Acid Base Action Plan Acid-Base Action Plan
Teacher Guide
5 Blocks (90-Minute Periods) • 3 Major Experiences
Core Objectives & TEKS
Primary Goals
Name and write formulas for acids and bases (C.12A)
Compare Arrhenius & Brønsted-Lowry models (C.12B)
Calculate pH from [H+] concentrations (C.12E)
Predict neutralization reaction products (C.12D)
Key Skills
Develop models for strong/weak acid behavior
Perform acid-base titration experiments
Analyze buffer system stabilization
Instructional Flow
Block 1 & 2: Experience 1 - Definitions
Engage Ocean Acidification Phenomenon.
Explore Lab: Measuring pH of Unknowns.
Explain Direct Instruction: Naming & Models.
Block 3 & 4: Experience 2 - Reactions
Engage Stomach Acid Neutralization.
Explore Lab: Vinegar Titration Experiment.
Explain Titration Curves and Calculations.
Block 5: Experience 3 - Buffers
Explore Lab: Buffer Ranges & Capacity.
Explain CER: Lake Acid Rain Resistance.
Evaluate Unit 12 DCA Summative.
Implementation Strategies
pH Math Scaffold
Focus on pH as an exponent. pH 4 is 10−4 M. Use the calculator [log] key explicitly during modeling.
Titration Mastery
Identify the endpoint as the first hint of permanent pink. Stress the importance of rinsing burets with titrant first.
60-Minute Essentials Session
Focus: pH Math & Neutralization Principles
00:00 - 05:00
Engagement Classification
Classify HCl, NaOH, and NH3 using the Arrhenius model.
05:00 - 20:00
Direct Instruction: pH Math
Formula derivation: pH = −log[H+]. Practice converting between [H+] and pH for strong acids.
20:00 - 35:00
Guided Application
Field Notes Sections I & II. naming binary acids and oxyacids. Calculating pOH from pH.
35:00 - 55:00
Reaction Modeling
Predicting Products: Write the balanced equation for the reaction of Nitric Acid and Magnesium Hydroxide.
55:00 - 60:00
Exit Check
Independent: Find the pH of 0.01 M HCl and 0.01 M NaOH.
pH Profile Reading Activity pH Profile Reading
Characteristics & The Power of Hydrogen
Scientist:
Read the following passage carefully. Use the insights to complete the pH Mini-Project on the final page.
Chemical Personalities
The Acidic Profile
Acids are often described by their sharp, reactive nature. Physically, many acids have a distinct sour taste . Citric acid in lemons and acetic acid in vinegar are classic examples. In terms of reactivity, acids are known for their ability to react with metals to produce hydrogen gas and their corrosive ability to "eat through" materials. In a laboratory setting, acids are identified by turning blue litmus paper red .
Key Fact: Acids increase the concentration of H+ ions in a solution.
The Basic Profile
Bases, also known as alkaline substances, provide a stark contrast. Many bases feel slippery or soapy to the touch because they react with the oils on your skin (saponification). They typically possess a bitter taste , like baking soda. Unlike acids, bases turn red litmus paper blue . While they don't react with most metals, strong bases are just as corrosive as strong acids and can cause severe chemical burns.
Key Fact: Bases increase the concentration of OH− ions in a solution.
The Power of Hydrogen (pH)
The pH scale is the universal ruler for measuring how acidic or basic a water-based solution is. The scale typically runs from 0 to 14 . A pH of 7 is neutral (pure water). Solutions with a pH less than 7 are acidic, while those greater than 7 are basic.
The Logarithmic Secret
The pH scale is not linear; it is logarithmic . This means each whole number change on the scale represents a ten-fold (10x) change in acidity or basicity. For example, a solution with a pH of 4 is 10 times more acidic than a solution with a pH of 5.
🍋
Stomach Acid
pH 1.0 - 2.0
💧
Pure Water
pH 7.0 (Neutral)
🧼
Bleach
pH 12.0 - 13.0
pH Profile Mini-Project
Choose one household substance and complete the profile below based on the reading and your chemical knowledge.
Chosen Substance:
Draw a diagram of your substance's molecules or its use in the real world
Is it Acidic, Basic, or Neutral?
Acidic
Basic
Neutral
Estimated pH Level:
Expected Properties (Litmus, Taste, Texture):
The 10x Math Challenge:
How many times more acidic/basic is your substance compared to neutral water (pH 7)?
Nomenclature Nexus Worksheet Nomenclature Nexus
Unit: Acid Base Investigation
Chemical Classification Protocol
Agent Name
Date
Class
Protocol: The Naming Rules
Binary Acids
H + Single Element
Hydro- [root] -ic acid
Ex: HCl
Hydrochloric acid
Oxyacids
H + Polyatomic Ion
-ate → -ic acid
-ite → -ous acid
Ex: \( \text{H}_2\text{SO}_4 \)
Sulfuric acid
Bases
Metal + Hydroxide
[Metal Name] hydroxide
Ex: NaOH
Sodium hydroxide
1
Formula Reconstruction
HF
KOH
\( \text{HNO}_3 \)
\( \text{Mg(OH)}_2 \)
\( \text{H}_2\text{CO}_3 \)
\( \text{H}_3\text{PO}_4 \)
\( \text{H}_2\text{SO}_3 \)
\( \text{Ba(OH)}_2 \)
2
Structural Synthesis
Hydrosulfuric acid
Lithium hydroxide
Nitrous acid
Calcium hydroxide
Hydroiodic acid
Aluminum hydroxide
Perchloric acid
Chloric acid
3
The Anomaly Analysis
Q1: Contrast the naming convention of HCl (aq) vs HCl (g).
Q2: Error identification: "Hydro-phosphoric acid" for \( \text{H}_3\text{PO}_4 \).
Q3: Why does base naming differ from specific acid prefix/suffix rules?
Q4: Name the following molecule: \( \text{H}_2\text{CrO}_4 \).
pH Powerhouse Worksheet pH Powerhouse
Scaffolded Calculation Mastery
Student:
Date: _________ Period: ___
I
The Calculation Circuit
[H+]
10−pH ← → −log[H+]
pH
Kw / [H+]
14 − pH
[OH−]
10−pOH ← → −log[OH−]
pOH
Sig Fig Protocol
Sig Figs in [Ion] = Dec Places in Result
Ex: 1.5 x 10−4 (2 sig figs) → 3.82 (2 dec places)
Kw = 1.0 x 10−14
II
Calculating pH & pOH from Concentration
EXAMPLE: Find pH if [H+] = 2.5 x 10−5 M
1. Formula: pH = −log[H+] → 2. Plug in: pH = −log(2.5 x 10−5) → 3. Solve: 4.60
A. Find pH if [H+] = 1.0 x 10−3 M
B. Find pOH if [OH−] = 4.2 x 10−5 M
III
Calculating Concentration from pH/pOH
EXAMPLE: Find [H+] if pH = 3.20
1. Formula: [H+] = 10−pH → 2. Plug in: [H+] = 10−3.20 → 3. Solve: 6.3 x 10−4 M
C. Find [H+] if pH = 8.45
IV
The pH & pOH Relationship
EXAMPLE: If pOH = 11.2, what is the pH?
1. Formula: pH + pOH = 14 → 2. Rearrange: pH = 14 − 11.2 → 3. Solve: 2.8
D. If pH = 2.3, what is the pOH?
pH Powerhouse // Mission Control
PAGE 02
V
Strong Acid & Base Systems
Acid Strategy:
[H+] = [Acid]. Apply −log formula.
Base Strategy:
[OH−] = [Base]. Find pOH, then pH.
1. Calculate the pH of a 0.0050 M solution of Nitric Acid (HNO3).
2. Calculate the pH of a 0.015 M solution of Sodium Hydroxide (NaOH).
VI
Advanced Scenarios
Physiology: Stomach Acid
A medical test shows a patient's stomach fluid has a pOH of 12.4. Calculate the pH. Is this within the normal digestive range (pH 1.0 - 2.0)?
Advanced: Molarity Integration
Calculate the final pH of a solution made by dissolving 0.40 grams of NaOH in 500.0 mL of water. (Molar Mass NaOH: 40.0 g/mol)
Step 1: grams → moles Step 2: [OH−] = mol / L Step 3: Find pOH Step 4: Find pH
Unit 12: Aqueous Systems - pH Powerhouse Mastery Circuit
© 2026 Chemistry Investigations Lab
Molecular Mechanics Slides PH POWER
Acid-Base Chemistry Essentials
Investigation
#12
Theme
Aqueous
Defining the Players
Arrhenius Model
Acid: H+ producer in water.
Base: OH− producer in water.
Brønsted-Lowry
Acid: Proton (H+) Donor.
Base: Proton (H+) Acceptor.
Acid & Base Strength
The Strong List
~100% Dissociation (Ionization)
Strong Acids
Strong Bases
The Weak List
<5% Dissociation (Equilibrium)
Weak Acids
Weak Bases
Naming Binary Acids
Rule: Hydrogen + 1 Non-metal
hydro- + [root] + -ic acid
Worked Examples
HCl: Hydrochloric acid
HBr: Hydrobromic acid
HI: Hydroiodic acid
Naming Tips
Only use "hydro-" if there is NO oxygen.
Identify the non-metal root first.
Always add the word "acid" at the end.
Naming Oxyacids
Rule 1: Anion ends in -ATE
[root] + -ic acid
"I ATE something ICky."
Rule 2: Anion ends in -ITE
[root] + -ous acid
"SPRITE is delicIOUS."
ATE Examples
HNO3 (Nitrate) → Nitric acid
H2SO4 (Sulfate) → Sulfuric acid
ITE Examples
HNO2 (Nitrite) → Nitrous acid
H2SO3 (Sulfite) → Sulfurous acid
Acid-Maker Toolbox
Ion Name Acid Result NO3− Nitrate Nitric Acid NO2− Nitrite Nitrous Acid SO42−
Domestic Detective Lab Guide Domestic Detective Lab
Classifying Household Substances by pH
Scientist:
Partners:
Briefing: The pH scale measures hydrogen ion (H+) concentration from 0 (acidic) to 14 (basic) . 7 is neutral . Each whole number shift is a 10-fold change in strength. Indicators change color to reveal the pH "code" of a substance.
Safety: Wear goggles. No tasting.
Goal: Classify samples using litmus & pH paper.
Equipment
• Well Plate, Litmus (R/B), pH Paper
• pH Color Chart & 10 Samples
Procedure
Test each sample with Litmus and pH Paper .
Record colors and estimate pH value.
Color and map findings on Page 2.
Investigation Log
Substance Red Litmus Blue Litmus pH Paper Color pH # Class Lemon Juice Ammonia Distilled Water Vinegar Liquid Soap Baking Soda Milk
<table class="w-full border-collapse border-2 border-black"><tbody><tr><td class="border-2 border-black p-1.5 text-[11px] font-bold bg-slate-50 w-1/4">Soda / Coke</td><td class="border-2 border-black h-12"></td><td class="border-2 border-black h-12 w-1/6"></td><td class="border-2 border-black h-12 w-1/6"></td><td class="border-2 border-black h-12 w-1/12"></td><td class="border-2 border-black h-12 w-1/6"></td></tr><tr><td class="border-2 border-black p-1.5 text-[11px] font-bold bg-slate-50 w-1/4">Antacid</td><td class="border-2 border-black h-12"></td><td class="border-2 border-black h-12"></td><td class="border-2 border-black h-12"></td><td class="border-2 border-black h-12"></td><td class="border-2 border-black h-12"></td></tr><tr><td class="border-2 border-black p-1.5 text-[11px] font-bold bg-slate-50 w-1/4">Coffee</td><td class="border-2 border-black h-12"></td><td class="border-2 border-black h-12"></td><td class="border-2 border-black h-12"></td><td class="border-2 border-black h-12"></td><td class="border-2 border-black h-12"></td></tr></tbody></table>
Nomenclature Nexus Key Nomenclature Nexus [KEY]
Unit: Acid Base Investigation
Teacher Evaluation Guide
1
Formula Reconstruction
HF
hydrofluoric acid
KOH
potassium hydroxide
\( HNO_3 \)
nitric acid
\( Mg(OH)_2 \)
magnesium hydroxide
\( H_2CO_3 \)
carbonic acid
\( H_3PO_4 \)
phosphoric acid
\( H_2SO_3 \)
sulfurous acid
\( Ba(OH)_2 \)
barium hydroxide
2
Structural Synthesis
Hydrosulfuric acid
\( H_2S \)
Lithium hydroxide
LiOH
Nitrous acid
\( HNO_2 \)
Calcium hydroxide
\( Ca(OH)_2 \)
Hydroiodic acid
HI
Aluminum hydroxide
\( Al(OH)_3 \)
Perchloric acid
\( HClO_4 \)
Chloric acid
\( HClO_3 \)
3
Analysis Annotations
Q1: Contrast HCl (aq) vs HCl (g).
HCl (g) is named hydrogen chloride (covalent gas). HCl (aq) is hydrochloric acid; (aq) indicates dissociation in water to donate protons (\( H^+ \)), activating specific acid naming protocols.
Q2: Error in "Hydro-phosphoric acid" for \( H_3PO_4 \).
"Hydro-" is for binary acids only. \( H_3PO_4 \) is an oxyacid. Standard protocol: "-ate" ions become "-ic" acids. Correct Name: Phosphoric acid.
Q3: Why base naming differs?
Bases are typically ionic hydroxides. They follow standard ionic nomenclature (Metal + Hydroxide) because they separate existing ions, while acids are molecules that react with water to create ions.
Q4: Molecule: \( H_2CrO_4 \).
Chromic acid (from chromate ion, \( CrO_4^{2-} \)).
pH Powerhouse Key pH Powerhouse
Official Teacher Key // Internal Use Only
Master Key
I
The Calculation Circuit
[H+]
10−pH ← → −log[H+]
pH
[OH−]
10−pOH ← → −log[OH−]
pOH
Sig figs in input = dec places in output.
II
Calculating pH & pOH from Concentration
A. Find pH if [H+] = 1.0 x 10−3 M
pH = 3.00
B. Find pOH if [OH−] = 4.2 x 10−5 M
pOH = 4.38
III
Calculating Concentration from pH/pOH
C. Find [H+] if pH = 8.45
[H+] = 10−8.45 = 3.5 x 10−9 M
IV
The pH & pOH Relationship
D. If pH = 2.3, what is the pOH?
pOH = 14 − 2.3 = 11.7
pH Powerhouse // Master Key
PAGE 02
V
Strong Acid & Base Systems
1. Calculate pH of 0.0050 M Nitric Acid (HNO3)
Strong acid: [H+] = 0.0050 M
pH = −log(0.0050) = 2.30
2. Calculate pH of 0.015 M Sodium Hydroxide (NaOH)
pOH = −log(0.015) = 1.82
pH = 14 − 1.82 = 12.18
VI
Advanced Scenarios (Validated)
Physiology Verdict
pH = 14 − 12.4 = 1.6
Verdict: Yes, within normal stomach range (pH 1.0 - 2.0).
Molarity Integration Key
Step 1: 0.40 g / 40.0 g/mol = 0.010 mol NaOH
Step 2: 0.010 mol / 0.5000 L = 0.020 M [OH−]
Step 3: pOH = −log(0.020) = 1.70
Final pH = 12.30
Unit 12: Aqueous Systems - pH Powerhouse Mastery Key
© 2026 Chemistry Investigations Lab - Internal Use Only
Anion Arsenal Reference Anion Arsenal
Unit: Acid Base Investigation
Essential Nomenclature Reference
Transformation Protocol: -ide anions often form binary acids (hydro-[root]-ic); -ate polyatomics form "-ic" acids; -ite polyatomics form "-ous" acids.
Monatomic Anions
Formula Name Formula Name F− Fluoride O2− Oxide Cl− Chloride S2− Sulfide Br− Bromide N3− Nitride I− Iodide P3− Phosphide
Polyatomic Ions (−1)
Formula Name Formula Name OH− Hydroxide ClO2− Chlorite NO3− Nitrate ClO− Hypochlorite NO2− Nitrite CN− Cyanide ClO4− Perchlorate C2H3O2− Acetate ClO3− Chlorate MnO4− Permanganate HCO3− Bicarbonate HSO4− Bisulfate
Polyatomic Ions (−2)
Formula Name Formula Name SO42− Sulfate Cr2O72− Dichromate SO32− Sulfite C2O42− Oxalate CO32− Carbonate O22− Peroxide CrO42− Chromate SiO32− Silicate
Charge (−3)
Formula Name PO43− Phosphate PO33− Phosphite
Domestic Detective Key Detective Briefing
Teacher Key & Lab Facilitation Guide
Teacher Resource
Expected Investigation Data
Note: Specific values vary by brand.
Substance Red Litmus Blue Litmus pH Paper pH # Class Lemon Juice Stays Red Turns Red Red/Orange 2.0 Acid Ammonia Turns Blue Stays Blue Purple 11.0 Base Distilled H2O No Change No Change Green 7.0 Neutral Vinegar Stays Red Turns Red Orange 3.0 Acid Liquid Soap Turns Blue Stays Blue Blue-Green 8.5 Base Baking Soda Turns Blue Stays Blue Blue 9.0 Base Milk Slight Red Stays Blue Yellow-Grn 6.5 Weak Acid Soda (Coke) Stays Red Turns Red Dk Orange 2.5 Acid Antacid Turns Blue Stays Blue Dk Blue 10.0 Base Coffee Stays Red Turns Red Yellow 5.0 Acid
Analysis Model Answers
Q2 (Lowest H+): Ammonia or Antacid. Basic substances have the lowest concentration of H+ ions (and highest OH−).
Q3 (Logarithm): Difference of 2 units (5 - 3 = 2). 102 = 100 times more acidic.
Q4 (Synthesis): Acids often protect foods from bacteria or provide flavor (tartness). Bases react with fats/oils to break them down, making them effective cleaners.
Lab Prep Checklist
Check expiration of litmus paper
Dilute thick soaps with water
Pre-label dropper bottles
pH Puzzle Pursuit Worksheet pH Puzzle Pursuit
Circuit Solving Challenge
Name:
Date: _________ Period: ___
Instructions: Solve each clue below and write your answer on the line. Once solved, find the corresponding number in the grid at the bottom and shade the circle. You must show your work for full credit!
I
The pH & pOH Relationship (pH + pOH = 14)
WORKED EXAMPLE: Find pOH if pH = 4.5
14 − 4.5 = 9.5
2. pOH when a solution is neutral
3. pH + pOH = ?
4. pOH when pH = 2.2
8. pOH of a solution whose pH is 12.80
9. pH when pOH is 9.16
20. pH when pOH is 10.01
22. pH = 4.52, pOH = ?
II
Log Power: pH = −log[H+]
WORKED EXAMPLE: pH when [H+] = 1.0 x 10−5 M
−log(1.0 x 10−5) = 5.00
1. pH when [H+] = 4.66 x 10−8 M
6. pH when [H+] = 0.034 M
11. pOH when [OH−] = 5.49 x 10−9 M
16. pH when [H+] = 9.89 x 10−7 M
18. pH when [H+] = 1 x 10−3 M
19. pH when [H+] = 1.25 x 10−11 M
pH Puzzle pursuit // Part 02
PAGE 02
III
Antilogs: [H+] = 10−pH
WORKED EXAMPLE: [H+] if pH = 2.0
10−2.0 = 1.0 x 10−2 M
5. Hydroxide ion concentration [OH−] when pOH = 0.5
13. Hydroxide ion concentration [OH−] when pOH = 0.76
12. [H+] when [OH−] = 1.01 x 10−14 M (Hint: Use Kw = 10−14)
IV
Solution Mastery: Multi-Step Logic
WORKED EXAMPLE: pOH when [H+] = 1.0 x 10−3 M
Step 1: Find pH = −log(1.0 x 10−3) = 3.0 → Step 2: pOH = 14 − 3 = 11.0
7. The [OH−] when the [H+] is 1.136 x 10−14 M
10. pOH when [H+] = 6.3 x 10−4 M
14. pH when there are 1.6 x 10−7 moles of H+ in 500 mL of solution.
15. pOH when there are 1.6 x 10−7 moles of H+ in 500 mL of solution.
17. pH of 350 mL of a solution that contains 2.64 x 10−2 mol H3O+
21. pH when [OH−] is 6.71 x 10−5 M
Puzzle Pursuit Grid
0.97
6.76
1.87
3.41
5.05
7.33
11.1
3.62
0.32
0.08
8.75
5.51
13.0
2.42
4.03
Character Card Sort Activity Chemical Card Sort
Unit 12 Classification Challenge
Instructions: Cut out all 20 cards. Sort into Acids, Bases, and Salts. Identify as Strong or Weak.
HCl Hydrochloric
NaOH Sodium Hydroxide
NH3 Ammonia
H2CO3 Carbonic Acid
CH3COOH Acetic Acid
NaCl Sodium Chloride
H2SO4 Sulfuric Acid
Mg(OH)2 Magnesium Hydroxide
KCl Potassium Chloride
LiOH Lithium Hydroxide
HI Hydroiodic Acid
Proton Donor
Proton Acceptor
Turns Indicator Pink
pH < 7
pH > 7
Slippery Feel
Sour Taste
MgSO4 Magnesium Sulfate
Acid or Base?
Print on Cardstock for Best Results
Acid Base Field Notes Acid-Base Field Notes
Unit 12: Aqueous Systems
Name: ________________________________
Date: __________________ Period: _______
I. Defining the Players
Arrhenius Model
Acid: Produces ions in H2O.
Base: Produces ions in H2O.
Brønsted-Lowry Model
Acid: Proton (H+) .
Base: Proton (H+) .
II. The Math of pH
pH = −log[H+]
pOH = −log[OH−]
14 = pH + pOH
1. Find pH if [H+] = 1.0 x 10-4 M.
2. Find pH if [H+] = 4.5 x 10-9 M.
3. If pH = 3.5, find pOH.
4. Find pH of 0.025 M NaOH.
III. Naming Acids
Anion Ending Acid Rule Example -ide hydro-______-ic acid HCl (Hydrochloric) -ate ______-ic acid H2SO4 (Sulfuric) -ite ______-ous acid HNO2 (Nitrous)
a) HI (Name):
b) Phosphoric Acid (Formula):
c) HNO3 (Name):
d) Acetic Acid (Formula):
IV. Neutralization Reactions
1. Hydrochloric Acid + Potassium Hydroxide
HCl + KOH →
2. Sulfuric Acid + Sodium Hydroxide
H2SO4 + NaOH →
3. Magnesium Hydroxide + Nitric Acid
Mg(OH)2 + HNO3 →
pH Puzzle Pursuit Key pH Puzzle Pursuit
Official Teacher Key // Internal Only
Master Key
I
The pH & pOH Relationship (Answers)
7.002. pOH when a solution is neutral
14.03. pH + pOH = ?
11.84. pOH when pH = 2.2
1.208. pOH of a solution whose pH is 12.80
4.849. pH when pOH is 9.16
3.9920. pH when pOH is 10.01
9.4822. pH = 4.52, pOH = ?
II
Log Power (Answers)
7.331. pH when [H+] = 4.66 x 10−8 M
1.476. pH when [H+] = 0.034 M
8.2611. pOH when [OH−] = 5.49 x 10−9 M
6.0016. pH when [H+] = 9.89 x 10−7 M
3.0018. pH when [H+] = 1 x 10−3 M
10.919. pH when [H+] = 1.25 x 10−11 M
pH Puzzle pursuit // Key Part 02
PAGE 02
III
Antilogs (Answers)
0.325. [OH−] when pOH = 0.5 (10−0.5)
0.1713. [OH−] when pOH = 0.76 (10−0.76)
0.9912. [H+] when [OH−] = 1.01 x 10−14 M
IV
Advanced Solutions
8.807. [OH−] (Actually 8.80 x 10−1 M or 0.88 M) → Value in grid: 0.88
10.810. pOH when [H+] = 6.3 x 10−4 M (pH=3.2, pOH=10.8)
6.4914. pH (mol/L = 1.6x10−7 / 0.5 = 3.2x10−7 → pH=6.49)
7.5115. pOH (14 − 6.49 = 7.51)
1.1217. pH (mol/L = 2.64x10−2 / 0.35 = 0.0754 → pH=1.12)
9.8321. pH (pOH=4.17 → pH=9.83)
Puzzle Solver Overlay
0.97
6.76
1.87
3.41
5.05
7.33
11.1
3.62
0.32
0.08
8.75
5.51
13.0
2.42
4.03
8.26
2.33
12.7
6.00
13.7
12.6
9.48
4.84
6.49
10.1
3.99
11.8
3.00
10.8
1.98
5.13
14.0
9.91
9.83
8.16
0.88
0.51
8.41
7.00
5.92
6.31
1.12
0.17
0.99
5.98
1.47
Strength Scale Slides Strength Scale
Unlocking the Secrets of Partial Dissociation and Equilibrium Constants
\( K_a \) & \( K_b \) Dynamics
The ICE Method
The Great Divide
01 / STRENGTH
Strong Acids/Bases
Dissociate completely in water. 100% of molecules turn into ions.
[HCl]initial = [H3O+]final
Weak Acids/Bases
Dissociate only partially. Most molecules remain intact. This creates an equilibrium!
HA + H2O ↔ H3O+ + A-
Strong (All Ions)
Weak (Mostly Molecules)
Molecular View
"The difference isn't concentration (Molarity), it's the tendency to release ions."
Measuring Mastery: \( K_a \)
02 / MATH
The Acid Dissociation Constant expression:
\[ K_a = \frac{[H_3O^+][A^-]}{[HA]} \]
Large \( K_a \)
High product yield. Stronger acid. More ions!
Small \( K_a \)
Low product yield. Weaker acid. Mostly molecules.
Pro-Tip
\( pK_a = -\log(K_a) \). Lower \( pK_a \) means stronger acid.
The ICE Strategy
03 / CALCULATION
I
Initial
Concentrations before reaction.
C
Change
The "x" shift.
E
Equilibrium
Final amounts for equation.
Rule: If \( K_a < 10^{-5} \), assume x is negligible compared to initial concentration.
Row [HA] [H3O+] [A-] Initial 0.100 M ~0 0 Change - x + x + x Equil. 0.100 - x x x
\[ K_a = \frac{x^2}{0.100 - x} \approx \frac{x^2}{0.100} \]
Equilibrium Quest Worksheet Equilibrium Quest
Acid & Base Strength Mastery Worksheet
Scientist:
Show all work
Assume 25°C
Sig Figs count!
1 Strength Identification
Rank the following 0.10 M acids in order of increasing acidity (Highest pH to Lowest pH) .
HF
Ka = 6.6 x 10-4
HCN
Ka = 6.2 x 10-10
HNO2
Ka = 4.6 x 10-4
Acetic
Ka = 1.8 x 10-5
Order:
#1
#2
#3
#4
2 Weak Acid Equilibrium
Calculate the pH of 0.150 M propanoic acid (HC3H5O2). Ka = 1.3 x 10-5.
Row
[HA]
[H3O+]
[A-]
I
C
E
3 Finding the Constant
A 0.100 M weak monoprotic acid has a pH of 3.25 . Calculate the Ka and the percent ionization of this acid.
Calculations for Ka:
Percent Ionization:
4 Base Behavior
Calculate the pH of 0.20 M Ammonia (NH3). Kb = 1.8 x 10-5.
pOH =
pH =
Equilibrium Quest Key Answer Key
Equilibrium Quest: Acid & Base Strength Mastery
Teacher Resource
1. Strength Identification
#1 (Highest pH)
HCN
6.2 x 10-10
#2
Acetic
1.8 x 10-5
#3
HNO2
4.6 x 10-4
#4 (Lowest pH)
HF
6.6 x 10-4
Logic: Increasing Acidity = Lowering pH. Higher Ka = Stronger Acid = More H3O+ = Lower pH.
2. Weak Acid Equilibrium
Step
[HA]
[H3O+]
[A-]
Initial
0.150
0
0
Change
-x
+x
+x
Equil.
0.150 - x
x
x
Ka = x2 / (0.150 - x) ≈ x2 / 0.150 = 1.3 x 10-5
x2 = 1.95 x 10-6 &implies; x = 1.40 x 10-3 M = [H3O+]
pH = -log(1.40 x 10-3) = 2.85
3. Finding the Constant
A) Determine Ka
[H3O+] = 10-3.25 = 5.62 x 10-4 M
Ka = (5.62 x 10-4)2 / (0.100 - 5.62 x 10-4)
Ka = 3.16 x 10-6
B) % Ionization
% = (5.62 x 10-4 / 0.100) x 100
% = 0.562%
4. Base Behavior (Ammonia)
Kb = x2 / 0.20 = 1.8 x 10-5 &implies; x = 1.90 x 10-3 M = [OH−]
pOH = -log(1.90 x 10-3) = 2.72
pH = 14.00 - 2.72 = 11.28
Strength Misconception Guide Strength Pitfalls
Teacher's Misconception & Strategy Guide
The Concentration Trap
Misconception: Students often believe that "Strong" means "Highly Concentrated" (e.g., 12M HCl) and "Weak" means "Dilute" (e.g., 0.01M HCl).
Teaching Fix:
Use the "Social vs. Introverted" analogy. A "Strong" acid is socially extroverted—it always gives away its proton, even if only a few molecules are present. A "Weak" acid is introverted—it holds onto its proton, even in a crowded room.
The pKa Paradox
Misconception: Students assume a high _pK_a means a high acid strength because the number is "bigger."
Teaching Fix:
Relate it back to pH. Just as a low pH (1) is more acidic than a high pH (6), a low _pK_a is "stronger" than a high _pK_a. The "p" stands for negative log—it physically flips the scale's direction.
The ICE Method Scaffolding
1
Visualizing the "x" Shift
When filling out the Change line, remind students that for every 1 HA that disappears, 1 H3O+ and 1 A- must appear. The "x" is the physical progress of the dissociation reaction.
2
The 5% Rule Verification
Teach students to always check their assumption. If (x / [HA]initial) × 100 < 5%, the simplification (dropping the -x) is valid. If higher, they need the quadratic formula.
3
Base Conversions Red Flag
The most common error in weak base problems is stopping at pOH. Force a "Reality Check" : If it's a base, the final pH must be above 7. If pH < 7 for Ammonia, they missed the (14 - pOH) step.
Conductivity Comparison Demo
Compare the light bulb brightness of 0.1 M HCl (Strong) vs. 0.1 M Acetic Acid (Weak).
HCl Result:
Blindingly bright bulb. Total ions = high conductivity.
CH3COOH Result:
Dim, flickering glow. Few ions = low conductivity.
Reaction Station Circuit Activity Reaction Station Circuit
Lab Guide
Station 1: pH Indicator Array
Task: Identify 3 unknown solutions using indicators.
Sample Color Change pH Paper Value Classification A B C
Station 2: Nomenclature Maze
Write formulas for these named species.
Hydroiodic Acid
Calcium Hydroxide
Chlorous Acid
Carbonic Acid
Station 3: Titration Curve Analysis
pH 14 pH 7 pH 0 Titrant Volume (mL) 25.0
Equivalence Volume:
pH at Equivalence:
Station 4: Buffer Battle
Solution Initial pH +1 Drop HCl ΔpH Water Buffer
Observation/Explanation:
End of Lab Rotation
Titration Tactics Key Titration Tactics Key
Teacher Resource
Quantitative Analysis of Acetic Acid in Commercial Vinegar
Instructional Context
This key uses a model value of 0.83 M for vinegar (approx. 5% acidity). Use these values to verify student calculations. Note: Student results will vary. Focus on the mathematical evidence chain.
Model Experimental Data
Trial Init. (mL) Final (mL) Vol. Used 1 0.00 82.85 82.85 2 0.00 83.15 83.15 3 1.00 84.00 83.00
Calculated Average
83.00 mL
Used for all subsequent math.
Mathematical Evidence Key
1 Convert Avg. NaOH Volume to Liters (L)
83.00 mL × (1 L / 1000 mL) = 0.0830 L
2 Moles of NaOH consumed (M × V)
n = (0.100 M) × (0.0830 L) = 0.00830 mol NaOH
3 Determined Molarity of Vinegar
Equation: CH3COOH + NaOH → NaCH3COO + H2O
MA = 0.00830 mol / 0.010 L = 0.83 M
Technique Check
Dark pink results usually indicate values >0.85 M. If <0.80 M, check for air bubbles in the buret tip.
Math Evidence
Ensure 3 sig figs (e.g., 0.830 M). Values must include the 'M' unit for full credit.
Key Conclusion
Target Range (5% Vinegar)
0.80 - 0.85 M
Neutralization Numbers Worksheet Neutralization Numbers
Stoichiometry & Equivalence Math
Student Worksheet
Name: ______________________ Date: _________
1. The Balanced Foundation (Predict & Balance)
Before performing neutralization math, you must have a balanced chemical equation. The coefficients provide the vital mole-to-mole ratio between the acid and base reactants. Complete and balance the reactions below.
Problem A: Nitric Acid (\(\text{HNO}_3\)) + Sodium Hydroxide (\(\text{NaOH}\))
Ratio: 1:1
\(\text{HNO}_3\) + \(\text{NaOH}\) → + \(\text{H}_2\text{O}\)
Salt Chemical Formula
Mole Ratio (\(\text{Acid} : \text{Base}\))
______ : ______
Problem B: Sulfuric Acid (\(\text{H}_2\text{SO}_4\)) + Potassium Hydroxide (\(\text{KOH}\))
Ratio: 1:2
\(\text{H}_2\text{SO}_4\) + \(\text{KOH}\) → + \(\text{H}_2\text{O}\)
Salt Chemical Formula
Mole Ratio (\(\text{Acid} : \text{Base}\))
______ : ______
2. The Equivalence Shortcut Formula
At the equivalence point of a titration, the total moles of hydrogen ions (\(\text{H}^+\)) from the acid must exactly equal the total moles of hydroxide ions (\(\text{OH}^-\)) from the base. We can express this with an elegant shortcut:
\(M_{\text{acid}} \times V_{\text{acid}} \times \boldsymbol{n}_{\text{acid}} = M_{\text{base}} \times V_{\text{base}} \times \boldsymbol{n}_{\text{base}}\)
M = Molarity (\(\text{mol/L}\))
V = Volume (any matching unit)
\(n\) = Number of reactive ions (\(\text{H}^+\) or \(\text{OH}^-\))
Worked Scaffolded Example
What volume of \(0.500\text{ M NaOH}\) is needed to fully neutralize \(25.0\text{ mL}\) of \(0.200\text{ M H}_2\text{SO}_4\)?
1. Identify variables:
Acid: \(\text{H}_2\text{SO}_4 \implies M_A = 0.200, \, V_A = 25.0, \, \boldsymbol{n_A = 2}\)
Base: \(\text{NaOH} \implies M_B = 0.500, \, V_B = ?, \, \boldsymbol{n_B = 1}\)
2. Solve the equation:
\((0.200) \times (25.0) \times (2) = (0.500) \times V_B \times (1)\)
\(10.0 = 0.500 \times V_B\)
\(V_B = 20.0\text{ mL}\) of \(\text{NaOH}\)
Page 1 of 2 • Continue on Page 2
Neutralization Numbers
Calculations & Stoichiometry Practice
Name: ______________________
3. Guided Math Practice
1 Determining Acid Molarity via Titration
Titration Tutorial Slides Lesson 2: Reactions
Titration
Tutorial
Quantifying the Unknown with Mathematical Precision
The Core Concept
What is Titration?
A laboratory technique where a solution of known concentration (titrant) is used to determine the concentration of an unknown solution (analyte).
Goal: Reach the Stoichiometric Equivalence Point
Key Roles
Visual Evidence: Indicators
Acidic
Colorless
Equivalence
Endpoint
Faint Pink
Basic
Magenta
Mathematical Evidence
Neutralization Equation
Stoichiometry 1:1
MA VA = MB VB
MA
Molarity Acid
VA
Volume Acid
MB
Molarity Base
VB
Volume Base
Execution Strategy
Trial Precision
1. Prep
10.0 mL Vinegar + Indicator
2. Zero
Fill buret with 0.10 M NaOH
3. Drip
Titrate to persistent pink
4. Record
Note final volume for math
Neutralization Numbers Key Neutralization Numbers
Stoichiometry & Equivalence Math
Teacher Answer Key
Name: TEACHER ANSWER KEY Date: _________
1. The Balanced Foundation (Predict & Balance) - Key
Before performing neutralization math, you must have a balanced chemical equation. The coefficients provide the vital mole-to-mole ratio between the acid and base reactants. Complete and balance the reactions below.
Problem A: Nitric Acid (\(\text{HNO}_3\)) + Sodium Hydroxide (\(\text{NaOH}\))
Ratio: 1:1
1 \(\text{HNO}_3\) + 1 \(\text{NaOH}\) → \(\text{NaNO}_3\) + 1 \(\text{H}_2\text{O}\)
Salt Chemical Formula
\(\text{NaNO}_3\)
Mole Ratio (\(\text{Acid} : \text{Base}\))
1 : 1
Problem B: Sulfuric Acid (\(\text{H}_2\text{SO}_4\)) + Potassium Hydroxide (\(\text{KOH}\))
Ratio: 1:2
1 \(\text{H}_2\text{SO}_4\) + 2 \(\text{KOH}\) → \(\text{K}_2\text{SO}_4\) + 2 \(\text{H}_2\text{O}\)
Salt Chemical Formula
\(\text{K}_2\text{SO}_4\)
Mole Ratio (\(\text{Acid} : \text{Base}\))
1 : 2
2. The Equivalence Shortcut Formula
At the equivalence point of a titration, the total moles of hydrogen ions (\(\text{H}^+\)) from the acid must exactly equal the total moles of hydroxide ions (\(\text{OH}^-\)) from the base. We can express this with an elegant shortcut:
\(M_{\text{acid}} \times V_{\text{acid}} \times \boldsymbol{n}_{\text{acid}} = M_{\text{base}} \times V_{\text{base}} \times \boldsymbol{n}_{\text{base}}\)
M = Molarity (\(\text{mol/L}\))
V = Volume (any matching unit)
\(n\) = Number of reactive ions (\(\text{H}^+\) or \(\text{OH}^-\))
Worked Scaffolded Example
What volume of \(0.500\text{ M NaOH}\) is needed to fully neutralize \(25.0\text{ mL}\) of \(0.200\text{ M H}_2\text{SO}_4\)?
1. Identify variables:
Acid: \(\text{H}_2\text{SO}_4 \implies M_A = 0.200, \, V_A = 25.0, \, \boldsymbol{n_A = 2}\)
Base: \(\text{NaOH} \implies M_B = 0.500, \, V_B = ?, \, \boldsymbol{n_B = 1}\)
2. Solve the equation:
\((0.200) \times (25.0) \times (2) = (0.500) \times V_B \times (1)\)
\(10.0 = 0.500 \times V_B\)
\(V_B = 20.0\text{ mL}\) of \(\text{NaOH}\)
Page 1 of 2 • Answer Key
Neutralization Numbers
Calculations & Stoichiometry Practice
Name: TEACHER ANSWER KEY
Titration Tactics Lab Guide Student: ___________________________ Date: ___________
Period: _____
Titration Tactics
Lab Guide
Quantitative Analysis of Acetic Acid in Commercial Vinegar
Core Principle
Titration uses a standardized solution to find an unknown concentration. At the endpoint, the moles of base added equal the moles of acid initially present (for 1:1 reactions).
MAVA = MBVB
Procedure
Fill buret with standardized 0.100 M NaOH.
Pipet exactly 10.00 mL of vinegar into an Erlenmeyer flask.
Add 3 drops of phenolphthalein.
Slowly titrate until a faint, persistent pink endpoint is reached.
Perform three trials for accuracy.
Experimental Data
Trial Init. (mL) Final (mL) 1 2 3
Avg. NaOH Used (mL):
Mathematical Evidence
1. Convert Average Volume of NaOH to Liters (L)
2. Calculate Moles of NaOH (M × V)
3. Calculate Unknown Molarity (nNaOH / 0.010 L)
Show your work using the MAVA = MBVB relationship.
Conclusion
Final Concentration of Vinegar
________ M
Neutralization Master Worksheet Neutralization Master
Predicting Products & Balancing Equations
Student Resource
Name: ___________________________ Date: _________
The Universal Pattern
Acid
H+ Donor
Base
OH- Donor
Salt
Ionic Solid
Water
H2O
Phase 1: The 4-Step Strategy
01
Identify Ions
Break reactants into their cations (+) and anions (-).
02
Swap Partners
The Cation from the Base joins the Anion from the Acid.
03
Criss-Cross
Write the Salt formula by balancing charges.
04
Balance
Add coefficients to satisfy conservation of mass.
Guided Example: H2SO4 + NaOH
1
Reactants: Sulfuric Acid (H2SO4) and Sodium Hydroxide (NaOH). Ions are H+, SO42- and Na+, OH-.
2
Swap: Na+ joins SO42-.
3
Criss-Cross: +1 and -2 charges means salt is Na2SO4.
4
Pre-Balance: H2SO4 + NaOH → Na2SO4 + H2O
Balanced Final Equation
H2SO4 + 2NaOH → Na2SO4 + 2H2O
Phase 2: Scaffolded Practice
Predict the products (Salt + Water) and balance each equation. Use the workspace for your scratch work.
1. Hydrochloric Acid + Potassium Hydroxide
Level: Introduction
HCl + KOH → + H2O
Identify Ions & Criss-Cross Salt
Atom Inventory (L vs R)
2. Nitric Acid + Calcium Hydroxide
Level: Moderate
HNO3 + Ca(OH)2 → + H2O
Identify Ions & Criss-Cross Salt
Atom Inventory (L vs R)
3. Phosphoric Acid + Sodium Hydroxide
Level: Challenging
H3PO4 + NaOH → + H2O
Identify Ions & Criss-Cross Salt
Atom Inventory (L vs R)
Independent Challenge
Write the complete balanced chemical equation for the neutralization of Acetic Acid (HC2H3O2) and Barium Hydroxide (Ba(OH)2).
_________________________________________________________
Neutralization Master Key Neutralization Master Key
Teacher Reference & Answer Key
Answer Key
Pedagogical Note
Focus on the "Swap and Drop" (Criss-Cross) method for the salt. Common student errors include failing to balance the salt formula before balancing the overall equation. Remind students that water is always a product!
1. HCl + KOH
1:1:1:1 Ratio
1HCl + 1KOH → KCl + 1H2O
Salt Calculation
K+1 and Cl-1 → KCl (Net Zero)
Atom Inventory
Balanced: 1 K, 1 Cl, 2 H, 1 O
2. HNO3 + Ca(OH)2
2:1:1:2 Ratio
2HNO3 + 1Ca(OH)2 → Ca(NO3)2 + 2H2O
Salt Calculation
Ca+2 and NO3-1 → Ca(NO3)2
Atom Inventory
Balanced: 1 Ca, 2 N, 4 H, 8 O
3. H3PO4 + NaOH
1:3:1:3 Ratio
1H3PO4 + 3NaOH → Na3PO4 + 3H2O
Salt Calculation
Na+1 and PO4-3 → Na3PO4
Atom Inventory
Balanced: 3 Na, 1 P, 6 H, 7 O
Independent Challenge Key
2HC2H3O2 + 1Ba(OH)2 → Ba(C2H3O2)2 + 2H2O
Salt Calculation
Ba+2 + C2H3O2-1 → Salt formula
Atom Inventory
1 Ba, 4 C, 8 H, 6 O Balanced
Acid Base Showdown Quiz Acid Base Showdown
Unit 12 Assessment • Summative
Name: ________________________________
Date: __________________ Period: _______
Part I: Concept Application
1. Identify the Brønsted-Lowry base in the following equilibrium:
A) NH3
B) H2O
C) NH4+
D) OH−
2. Calculate the pH of a solution where [H+] = 1.0 × 10−10 M.
A) 1
B) 4
C) 7
D) 10
3. Buffers resist pH change because they contain which system?
A) Strong acid + Salt
B) Weak acid + Conjugate base
C) Equal HCl + NaOH
D) Pure Distilled Water
Part II: Technical Response
4. Reaction Completion: Balance and write products for:
HNO3 + Ba(OH)2 →
5. Stoichiometry: 25.0 mL of 0.50 M NaOH neutralizes 50.0 mL of HCl. Find the molarity of HCl.
6. Theory: Compare strong vs weak acids using dissociation percentage. Why do strong acids conduct better?
Teacher Answer Key
1. A: NH3 accepts a proton to become NH4+.
2. D: −log(10−10) = 10.
3. B: Conjugate pairs neutralize added H+ and OH−.
4: 2 HNO3 + Ba(OH)2 → 2 H2O + Ba(NO3)2
5: 0.25 M HCl. (MAVA = MBVB → MA*50 = 0.5*25 → MA = 12.5/50 = 0.25).
6: Strong acids dissociate ~100%, weak acids <5%. More ions = higher conductivity.
Vocab Vault Cards Vocab Vault Cards
Unit Review Matching & Sort Activity
Scientist:
Cut along dashed lines. Match the Term cards to their Definition cards.
Term
Arrhenius Acid
Definition
A substance that increases the concentration of hydrogen ions (H+) when dissolved in water.
Term
Brønsted-Lowry Base
Definition
A molecule or ion that acts as a proton acceptor in a chemical reaction.
Term
Amphoteric
Definition
A substance (like water) that can behave as either an acid or a base depending on the environment.
Term
Conjugate Pair
Definition
Two substances related by the loss or gain of a single hydrogen ion (proton).
Term
Titration
Definition
A laboratory technique used to determine the unknown concentration of a solution using a standard solution.
Vocab Vault Cards (Continued)
Term
Equivalence Point
Definition
The point in a titration where moles of H+ from the acid exactly equal moles of OH− from the base.
Term
Indicator
Definition
A weak acid or base that changes color based on the pH, signaling the end point of a titration.
Term
_K_a
Definition
The equilibrium constant for the dissociation of an acid; a quantitative measure of acid strength.
Term
Buffer
Definition
A solution made of a weak acid and its conjugate base that resists changes in pH when acid or base is added.
Term
ICE Table
Definition
A systematic way to calculate equilibrium concentrations by tracking Initial, Change, and Equilibrium values.
Solution Secrets Study Guide v4 Solution Secrets Study Guide
Essential Concepts for Acid-Base Mastery
UNIT REVIEW
1. Property Profiles
Acids (The H+ Donors)
Taste: Sour
Arrhenius: Produces H+ ions
Brønsted-Lowry: Proton (H+) Donor
pH Range: 0 ≤ pH < 7
Litmus: Turns Red
Bases (The H+ Acceptors)
Taste/Feel: Bitter & Slippery
Arrhenius: Produces OH− ions
Brønsted-Lowry: Proton (H+) Acceptor
pH Range: 7 < pH ≤ 14
Litmus: Turns Blue
2. The Name Game
Acid Type Suffix Rule Example Binary (H + Element) Hydro- [root] -ic acid HCl = Hydrochloric Oxyacid (-ate ion) [root] -ic acid HNO3 = Nitric acid Oxyacid (-ite ion) [root] -ous acid HNO2 = Nitrous acid
Worked Example: Naming
Problem: Name the compound H2SO4.
Identify the anion: SO42− is the Sulfate ion.
Apply rule: Ions ending in -ate become -ic acids.
Final Result: Sulfuric Acid.
3. The Proton Swap: Brønsted-Lowry
Understanding "Conjugates"
In Brønsted-Lowry theory, acids and bases are defined by the transfer of a proton (H+) .
• ACID: The Donor. Gives away an H+.
Conjugate Pairs: Two species that differ by exactly ONE hydrogen atom and a charge difference of 1. When an acid loses H, its partner is the Conjugate Base .
Cheat Sheet Diagram
The Twin Rule
"Look for the twin with MORE H atoms. That twin is ALWAYS the acid."
The Charge Logic
"Add H+ → Charge +1.
NH3
Base
+H+
NH4+
Conj. Acid
|
H2O
Acid
−H+
OH−
Conj. Base
Equation Strategy
Acid: Reactant that has one less H on the product side.
Conj. Base: The product twin that the acid turned into.
Solution Showdown Quiz Solution Showdown Quiz
Unit Review Assessment
Name: __________________________
Date: __________________________
Instructions: Select the best answer for each question. Show calculation work in provided boxes to receive full credit.
1. Which set of characteristics best describes an Arrhenius base?
Sour taste, turns litmus red, produces hydrogen ions
Slippery feel, turns litmus blue, produces hydroxide ions
Salty taste, no effect on litmus, produces water
No taste, turns litmus purple, produces salt
2. What is the correct IUPAC name for the acid with the formula H2CO3?
Hydrocarbonic acid
Carbonous acid
Carbonic acid
Dihydrogen carbonate
3. A solution has a hydrogen ion concentration of 1.0 × 10−11 M. What is its pH and classification?
pH = 11, Basic
pH = 11, Acidic
pH = 3, Basic
pH = 3, Acidic
Calculation work:
4. In the reaction HCN + H2O → CN− + H3O+, which species acts as the conjugate base?
HCN
H2O
CN−
H3O+
5. Stoichiometry Challenge
A student neutralizes 100 mL of 0.40 M H2SO4 using a 0.20 M NaOH solution. What volume of NaOH is required?
100 mL
200 mL
400 mL
800 mL
Stoichiometry work (Eq → Moles → Volume):
6. Which statement correctly describes the difference between a strong acid and a weak acid?
Strong acids have a higher pH than weak acids at the same concentration.
Strong acids only partially dissociate, while weak acids fully dissociate.
Strong acids dissociate completely into ions, while weak acids remain mostly as neutral molecules.
Weak acids are more likely to lose protons than strong acids.
7. If the hydrogen ion concentration of a solution is decreased by a factor of 10, how does the pH change?
The pH increases by 1 unit.
The pH decreases by 1 unit.
The pH increases by 10 units.
The pH does not change.
8. What are the products when hydrobromic acid (HBr) reacts with calcium hydroxide (Ca(OH)2)?
H2 and CaBr2
H2O and CaBr2
H2O and CaBr
O2 and CaH2
Acid Base Showdown Key Showdown Key
TEACHER VERSION: Unit 12 Quiz
Answer Key
Part I: Multiple Choice
1
B) \( H_2O \) — In the forward reaction, water donates a proton to ammonia.
2
D) 10 — \( pH = -\log(1.0 \times 10^{-10}) = 10 \).
3
B) A weak acid and its conjugate base — This allows the system to neutralize both added acid and base.
4
C) Sulfuric Acid — Sulfate anion (-ate) changes to -ic acid.
Part II: Free Response
5. Reaction Prediction
\( 2 HNO_3 + Ba(OH)_2 \rightarrow 2 H_2O + Ba(NO_3)_2 \)
6. Titration Math
\( M_A V_A = M_B V_B \)
\( M_A (50.0 \text{ mL}) = (0.50 \text{ M}) (25.0 \text{ mL}) \)
\( M_A = \frac{12.5}{50.0} \)
Result: 0.25 M HCl
7. Theoretical Explanation
"A Strong Acid undergoes 100% dissociation (ionization) in water, meaning all molecules break into ions. A Weak Acid only partially dissociates, meaning most molecules stay together and only a small fraction produce \( H^+ \) ions, establishing an equilibrium."
Buffer System Organizer The Buffer Balance
Unit 12 Equilibrium Graphic Organizer
Acidic Stress
Stress: [H+] Increase
The Responder:
____________________
Net Reaction:
H+ + A− →
Basic Stress
Stress: [OH−] Increase
The Responder:
____________________
Net Reaction:
OH− + HA →
Le Châtelier Connection
A buffer creates a Dynamic Equilibrium. When stress is applied, the system shifts to consume the added species, restoring the [H+] balance.
HA + H2O ⇌ H3O+ + A−
If [H3O+] is added, shift is:
LEFT
RIGHT
Buffer Analysis Tool Name: ________________________________
Vocab Web Map Activity Vocab Connection Map
Connecting Unit Concepts
Scientist:
Arrange your Vocab Cards on the map below. Connect touching cards by drawing lines and labeling the relationship. Ensure the logic of your map reflects chemical principles.
Arrhenius
Brønsted-Lowry
Amphoteric
Conjugate
Titration
Equivalence
Indicator
_K_a
Buffer
ICE Table
Connection Explanations
Relationship 1:
Relationship 2:
Relationship 3:
Solution Showdown Key v4 Solution Showdown Key
Teacher Reference & Answer Rationales
TEACHER ANSWER KEY
1. Which set of characteristics best describes an Arrhenius base?
Correct: B Slippery feel, turns litmus blue, produces hydroxide ions
Rationale Arrhenius bases are defined by their ability to produce hydroxide (OH−) ions in solution. Physically, they are characterized by a slippery feel and bitter taste, and they turn red litmus paper blue.
2. What is the correct IUPAC name for the acid with the formula H2CO3?
Correct: C Carbonic acid
Rationale The anion is carbonate (CO32−). According to oxyacid naming rules, anions ending in −ate become −ic acids. Carbonate → Carbonic acid.
3. A solution has a hydrogen ion concentration of 1.0 × 10−11 M. pH and classification?
Correct: A pH = 11, Basic
Detailed Solution pH = −log(1.0 × 10−11) = 11. Since pH is greater than 7, the solution is basic (alkaline).
4. In the reaction HCN + H2O → CN− + H3O+, which is the conjugate base?
Correct: C CN−
Rationale HCN is the acid because it donates a proton. The resulting ion CN− is the conjugate base.
5. Stoichiometry Challenge: H2SO4 and NaOH
Correct: C (400 mL)
Worked Solution Path
Balanced Equation: H2SO4 + 2NaOH → Na2SO4 + 2H2O
Step 1 (Moles Acid): 0.100 L × 0.40 M = 0.04 moles H2SO4.
Step 2 (Stoichiometry): From the 1:2 ratio, moles NaOH = 0.04 × 2 = 0.08 moles.
Step 3 (Volume Base): Volume = 0.08 moles / 0.20 M = 0.40 L = 400 mL.
6. Difference between strong and weak acids?
Correct: C Strong acids dissociate completely into ions...
Rationale Strength refers to the degree of dissociation. Strong acids split 100% into ions in water. Weak acids only partially dissociate, remaining mostly as neutral molecules in equilibrium with their ions.
7. Decreasing concentration by 10x?
Correct: A The pH increases by 1 unit.
Rationale Lower concentration means less acidic, which moves the pH up the scale (toward 7 or 14). Each factor of 10 is 1 unit on the logarithmic scale.
8. Products of HBr and Ca(OH)2?
Correct: B H2O and CaBr2
Rationale Acid-base reactions yield water and salt. Charge balance for calcium bromide requires 2 bromines: CaBr2.
