁺ ions = less acidic = HIGHER pH number
| [H⁺] Concentration | pH Value | Solution Type | Relationship Pattern |
|---|---|---|---|
| 1.0 × 10⁻¹ M | 1.0 | Very Acidic | HIGH [H⁺] → LOW pH |
| 1.0 × 10⁻³ M | 3.0 | Acidic | Medium [H⁺] → Medium-low pH |
| 1.0 × 10⁻⁷ M | 7.0 | Neutral | Balanced point |
| 1.0 × 10⁻¹¹ M | 11.0 | Basic | LOW [H⁺] → HIGH pH |
Given: pH = 3.25, Find: [H⁺] = ?
✅ CORRECT Button Sequence:
❌ What NOT to do:
Forgetting the [+/-] step!
Result would be 1778 M (impossible!)
Given: [H⁺] = 2.5 × 10⁻⁴ M, Find: pH = ?
✅ CORRECT Button Sequence:
❌ What NOT to do:
Entering 0.00025 instead of using EE
Can lead to rounding errors!
Apply your understanding of inverse logarithmic relationships to these conceptual questions.
CHALLENGE QUESTION 1 If the pH decreases from 5 to 4, by what factor did [H⁺] increase?
Click for Answer & Explanation
Answer: [H⁺] increased by a factor of 10
Not what you got? Study this walk-through to understand the inverse logarithmic relationship.
Step 1: Calculate [H⁺] at pH 5
pH 5: [H⁺] = 10⁻⁵ = 1 × 10⁻⁵ M
Step 2: Calculate [H⁺] at pH 4
pH 4: [H⁺] = 10⁻⁴ = 1 × 10⁻⁴ M
Step 3: Find the factor of increase
Factor = (1 × 10⁻⁴)/(1 × 10⁻⁵) = 10
Key Insight: Each 1-unit pH decrease = 10× increase in [H⁺]. This is the fundamental inverse logarithmic relationship!
CHALLENGE QUESTION 2 Which solution is more acidic: pH 2.3 or pH 2.7? How much more acidic?
Click for Answer & Explanation
Answer: pH 2.3 is more acidic (2.5× more H⁺ ions)
Not what you got? Remember: LOWER pH = MORE acidic = HIGHER [H⁺].
Step 1: Calculate [H⁺] at pH 2.3
pH 2.3: [H⁺] = 10⁻²·³ = 5.0 × 10⁻³ M
Step 2: Calculate [H⁺] at pH 2.7
pH 2.7: [H⁺] = 10⁻²·⁷ = 2.0 × 10⁻³ M
Step 3: Find the ratio
Factor = (5.0 × 10⁻³)/(2.0 × 10⁻³) = 2.5×
Key Insight: Lower pH = more acidic = higher [H⁺]. The pH difference is only 0.4 units, but that translates to 2.5 times more H⁺ ions!
Critical Foundation for LO5.1.2: Master the seven strong acids from Video 006. Knowing these by heart is essential - everything else is a weak acid! Click each card to reveal the acid formula and key memory device.
STUDY STRATEGY: Practice these cards multiple times until you can recall all seven acids instantly. This knowledge determines success in every subsequent acid-base problem!
The acid that helps you digest food in your stomach
Hydrochloric Acid
• Most common strong acid
• Essential for digestion
• 100% ionization in water
Used to manufacture explosives and fertilizers
Nitric Acid
• Powerful oxidizing agent
• Industrial importance
• Strong acid with NO₃⁻ anion
Powers your car battery and many industrial processes
Sulfuric Acid
• Diprotic acid (2 H⁺ ions)
• Most produced chemical
• Battery electrolyte
Stronger than HCl as you go down Group 7
Hydrobromic Acid
• Hydrohalic acid series
• Stronger than HCl
• Follows periodic trends
The strongest of all the HX acids
Hydroiodic Acid
• Strongest hydrohalic acid
• Largest halide = weakest bond
• Easiest H⁺ release
The most powerful oxidizing acid
Perchloric Acid
• Strongest known acid
• Powerful oxidizing agent
• ClO₄⁻ is very stable
Strong oxyacid with three oxygen atoms
Chloric Acid
• Strong oxyacid
• Three oxygen atoms
• ClO₃⁻ is chlorate ion
Remember the Strong Seven:
Study Strategy:
⚠️ MASTER THIS FIRST: Before solving ANY acid-base problem, you must correctly identify the problem type. This prevents the most common calculation errors!
Is this acid one of the
"Strong Seven"?
(HCl, HBr, HI, HNO₃,
H₂SO₄, HClO₄, HClO₃)
0.01 M HCl → [H⁺] = 0.01 M → pH = 2.0
0.01 M CH₃COOH → Use Ka = 1.8×10⁻⁵ + ICE
Looking for Ka values for HCl, HNO₃, HClO₃, etc. Strong acids don't have Ka values in your reference table!
Setting up ICE tables for the Strong Seven. They ionize 100%—no equilibrium needed!
Apply LO5.1.2: Use the decision tree above to identify problem types FIRST, then solve. Each problem now requires you to make the critical decision before calculating. Try each problem before checking the answer!
DECISION PRACTICE Before solving the problems below, identify which calculation method you'll use:
Method needed: ________________
Method needed: ________________
Click to Check Your Decision-Making
Acid A (HI): Strong acid pathway - HI is one of the Strong Seven! Use [H⁺] = 0.050 M, pH = 1.30
Acid B (HF): Weak acid pathway - HF is NOT in the Strong Seven! Use Ka = 7.2×10⁻⁴ and ICE table
If you got these right, you're ready for the calculation problems below!
PROBLEM 1 - STRONG ACID Calculate the pH of a 0.010 M HCl solution.
Answer & Step-by-Step Solution (Decision Tree Approach)
Answer: pH = 2.00
Not what you got? Study this walk-through to understand the decision-making process.
Step 0 (CRITICAL): Decision Tree - Is HCl one of the Strong Seven? YES! → Use strong acid pathway
Step 1: Recognize that HCl completely ionizes: HCl → H⁺ + Cl⁻ (no equilibrium)
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
✓ Decision Check: No Ka lookup needed, no ICE table needed - this confirms we chose the right pathway!
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 0 (DECISION): Strong bases follow the same logic as strong acids - complete dissociation means simple stoichiometry!
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
✓ Decision Check: No Kb lookup needed, no ICE table needed - strong bases use simple stoichiometry just like strong acids!
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.
📝 Decision Note: The decision tree doesn't apply here! You're given pH and asked to find [H⁺] - this is just math (antilog), not an equilibrium vs. stoichiometry decision.
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.
📝 Decision Note: Decision tree not needed! This is a simple conversion between two ways of expressing the same thing (pKₐ ↔ Kₐ). Save the decision tree for when you need to choose between calculation methods.
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 0 (CRITICAL): Decision Tree - Is HCOOH one of the Strong Seven? NO! → Use weak acid pathway (equilibrium calculations)
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
✓ Decision Check: We used Ka and ICE table - this confirms we chose the weak acid pathway correctly!
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.
📝 Decision Note: Decision tree not applicable! You're given [OH⁻] and using the Kw relationship to find [H⁺]. The strong vs. weak decision only matters when you're starting from an acid or base formula and concentration.
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 ✓
⚡ Master the Decision Tree: These problems are intentionally mixed! You MUST use your decision tree first. Some are strong acids, some are weak - identify which pathway to use before calculating.
🧠 CHALLENGE 1 Calculate the pH of a 0.050 M HF solution. (Ka = 7.2 × 10⁻⁴)
⚠️ DECISION CHECKPOINT: Before you calculate anything, use your decision tree! Is HF one of the Strong Seven acids?
Decision Analysis & Complete Solution
Answer: pH = 2.05
🎯 DECISION ANALYSIS: HF is NOT one of the Strong Seven (HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄, HClO₃). Therefore, use the WEAK ACID pathway!
Did you choose the weak acid pathway? If not, review the decision tree above!
Step 1: ICE table for HF ⇌ H⁺ + F⁻:
I: [HF]=0.050, [H⁺]=0, [F⁻]=0
C: [HF]=-x, [H⁺]=+x, [F⁻]=+x
E: [HF]=0.050-x, [H⁺]=x, [F⁻]=x
Step 2: Ka = [H⁺][F⁻]/[HF] = x²/(0.050-x) = 7.2 × 10⁻⁴
Step 3: Since Ka is relatively large, don't assume x << 0.050. Use quadratic formula.
Step 4: x² + 7.2 × 10⁻⁴x - 3.6 × 10⁻⁵ = 0
Step 5: x = 5.9 × 10⁻³ M = [H⁺]
Step 6: pH = -log(5.9 × 10⁻³) = 2.23
🧠 CHALLENGE 2 Calculate the pH of a 0.025 M HNO₃ solution.
🎯 DECISION CHECKPOINT: Apply your decision tree carefully. What type of acid is HNO₃?
Decision Analysis & Complete Solution
Answer: pH = 1.60
✅ DECISION ANALYSIS: HNO₃ IS one of the Strong Seven! Use the STRONG ACID pathway - simple stoichiometry only!
Did you immediately recognize this as strong? Good decision-making!
Step 1: HNO₃ → H⁺ + NO₃⁻ (complete ionization)
Step 2: [H⁺] = [HNO₃] = 0.025 M
Step 3: pH = -log(0.025) = -log(2.5 × 10⁻²) = 1.60
⚠️ Avoid This Error: Don't look for a Ka value for HNO₃ - it doesn't exist because it's 100% ionized!
🧠 CHALLENGE 3 Calculate the pH of a 0.10 M CH₃COOH (acetic acid) solution. (Ka = 1.8 × 10⁻⁵)
🤔 DECISION CHECKPOINT: Acetic acid (vinegar) - is this one of the Strong Seven acids?
Decision Analysis & Complete Solution
Answer: pH = 2.87
🎯 DECISION ANALYSIS: CH₃COOH is NOT one of the Strong Seven! This is a weak acid - use equilibrium calculations.
Correct pathway choice! Now for the equilibrium math...
Step 1: ICE table for CH₃COOH ⇌ H⁺ + CH₃COO⁻
Step 2: Ka = x²/(0.10-x) = 1.8 × 10⁻⁵
Step 3: Since Ka is small, assume x << 0.10: x²/0.10 = 1.8 × 10⁻⁵
Step 4: x² = 1.8 × 10⁻⁶, so x = 1.34 × 10⁻³ M = [H⁺]
Step 5: Check assumption: 1.34 × 10⁻³/0.10 = 1.3% < 5% ✓
Step 6: pH = -log(1.34 × 10⁻³) = 2.87
If you successfully identified the acid type in each challenge problem above, you've mastered the most critical skill in acid-base calculations!
WEAK acid pathway
STRONG acid pathway
WEAK acid pathway
You can now confidently identify acid types and perform accurate pH calculations using the appropriate method. The decision tree is your most powerful tool - master it!