Acid-Base Theories and pH Scale

Understand Arrhenius, Brønsted-Lowry, and Lewis theories; master pH, pOH, and the pH scale.

Acid-Base Theories and pH Scale

Arrhenius Theory

Simplest definition (aqueous solutions):

Acid: Produces H⁺ in water

HCl → H⁺ + Cl⁻

Base: Produces OH⁻ in water

NaOH → Na⁺ + OH⁻

Limitation: Only for aqueous solutions

Brønsted-Lowry Theory

More general definition:

Acid: Proton (H⁺) donor Base: Proton (H⁺) acceptor

Example:

$\ce{HCl + H2O -> H3O^+ + Cl^-}$

HCl: acid (donates H⁺)
H₂O: base (accepts H⁺)

Conjugate acid-base pairs:

$\ce{HA + B <=> A^- + HB^+}$

HA/A⁻: conjugate pair (differ by H⁺)
B/HB⁺: conjugate pair

Example: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

NH₃/NH₄⁺: conjugate pair
H₂O/OH⁻: conjugate pair

Lewis Theory

Most general:

Acid: Electron pair acceptor Base: Electron pair donor

Example: BF₃ + NH₃ → F₃B-NH₃

BF₃: Lewis acid (accepts electron pair)
NH₃: Lewis base (donates electron pair)

All Brønsted acids/bases are Lewis, but not vice versa

Autoionization of Water

Water self-ionizes:

$\ce{2H2O <=> H3O^+ + OH^-}$

Simplified: H₂O ⇌ H⁺ + OH⁻

Ion product constant (K_w):

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

Key relationships:

Pure water: [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ M
Acidic: [H⁺] > [OH⁻]
Basic: [H⁺] < [OH⁻]
Always: [H⁺][OH⁻] = 1.0 × 10⁻¹⁴

pH Scale

pH definition:

$\text{pH} = -\log[H^+]$

pOH definition:

$\text{pOH} = -\log[OH^-]$

Relationship:

$\text{pH} + \text{pOH} = 14.00 \text{ at 25°C}$

pH Scale interpretation:

| pH | [H⁺] (M) | Type | |----|----------|------| | 0 | 1 | Very acidic | | 1 | 0.1 | Strong acid | | 7 | 1.0 × 10⁻⁷ | Neutral | | 13 | 1.0 × 10⁻¹³ | Strong base | | 14 | 1.0 × 10⁻¹⁴ | Very basic |

Ranges:

pH < 7: Acidic
pH = 7: Neutral
pH > 7: Basic

Calculating pH

From [H⁺]:

$\text{pH} = -\log[H^+]$

Example: [H⁺] = 1.0 × 10⁻³ M

pH = -log(1.0 × 10⁻³) = 3.00

From [OH⁻]:

Calculate pOH = -log[OH⁻]
pH = 14.00 - pOH

Example: [OH⁻] = 1.0 × 10⁻⁴ M

pOH = -log(1.0 × 10⁻⁴) = 4.00 pH = 14.00 - 4.00 = 10.00

Calculating [H⁺] from pH

Inverse relationship:

$[H^+] = 10^{-\text{pH}}$

Example: pH = 5.00

[H⁺] = 10⁻⁵·⁰⁰ = 1.0 × 10⁻⁵ M

Strong Acids and Bases

Strong acids (completely ionize):

HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄

For strong monoprotic acid:

[H⁺] = [acid]

Example: 0.010 M HCl

[H⁺] = 0.010 M
pH = -log(0.010) = 2.00

Strong bases (completely dissociate):

Group 1A hydroxides: LiOH, NaOH, KOH
Group 2A hydroxides: Ca(OH)₂, Ba(OH)₂, Sr(OH)₂

For strong base:

[OH⁻] = [base] × (number of OH⁻)

Example: 0.010 M NaOH

[OH⁻] = 0.010 M
pOH = 2.00
pH = 12.00

Example: 0.010 M Ca(OH)₂

[OH⁻] = 2 × 0.010 = 0.020 M
pOH = -log(0.020) = 1.70
pH = 14.00 - 1.70 = 12.30

Significant Figures in pH

pH has decimal places = sig figs in [H⁺]

Example: [H⁺] = 2.5 × 10⁻³ M (2 sig figs)

pH = 2.60 (2 decimal places)

The digits before decimal point come from exponent

📚 Practice Problems

1Problem 1easy

❓ Question:

Identify the conjugate acid-base pairs in: HF(aq) + NH₃(aq) ⇌ F⁻(aq) + NH₄⁺(aq)

💡 Show Solution

Reaction: HF(aq) + NH₃(aq) ⇌ F⁻(aq) + NH₄⁺(aq)

Identify acids and bases:

Left side (reactants):

HF: Has H⁺ to donate → acid
NH₃: Can accept H⁺ → base

Right side (products):

F⁻: Accepted H⁺ to become HF → base
NH₄⁺: Donated H⁺ to become NH₃ → acid

Conjugate pairs differ by one H⁺:

Pair 1: HF and F⁻

HF → F⁻ + H⁺
HF is acid, F⁻ is its conjugate base
HF/F⁻ conjugate acid-base pair

Pair 2: NH₃ and NH₄⁺

NH₃ + H⁺ → NH₄⁺
NH₃ is base, NH₄⁺ is its conjugate acid
NH₃/NH₄⁺ conjugate acid-base pair

Summary:

| Species | Role | Conjugate | |---------|------|-----------| | HF | Acid | F⁻ (conjugate base) | | NH₃ | Base | NH₄⁺ (conjugate acid) | | F⁻ | Conjugate base | of HF | | NH₄⁺ | Conjugate acid | of NH₃ |

Answers:

Pair 1: HF/F⁻
Pair 2: NH₃/NH₄⁺

Pattern: Conjugate pairs always differ by exactly one proton (H⁺)

2Problem 2easy

❓ Question:

(a) Calculate the pH of a solution with [H⁺] = 2.5 × 10⁻⁴ M. (b) Calculate the pOH. (c) Is this solution acidic, basic, or neutral?

💡 Show Solution

Solution:

(a) Calculate pH: pH = -log[H⁺] pH = -log(2.5 × 10⁻⁴) pH = 3.60

(b) Calculate pOH: pH + pOH = 14.00 pOH = 14.00 - 3.60 = 10.40

3Problem 3medium

❓ Question:

Calculate the pH and pOH of a solution with [OH⁻] = 3.5 × 10⁻⁴ M at 25°C.

💡 Show Solution

Given:

[OH⁻] = 3.5 × 10⁻⁴ M
T = 25°C (K_w = 1.0 × 10⁻¹⁴)

Calculate pOH:

$\text{pOH} = -\log[OH^-]$

$\text{pOH} = -\log(3.5 \times 10^{-4})$

Using calculator:

$\text{pOH} = -(\log 3.5 + \log 10^{-4})$

$\text{pOH} = -(0.544 - 4)$

$\text{pOH} = -(-3.456)$

$\text{pOH} = 3.456$

Round to 2 decimal places (2 sig figs in 3.5):

pOH = 3.46

Calculate pH:

Use relationship:

$\text{pH} + \text{pOH} = 14.00$

$\text{pH} = 14.00 - \text{pOH}$

$\text{pH} = 14.00 - 3.46$

$\text{pH} = 10.54$

Verify using [H⁺]:

From K_w:

$[H^+] = \frac{K_w}{[OH^-]} = \frac{1.0 \times 10^{-14}}{3.5 \times 10^{-4}}$

$[H^+] = 2.86 \times 10^{-11} \text{ M}$

pH:

$\text{pH} = -\log(2.86 \times 10^{-11}) = 10.54$ ✓

Answers:

pOH = 3.46
pH = 10.54

Interpretation: pH > 7, solution is basic

4Problem 4medium

❓ Question:

Calculate the pH of a 0.025 M HCl solution. Assume HCl is a strong acid that completely dissociates.

💡 Show Solution

Solution:

HCl dissociation: HCl → H⁺ + Cl⁻

Since HCl is a strong acid, it completely dissociates: [H⁺] = 0.025 M

Calculate pH: pH = -log[H⁺] pH = -log(0.025) pH = -log(2.5 × 10⁻²) pH = 1.60

Note: For strong acids (HCl, HNO₃, H₂SO₄, HBr, HI, HClO₄), pH calculation is straightforward because [H⁺] = initial acid concentration.

5Problem 5hard

❓ Question:

A solution of Ba(OH)₂ has pH = 12.60 at 25°C. Calculate the concentration of Ba(OH)₂.

💡 Show Solution

Given:

Compound: Ba(OH)₂ (strong base)
pH = 12.60
T = 25°C

Note: Ba(OH)₂ → Ba²⁺ + 2OH⁻ (2 OH⁻ per formula unit)

Calculate pOH:

$\text{pH} + \text{pOH} = 14.00$

$\text{pOH} = 14.00 - 12.60 = 1.40$

Calculate [OH⁻]:

$[OH^-] = 10^{-\text{pOH}}$

$[OH^-] = 10^{-1.40}$

$[OH^-] = 3.98 \times 10^{-2} \text{ M}$

$[OH^-] = 0.0398 \text{ M}$

Calculate [Ba(OH)₂]:

Stoichiometry: Ba(OH)₂ → Ba²⁺ + 2OH⁻

Each Ba(OH)₂ produces 2 OH⁻:

$[Ba(OH)_2] = \frac{[OH^-]}{2}$

$[Ba(OH)_2] = \frac{0.0398}{2}$

$[Ba(OH)_2] = 0.0199 \text{ M}$

Answer: [Ba(OH)₂] = 0.020 M or 2.0 × 10⁻² M

Verify:

If [Ba(OH)₂] = 0.020 M:

[OH⁻] = 2 × 0.020 = 0.040 M
pOH = -log(0.040) = 1.40
pH = 14.00 - 1.40 = 12.60 ✓

Key point: Remember to account for stoichiometry!

Ba(OH)₂ produces 2 moles OH⁻ per mole compound.

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