🧪 Arrhenius Acids and Bases

Part 1 of 7 — The First Modern Definition

The study of acids and bases is central to chemistry — from biochemistry to industrial processes, these substances shape reactions across every field. We begin with the oldest modern definition: the Arrhenius model.

The Arrhenius Definition

In 1884, Svante Arrhenius proposed a simple classification:

Key Features

• Acids increase $[H^+]$ in water
• Bases increase $[OH^-]$ in water
• Neutralization produces water: $H^+(aq) + OH^-(aq) \rightarrow H_2O(l)$

Limitations

The Arrhenius model only works in aqueous solutions and cannot explain:

• Why $NH_3$ acts as a base (it doesn't contain $OH^-$)
• Acid-base behavior in non-aqueous solvents
• Reactions between gases that show acid-base character

Common Arrhenius Acids

Strong Acids (Complete Dissociation)

Common Strong Bases

Memorize the 6 strong acids and 4 strong bases — everything else is weak!

The Hydronium Ion

In reality, free $H^+$ ions (bare protons) don't exist in water. Instead, they bond to water molecules:

$H^+(aq) + H_2O(l) \rightarrow H_3O^+(aq)$

The hydronium ion $H_3O^+$ is a more accurate representation. In AP Chemistry:

• $H^+(aq)$ and $H_3O^+(aq)$ are used interchangeably
• Both notations are acceptable on the AP exam
• $H_3O^+$ is technically more correct
• $H^+$ is a convenient shorthand

Autoionization of Water

Pure water undergoes self-ionization:

$2H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)$

The equilibrium constant for this process is:

$K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \text{ at 25°C}$

In pure water: $[H^+] = [OH^-] = 1.0 \times 10^{-7}$ M

Arrhenius Concept Check 🎯

Arrhenius Neutralization

When an Arrhenius acid reacts with an Arrhenius base, they undergo neutralization:

$\text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water}$

Examples

$HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$

Net ionic equation:

$H^+(aq) + OH^-(aq) \rightarrow H_2O(l)$

This net ionic equation is the same for all strong acid–strong base neutralizations!

Double Replacement Pattern

$H_2SO_4(aq) + 2KOH(aq) \rightarrow K_2SO_4(aq) + 2H_2O(l)$

Note: sulfuric acid is diprotic — it has 2 acidic protons, so it requires 2 moles of $KOH$.

Arrhenius Classification 🔍

Exit Quiz — Arrhenius Acids & Bases ✅

🔄 Brønsted-Lowry Acids and Bases

Part 2 of 7 — Proton Donors and Acceptors

The Brønsted-Lowry model expands our understanding of acids and bases beyond aqueous solutions. Instead of focusing on $H^+$ and $OH^-$ production, it centers on proton transfer.

The Brønsted-Lowry Definition

Key Advantage

This definition works in any solvent — not just water!

Example: $HCl$ in Water

$HCl(aq) + H_2O(l) \rightarrow H_3O^+(aq) + Cl^-(aq)$

• $HCl$ donates a proton → acid
• $H_2O$ accepts a proton → base

Example: $NH_3$ in Water

$NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)$

• $NH_3$ accepts a proton → base
• $H_2O$ donates a proton → acid

Notice: water can act as either an acid or a base! This is called being amphoteric (or amphiprotic).

Conjugate Acid-Base Pairs

When an acid donates a proton, the product is its conjugate base. When a base accepts a proton, the product is its conjugate acid.

$\underbrace{HA}_{\text{acid}} + \underbrace{B}_{\text{base}} \rightleftharpoons \underbrace{A^-}_{\text{conjugate base}} + \underbrace{BH^+}_{\text{conjugate acid}}$

Examples

Critical Rule

A conjugate pair always differs by exactly one proton ($H^+$).

Strength Relationship

Strong acid → very weak conjugate base (and vice versa)

• $HCl$ is strong → $Cl^-$ is a negligible base (does not accept protons)
• $CH_3COOH$ is weak → $CH_3COO^-$ is a moderate conjugate base

Brønsted-Lowry Concept Check 🎯

Identifying Conjugate Pairs in Reactions

For any Brønsted-Lowry reaction, there are always two conjugate pairs:

$\underbrace{HF}_{\text{acid}_1} + \underbrace{H_2O}_{\text{base}_2} \rightleftharpoons \underbrace{F^-}_{\text{conj. base}_1} + \underbrace{H_3O^+}_{\text{conj. acid}_2}$

Pair 1: $HF / F^-$

Pair 2: $H_2O / H_3O^+$

Steps to Identify

1. Find the species that lost a proton → that's the acid; its product is the conjugate base
2. Find the species that gained a proton → that's the base; its product is the conjugate acid
3. Each acid is paired with its conjugate base (they differ by one $H^+$)

Conjugate Pair Identification 🔍

For the reaction: $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$

Conjugate Pair Practice 🧮

Identify the conjugate partners:

1. What is the conjugate base of $H_2CO_3$? (Enter the chemical formula, e.g. HCO3-)

2. What is the conjugate acid of $PO_4^{3-}$? (Enter the chemical formula, e.g. HPO42-)

3. What is the conjugate base of $H_2O$? (Enter the chemical formula, e.g. OH-)

Exit Quiz — Brønsted-Lowry Theory ✅

🔬 Lewis Acids and Bases

Part 3 of 7 — Electron Pair Donors and Acceptors

The Lewis definition is the broadest acid-base theory. It doesn't require protons at all — it focuses on electron pairs.

The Lewis Definition

Comparison of All Three Theories

Key Insight

Every Arrhenius acid is a Brønsted-Lowry acid, and every Brønsted-Lowry acid involves a Lewis acid interaction. The Lewis definition is the most inclusive.

$\text{Arrhenius} \subset \text{Brønsted-Lowry} \subset \text{Lewis}$

Common Lewis Acids

1. Metal Cations

Metal ions have empty orbitals and accept electron pairs from ligands:

$Cu^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4]^{2+}$

• $Cu^{2+}$: Lewis acid (accepts electron pairs)
• $NH_3$: Lewis base (donates lone pair)

2. Molecules with Incomplete Octets

$BF_3 + NH_3 \rightarrow F_3B\text{-}NH_3$

• $BF_3$: Lewis acid (boron has only 6 electrons, empty p orbital)
• $NH_3$: Lewis base (nitrogen has a lone pair)

3. Protons ($H^+$)

The proton itself is a Lewis acid — it accepts an electron pair:

$H^+ + OH^- \rightarrow H_2O$

This shows how the Lewis definition encompasses the Brønsted-Lowry definition.

Common Lewis Bases

Any species with a lone pair can be a Lewis base:

• $NH_3$, $H_2O$, $OH^-$, $F^-$, $CN^-$
• Molecules with lone pairs on N, O, S, or halide ions

Lewis Acid-Base Concept Check 🎯

Coordinate Covalent Bonds

When a Lewis base donates an electron pair to a Lewis acid, the resulting bond is called a coordinate covalent bond (or dative bond).

$F_3B + :NH_3 \rightarrow F_3B\text{←}NH_3$

The arrow ← shows that both electrons in the bond came from the nitrogen of $NH_3$.

In Coordination Chemistry

Metal ions form coordination compounds with Lewis bases (called ligands):

$Fe^{3+} + 6CN^- \rightarrow [Fe(CN)_6]^{3-}$

Lewis Acid-Base Classification 🔍

Theory Comparison 🧮

For each species, identify which acid-base theory can explain its behavior as an acid or base:

1. $NaOH$ acting as a base — which is the simplest theory that explains this? (Enter: Arrhenius, Bronsted-Lowry, or Lewis)

2. $NH_3$ acting as a base (no $OH^-$ in its formula) — simplest theory? (Enter: Arrhenius, Bronsted-Lowry, or Lewis)

3. $BF_3$ acting as an acid (no $H^+$ to donate) — simplest theory? (Enter: Arrhenius, Bronsted-Lowry, or Lewis)

Exit Quiz — Lewis Acids & Bases ✅

📊 The pH Scale

Part 4 of 7 — Measuring Acidity and Basicity

The pH scale provides a convenient way to express the acidity or basicity of a solution. It converts the wide range of $[H^+]$ values (from $10^0$ to $10^{-14}$ M) into a simple 0–14 scale.

pH, pOH, and Their Relationship

pH Definition

$pH = -\log[H^+]$

pOH Definition

$pOH = -\log[OH^-]$

The Key Relationship

At 25°C:

$pH + pOH = 14$

This comes from $K_w$:

$[H^+][OH^-] = 1.0 \times 10^{-14}$

Taking $-\log$ of both sides:

$-\log[H^+] + (-\log[OH^-]) = -\log(1.0 \times 10^{-14})$

$pH + pOH = 14$

Interpreting pH

pH Calculations

From $[H^+]$ to pH

Example: $[H^+] = 3.2 \times 10^{-4}$ M

$pH = -\log(3.2 \times 10^{-4}) = -(-3.49) = 3.49$

From pH to $[H^+]$

Example: $pH = 5.60$

$[H^+] = 10^{-pH} = 10^{-5.60} = 2.5 \times 10^{-6} \text{ M}$

From $[OH^-]$ to pH

Example: $[OH^-] = 4.0 \times 10^{-3}$ M

Step 1: $pOH = -\log(4.0 \times 10^{-3}) = 2.40$

Step 2: $pH = 14 - pOH = 14 - 2.40 = 11.60$

The "p" Notation

The prefix "p" always means $-\log$:

$pX = -\log X$

So $pK_a = -\log K_a$, $pK_b = -\log K_b$, $pK_w = -\log K_w = 14$

pH Concept Check 🎯

Significant Figures in pH

An important AP Chemistry rule:

The number of decimal places in the pH equals the number of significant figures in $[H^+]$.

Examples

The digits before the decimal in pH only indicate the order of magnitude — they don't count as sig figs!

pH Calculation Drill 🧮

1. What is the pH of a solution with $[H^+] = 5.0 \times 10^{-9}$ M? (2 decimal places)

2. What is the $[H^+]$ in a solution with $pH = 4.30$? (Enter in scientific notation, e.g. 5.0e-5)

3. What is the pH of a solution with $[OH^-] = 2.0 \times 10^{-4}$ M? (2 decimal places)

pH Scale Understanding 🔍

Exit Quiz — pH Scale ✅

💪 Strong Acids and Bases — pH Calculations

Part 5 of 7 — Complete Dissociation Means Easy Math

Strong acids and bases dissociate completely in water. This makes pH calculations straightforward — the concentration of $H^+$ or $OH^-$ equals the initial concentration of the acid or base.

pH of Strong Acids

For a strong acid $HA$ at concentration $C$:

$HA \rightarrow H^+ + A^-$

Since dissociation is 100% complete: $[H^+] = C$

$pH = -\log C$

Example 1

What is the pH of 0.025 M $HCl$?

$[H^+] = 0.025 \text{ M}$ $pH = -\log(0.025) = 1.60$

Example 2

What is the pH of 0.0040 M $HNO_3$?

$[H^+] = 0.0040 \text{ M}$ $pH = -\log(0.0040) = 2.40$

Diprotic Strong Acid ($H_2SO_4$)

For the first dissociation (strong): $H_2SO_4 \rightarrow H^+ + HSO_4^-$

For dilute solutions, each mole of $H_2SO_4$ produces approximately 2 moles of $H^+$:

$[H^+] \approx 2C \text{ (for dilute solutions)}$

Note: The second dissociation of $H_2SO_4$ ($HSO_4^-$) is weak ($K_a = 0.012$), so at higher concentrations the approximation $[H^+] = 2C$ may not hold exactly.

pH of Strong Bases

For a strong base like $NaOH$ at concentration $C$:

$NaOH \rightarrow Na^+ + OH^-$

$[OH^-] = C$, then:

$pOH = -\log C$ $pH = 14 - pOH$

Example 1

What is the pH of 0.010 M $NaOH$?

$[OH^-] = 0.010 \text{ M}$ $pOH = -\log(0.010) = 2.00$ $pH = 14 - 2.00 = 12.00$

Group 2 Hydroxides

For $Ba(OH)_2$ or $Ca(OH)_2$:

$Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-$

$[OH^-] = 2C$

Example 2

What is the pH of 0.0050 M $Ba(OH)_2$?

$[OH^-] = 2(0.0050) = 0.010 \text{ M}$ $pOH = -\log(0.010) = 2.00$ $pH = 14 - 2.00 = 12.00$

Strong Acid/Base pH Check 🎯

Mixing and Dilution

Diluting a Strong Acid

Use $M_1V_1 = M_2V_2$:

Example: 25.0 mL of 0.10 M $HCl$ is diluted to 100.0 mL. What is the new pH?

$M_2 = \frac{M_1V_1}{V_2} = \frac{(0.10)(25.0)}{100.0} = 0.025 \text{ M}$

$pH = -\log(0.025) = 1.60$

Mixing Strong Acid and Strong Base

Example: 50.0 mL of 0.10 M $HCl$ + 30.0 mL of 0.10 M $NaOH$

Moles $H^+$ = $0.050 \times 0.10 = 0.0050$ mol

Moles $OH^-$ = $0.030 \times 0.10 = 0.0030$ mol

Excess $H^+$ = $0.0050 - 0.0030 = 0.0020$ mol

Total volume = $80.0$ mL = $0.0800$ L

$[H^+] = \frac{0.0020}{0.0800} = 0.025 \text{ M}$

$pH = -\log(0.025) = 1.60$

Strong Acid/Base Calculation Drill 🧮

1. What is the pH of 0.0020 M $HClO_4$? (2 decimal places)

2. What is the pH of 0.050 M $Ba(OH)_2$? (2 decimal places)

3. 40.0 mL of 0.15 M $HNO_3$ is mixed with 20.0 mL of 0.15 M $NaOH$. What is the pH? (2 decimal places)

Strong Acid/Base Reasoning 🔍

Exit Quiz — Strong Acid/Base pH ✅

🛠️ Problem-Solving Workshop

Part 6 of 7 — Acid-Base Theories and pH

Let's put everything together with multi-step problems that mirror the AP Chemistry exam. Each problem integrates acid-base theory identification, pH calculations, and conceptual reasoning.

Problem 1: Identifying Acid-Base Behavior

Consider these reactions:

Reaction A: $HF(aq) + H_2O(l) \rightleftharpoons F^-(aq) + H_3O^+(aq)$

Reaction B: $BF_3 + F^- \rightarrow BF_4^-$

Reaction C: $NaOH(aq) \rightarrow Na^+(aq) + OH^-(aq)$

For each reaction, the acid-base theory required is:

• Reaction A: Brønsted-Lowry (proton transfer from $HF$ to $H_2O$)
• Reaction B: Lewis ($BF_3$ accepts electron pair from $F^-$)
• Reaction C: Arrhenius ($NaOH$ produces $OH^-$ in water)

Problem 1 Practice 🎯

Problem 2: Multi-Step pH Calculation

A chemist prepares the following solutions:

• Solution A: 0.035 M $HCl$
• Solution B: 0.035 M $NaOH$
• Solution C: 50.0 mL of Solution A mixed with 30.0 mL of Solution B

Solution A pH

$[H^+] = 0.035$ M → $pH = -\log(0.035) = 1.46$

Solution B pH

$[OH^-] = 0.035$ M → $pOH = -\log(0.035) = 1.46$ → $pH = 14 - 1.46 = 12.54$

Solution C pH

Moles $H^+$ = $(0.050)(0.035) = 1.75 \times 10^{-3}$ mol

Moles $OH^-$ = $(0.030)(0.035) = 1.05 \times 10^{-3}$ mol

Excess $H^+$ = $1.75 \times 10^{-3} - 1.05 \times 10^{-3} = 7.0 \times 10^{-4}$ mol

$[H^+] = \frac{7.0 \times 10^{-4}}{0.080} = 8.75 \times 10^{-3}$ M

$pH = -\log(8.75 \times 10^{-3}) = 2.06$

Problem 2 Practice 🧮

A student mixes 25.0 mL of 0.080 M $HNO_3$ with 15.0 mL of 0.080 M $KOH$.

1. How many moles of excess $H^+$ remain? (Enter in scientific notation, e.g. 8.0e-4)

2. What is the total volume in liters? (3 decimal places)

3. What is the pH of the resulting solution? (2 decimal places)

Problem 3: Conceptual Reasoning

The pH of Very Dilute Strong Acids

When a strong acid is extremely dilute (e.g., $10^{-8}$ M $HCl$), you cannot simply say $pH = 8$. An acid solution can never have $pH > 7$!

The autoionization of water contributes $[H^+] = 10^{-7}$ M, which is much larger than the acid's contribution.

Correct approach:

$[H^+]_{\text{total}} = [H^+]_{\text{acid}} + [H^+]_{\text{water}} \approx 10^{-8} + 10^{-7} = 1.1 \times 10^{-7} \text{ M}$

$pH = -\log(1.1 \times 10^{-7}) = 6.96$

This is slightly below 7, as expected for an acidic solution.

Conceptual Check 🎯

Workshop Synthesis 🔍

🎓 Synthesis & AP Review

Part 7 of 7 — Acid-Base Theories and pH

This final part ties together all the concepts: Arrhenius, Brønsted-Lowry, and Lewis definitions, conjugate pairs, the pH scale, and strong acid/base calculations. These are high-yield AP exam topics!

Complete Summary

Three Acid-Base Theories

Key Equations

$pH = -\log[H^+] \qquad pOH = -\log[OH^-]$

$pH + pOH = 14 \qquad K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$

$[H^+] = 10^{-pH} \qquad [OH^-] = 10^{-pOH}$

Strong Acid/Base Rules

• Strong acids: $HCl, HBr, HI, HNO_3, H_2SO_4, HClO_4$
• Strong bases: Group 1 hydroxides + $Ca(OH)_2, Sr(OH)_2, Ba(OH)_2$
• $[H^+] = C_{acid}$ for monoprotic strong acids
• $[OH^-] = nC_{base}$ where $n$ = number of $OH^-$ per formula unit

AP-Style Questions — Set 1 🎯

AP Calculation Practice 🧮

1. What is the pH of a solution made by mixing 100.0 mL of 0.15 M $HCl$ with 75.0 mL of 0.15 M $NaOH$? (2 decimal places)

2. A solution has a pH of 11.50. What is $[H^+]$? (Enter in scientific notation, e.g. 3.2e-12)

3. What volume (mL) of 0.20 M $NaOH$ is needed to exactly neutralize 50.0 mL of 0.10 M $H_2SO_4$? (Enter as whole number)

AP-Style Questions — Set 2 🎯

Comprehensive Review 🔍

Final Exit Quiz — Acid-Base Theories & pH ✅