Acid–Base Reactions

Acid-Base Reaction

An acid releases H⁺ in water, and a base releases OH⁻. pH is defined from the hydrogen-ion concentration: below 7 is acidic, 7 is neutral, and above 7 is basic. When acid and base meet, H⁺ and OH⁻ form water in a neutralization reaction. Here you change [H⁺] and acid volume with the sliders to watch the pH scale and the neutralization curve.

What Are Acids and Bases?

💡 Arrhenius Definition

①Acid: substance that releases H⁺ (hydrogen ion) in water

②Base: substance that releases OH⁻ (hydroxide ion) in water

③HCl → H⁺ + Cl⁻ (acid)

④NaOH → Na⁺ + OH⁻ (base)

pH Scale

[H⁺] exponent (10^-n)3

Definition of pH

pH = -log[H⁺]

[H⁺]: hydrogen ion concentration [M]

Autoionization of Water

[H⁺] × [OH⁻] = K_w = 1.0 × 10⁻¹⁴ (25℃)

Pure water: [H⁺] = [OH⁻] = 10⁻⁷ M → pH = 7

🔍 Try the Slider!

①pH < 7: acidic ([H⁺] > [OH⁻])

②pH = 7: neutral ([H⁺] = [OH⁻])

③pH > 7: basic ([H⁺] < [OH⁻])

④A change of 1 in pH means a 10× change in [H⁺]

Neutralization Reaction

HCl volume5 mL

Neutralization

H⁺ + OH⁻ → H₂O

H⁺ from acid and OH⁻ from base form water

Equivalence Condition

n_acid × valence = n_base × valence

CV(acid) × a = CV(base) × b

💡 Key Points of Titration

①10 mL NaOH + 10 mL HCl → complete neutralization (pH=7)

②Excess acid → pH < 7; excess base → pH > 7

③Temperature peaks at equivalence point (exothermic)

④Titration: determine unknown concentration via known reagent

Work It Out

Example 1

Find the pH of a 0.01 M hydrochloric acid (HCl) solution. (HCl is a strong acid, fully ionized)

HCl is a strong acid, so it ionizes completely and [H⁺] equals the concentration.

[H⁺] = 10⁻² M

Substitute into pH = −log[H⁺].

pH = −log(10⁻²) = 2

▸ pH = 2

For a strong acid the concentration is [H⁺], so just flip the sign of the exponent to get pH.

Example 2

Find the pH of a 0.001 M sodium hydroxide (NaOH) solution. (25℃)

NaOH is a strong base, so [OH⁻]=10⁻³ M and pOH=3.

[OH⁻]=10⁻³ M, pOH=3

Use the relation pH + pOH = 14.

pH = 14 − 3 = 11

▸ pH = 11

For a base, find pOH first, then subtract from 14 to get pH.

Summary

pH Calculation

pH = -log[H⁺]

pH + pOH = 14 (25℃)

2023 KICE mock Chemistry I type, adapted

What volume of 0.1 M NaOH solution is needed to neutralize 100 mL of 0.1 M HCl solution?

①50 mL

②100 mL

③150 mL

④200 mL

⑤250 mL

▸ ② 100 mL

At the neutralization point, n(H⁺)=n(OH⁻) holds.

Find the moles of HCl, then compute the volume of NaOH with the same moles.

n(HCl)=0.1×0.1=0.01 mol → V(NaOH)=0.01/0.1=0.1 L

🎯 Exam Points

①Memorize pH = -log[H⁺]

②Δ pH = 1 → [H⁺] differs by 10×

③Neutralization: H⁺ + OH⁻ → H₂O (net ionic)

④Equivalence: n_acid × valence = n_base × valence

⑤Strong+strong → pH=7, weak acid+strong base → pH>7

← Previous

Redox Reactions

Chemistry II · Gas Laws

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