Acids & Bases — Study Guide

1

Acid & Base Definitions

Arrhenius and Brønsted-Lowry models

There are two main definitions of acids and bases at the Grade 11 level. Both are valid; the Brønsted-Lowry definition is broader and more useful.

Model	Acid	Base
Arrhenius	Produces H³O&sup+; (or H&sup+;) in water	Produces OH&sup-; in water
Brønsted-Lowry	Proton (H&sup+;) donor	Proton (H&sup+;) acceptor

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Why is Brønsted-Lowry more useful?
The Arrhenius model only applies to water solutions. Brønsted-Lowry works in any solvent. More importantly, it focuses on what actually happens at the molecular level: one species donates an H&sup+; and another accepts it. It also introduces conjugate pairs, which are essential for understanding weak acids and buffers.

Conjugate Acid-Base Pairs

When an acid donates H&sup+;, the species it becomes is its conjugate base. When a base accepts H&sup+;, it becomes its conjugate acid. The pair differing by one H&sup+; is a conjugate pair.

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Example: HCl + H⊂2;O → Cl&sup-; + H⊂3;O&sup+;
HCl is the acid (donates H&sup+;); Cl&sup-; is its conjugate base.
H⊂2;O is the base (accepts H&sup+;); H⊂3;O&sup+; is its conjugate acid.
Conjugate pairs: (HCl / Cl&sup-;) and (H⊂3;O&sup+; / H⊂2;O).

HCl donates H&sup+; to water. The conjugate pairs are: (HCl, Cl&sup-;) shown in red, and (H⊂2;O, H⊂3;O&sup+;) shown in green.

✓ Checkpoint 1

a) Identify the conjugate acid-base pairs in: NH⊂3; + H⊂2;O → NH⊂4;&sup+; + OH&sup-;
b) Is H⊂2;O acting as an acid or a base in this reaction? Compare to how it acted in the HCl reaction above.
c) Write the conjugate base of H⊂2;SO⊂4; and the conjugate acid of CO⊂3;².

a) NH⊂3; accepts H&sup+; (base) → NH⊂4;&sup+; (its conjugate acid). H⊂2;O donates H&sup+; (acid) → OH&sup-; (its conjugate base). Pairs: (NH⊂4;&sup+;, NH⊂3;) and (H⊂2;O, OH&sup-;).

b) Here H⊂2;O acts as an acid (donates H&sup+; to NH⊂3;). In the HCl reaction, H⊂2;O acted as a base (accepted H&sup+; from HCl). This is called amphoterism — water can act as either acid or base depending on its reaction partner.

c) Conjugate base of H⊂2;SO⊂4;: HSO⊂4;&sup-; (loses one H&sup+;). Conjugate acid of CO⊂3;²: HCO⊂3;&sup-; (gains one H&sup+;).

2

pH, pOH, and Kw

Measuring acidity on a logarithmic scale

The pH scale measures the concentration of H⊂3;O&sup+; (or H&sup+;) ions in solution. Because concentrations range over many powers of 10, a logarithmic scale is used.

pH definition

pH = −log[H&sup+;]

pOH definition

pOH = −log[OH&sup-;]

Kw (at 25°C)

[H&sup+;][OH&sup-;] = 10&sup-;¹&sup4;

pH + pOH

pH + pOH = 14 (at 25°C)

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Why is pH = 7 neutral?
In pure water at 25°C, water self-ionises: H⊂2;O ⇋ H&sup+; + OH&sup-;. The equilibrium constant Kw = [H&sup+;][OH&sup-;] = 10&sup-;¹&sup4;. In pure water, [H&sup+;] = [OH&sup-;], so each is 10&sup-;&sup7; M. pH = −log(10&sup-;&sup7;) = 7. Any [H&sup+;] above 10&sup-;&sup7; gives pH < 7 (acidic); below 10&sup-;&sup7; gives pH > 7 (basic).

The pH scale runs from 0 (most acidic) to 14 (most basic). Each unit represents a 10-fold change in [H&sup+;].

★ Easy

Calculate pH from [H&sup+;]

A solution has [H&sup+;] = 3.5 × 10&sup-;&sup4; mol/L. Find the pH. Is it acidic or basic?

Show solution

1

Apply pH = −log[H&sup+;]

pH = −log(3.5 × 10&sup-;&sup4;)
pH = −(log 3.5 + log 10&sup-;&sup4;)
pH = −(0.544 + (−4)) = −(0.544 − 4) = 3.46

2

Interpret

pH = 3.46 < 7 → acidic. This is approximately the pH of orange juice.

Answer: pH = 3.46 (acidic)

★★ Medium

Find [H&sup+;] from pH, then [OH&sup-;] from Kw

A solution has pH = 9.30. Find [H&sup+;] and [OH&sup-;] at 25°C.

Show solution

1

Find [H&sup+;] using antilog

[H&sup+;] = 10&sup-;pH = 10&sup-;9·30 = 5.01 × 10&sup-;¹&sup0; mol/L

2

Find [OH&sup-;] from Kw

pOH = 14 − pH = 14 − 9.30 = 4.70
[OH&sup-;] = 10&sup-;4·70 = 2.0 × 10&sup-;5 mol/L

3

Verify: [H&sup+;][OH&sup-;] = Kw

(5.01 × 10&sup-;10)(2.0 × 10&sup-;5) = 1.0 × 10&sup-;14 ✓

Answer: [H&sup+;] = 5.0 × 10&sup-;10 mol/L [OH&sup-;] = 2.0 × 10&sup-;5 mol/L

✓ Checkpoint 2

a) Find the pH of a solution with [H&sup+;] = 1.0 × 10&sup-;2 mol/L.
b) A solution has pH = 11.6. Find [H&sup+;], [OH&sup-;], and pOH.
c) If [OH&sup-;] = 5.0 × 10&sup-;4 mol/L, find the pH.

a) pH = −log(1.0 × 10&sup-;2) = 2.0

b) [H&sup+;] = 10&sup-;11·6 = 2.51 × 10&sup-;12 mol/L. pOH = 14 − 11.6 = 2.4. [OH&sup-;] = 10&sup-;2·4 = 3.98 × 10&sup-;3 mol/L

c) pOH = −log(5.0 × 10&sup-;4) = 3.30. pH = 14 − 3.30 = 10.70 (basic)

3

Strong vs Weak Acids & Bases

Complete vs partial dissociation and the Ka/Kb

A strong acid dissociates completely in water: every molecule donates its H&sup+;. A weak acid only partially dissociates; most molecules remain intact.

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Common strong acids (memorize): HCl, HBr, HI, HNO⊂3;, H⊂2;SO⊂4;, HClO⊂4;
Common strong bases (memorize): NaOH, KOH, LiOH, Ba(OH)⊂2;, Ca(OH)⊂2;
Everything else: treat as weak unless told otherwise.

Calculating pH of a Strong Acid

For a strong acid, complete dissociation means [H&sup+;] = initial concentration of acid.

★ Easy

pH of strong acid

Find the pH of 0.025 mol/L HNO⊂3;(aq).

Show solution

1

HNO⊂3; is a strong acid: dissociates completely

HNO⊂3; → H&sup+; + NO⊂3;&sup-;
[H&sup+;] = 0.025 mol/L

2

Calculate pH

pH = −log(0.025) = −log(2.5 × 10&sup-;2) = 1.60

Answer: pH = 1.60

Ka and Kb for Weak Acids & Bases

Weak acids establish an equilibrium. The acid dissociation constant Ka measures how far the equilibrium lies toward dissociation (larger Ka = stronger weak acid).

Weak acid equilibrium

HA ⇋ H&sup+; + A&sup-; Ka = [H&sup+;][A&sup-;] / [HA]

For a simple Ka calculation: if Ka << initial concentration, use approximation [H&sup+;] ≈ √(Ka × C)

★★ Medium

pH of a weak acid

Find the pH of 0.100 mol/L acetic acid (CH⊂3;COOH), Ka = 1.8 × 10&sup-;5.

Show solution

1

Write ICE table (Initial, Change, Equilibrium)

CH⊂3;COOH ⇋ H&sup+; + CH⊂3;COO&sup-;
I: 0.100 0 0
C: −x +x +x
E: 0.100−x x x

2

Substitute into Ka expression

Ka = x² / (0.100 − x) ≈ x² / 0.100
(approximation valid since Ka << 0.100)

3

Solve for x = [H&sup+;]

x² = Ka × 0.100 = 1.8 × 10&sup-;5 × 0.100 = 1.8 × 10&sup-;6
x = √(1.8 × 10&sup-;6) = 1.34 × 10&sup-;3 mol/L

4

Calculate pH

pH = −log(1.34 × 10&sup-;3) = 2.87

Answer: pH = 2.87 (compare: strong acid at same concentration would give pH = 1.0)

✓ Checkpoint 3

a) 0.050 mol/L HCl vs 0.050 mol/L acetic acid (Ka = 1.8 × 10&sup-;5). Which has lower pH and why?
b) Find the pH of 0.20 mol/L HCl and compare to 0.20 mol/L HF (Ka = 6.8 × 10&sup-;4).

a) HCl has the lower pH. HCl is a strong acid → [H&sup+;] = 0.050 → pH = 1.30. Acetic acid is weak: [H&sup+;] = √(1.8×10&sup-;5 × 0.050) = √(9.0×10&sup-;7) = 9.49×10&sup-;4 → pH = 3.02. Lower pH = more acidic = HCl. Weak acids only partially dissociate, so fewer H&sup+; ions are released.

b) HCl: pH = −log(0.20) = 0.70. HF: [H&sup+;] = √(6.8×10&sup-;4 × 0.20) = √(1.36×10&sup-;4) = 0.01166 → pH = 1.93.

4

Neutralization Reactions

Acid + base → salt + water

When an acid and a base react, the H&sup+; from the acid combines with the OH&sup-; from the base to form water. The remaining ions form a salt.

General Neutralization

acid + base → salt + water

H&sup+; + OH&sup-; → H⊂2;O is the net ionic equation for strong acid + strong base

Reaction type	Example	Product salt	Salt pH
Strong + Strong	HCl + NaOH	NaCl	7 (neutral)
Strong acid + Weak base	HCl + NH⊂3;	NH⊂4;Cl	< 7 (acidic)
Weak acid + Strong base	CH⊂3;COOH + NaOH	CH⊂3;COONa	> 7 (basic)

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Why does NH⊂4;Cl make an acidic solution?
NH⊂4;&sup+; (ammonium) is the conjugate acid of a weak base (NH⊂3;). In solution, it donates H&sup+; back to water: NH⊂4;&sup+; + H⊂2;O ⇋ NH⊂3; + H⊂3;O&sup+;. This releases extra H&sup+;, making the solution acidic. Strong acid anions (like Cl&sup-;) are neutral spectators — the conjugate base of a strong acid has no significant tendency to accept H&sup+;.

★★ Medium

Neutralization stoichiometry

What volume of 0.500 mol/L NaOH is needed to completely neutralize 25.0 mL of 0.300 mol/L H⊂2;SO⊂4;? Reaction: H⊂2;SO⊂4; + 2NaOH → Na⊂2;SO⊂4; + 2H⊂2;O

Show solution

1

Find moles of H⊂2;SO⊂4;

n(H⊂2;SO⊂4;) = C × V = 0.300 mol/L × 0.0250 L = 0.00750 mol

2

Use mole ratio (1:2)

n(NaOH) = 0.00750 × 2 = 0.01500 mol

3

Find volume of NaOH

V(NaOH) = n/C = 0.01500 / 0.500 = 0.0300 L = 30.0 mL

Answer: 30.0 mL of 0.500 mol/L NaOH

✓ Checkpoint 4

a) 15.0 mL of 0.200 mol/L HCl is mixed with 15.0 mL of 0.200 mol/L NaOH. What is the pH of the resulting solution?
b) 20.0 mL of 0.100 mol/L HCl is mixed with 10.0 mL of 0.100 mol/L NaOH. Is the resulting solution acidic, basic, or neutral? Find the pH.

a) Equal moles HCl and NaOH. n(HCl) = 0.0150 × 0.200 = 0.00300 mol. n(NaOH) = 0.00300 mol. Complete neutralization → salt NaCl solution. pH = 7.0

b) n(HCl) = 0.0200 × 0.100 = 0.00200 mol. n(NaOH) = 0.0100 × 0.100 = 0.00100 mol. Excess HCl = 0.00100 mol. Total volume = 30.0 mL = 0.0300 L. [H&sup+;] = 0.00100/0.0300 = 0.0333 mol/L. pH = −log(0.0333) = 1.48 (acidic)

5

Acid-Base Titration

Equivalence point, indicators, and titration calculations

A titration is a procedure to determine the unknown concentration of an acid or base by reacting it with a standard solution of known concentration. The equivalence point is where moles of acid equal moles of base (stoichiometrically).

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At the equivalence point for a strong acid – strong base titration: moles of H&sup+; = moles of OH&sup-;. The key equation:
C⊂acid; × V⊂acid; × n⊂acid; = C⊂base; × V⊂base; × n⊂base;
where n = number of H&sup+; or OH&sup-; per formula unit.

Indicators

An indicator is a weak acid or base that changes colour at a specific pH. Choose an indicator whose colour change range includes the equivalence point pH.

Indicator	Acid colour	Base colour	pH range
Methyl orange	Red	Yellow	3.1–4.4
Methyl red	Red	Yellow	4.4–6.2
Litmus	Red	Blue	5.0–8.0
Phenolphthalein	Colourless	Pink	8.2–10.0

✔

Choosing an indicator: Strong acid + strong base → equivalence at pH 7 → use phenolphthalein or litmus. Strong acid + weak base → equivalence at pH < 7 → use methyl orange. Weak acid + strong base → equivalence at pH > 7 → use phenolphthalein.

Titration curve for a strong acid (e.g. HCl) titrated with a strong base (e.g. NaOH). The pH rises slowly at first, then shoots up steeply at the equivalence point (pH 7 for strong-strong). Phenolphthalein turns pink here.

★★ Medium

Titration calculation

20.00 mL of HCl solution requires 28.50 mL of 0.150 mol/L NaOH to reach the equivalence point. Find the concentration of the HCl.

Show solution

1

Find moles of NaOH used

n(NaOH) = 0.150 mol/L × 0.02850 L = 4.275 × 10&sup-;3 mol

2

At equivalence: n(HCl) = n(NaOH)

n(HCl) = 4.275 × 10&sup-;3 mol

3

Find concentration of HCl

C(HCl) = n/V = 4.275 × 10&sup-;3 / 0.02000 = 0.214 mol/L

Answer: [HCl] = 0.214 mol/L

✓ Checkpoint 5

a) 15.00 mL of Ba(OH)⊂2; solution is titrated with 0.200 mol/L HCl. The equivalence point requires 36.0 mL of HCl. Find the concentration of Ba(OH)⊂2;. (Ba(OH)⊂2; provides 2 OH&sup-; per formula unit.)
b) Would you use phenolphthalein or methyl orange for a titration of acetic acid (weak acid) vs NaOH (strong base)? Why?

a) n(HCl) = 0.200 × 0.0360 = 0.00720 mol. n(OH&sup-;) = n(HCl) = 0.00720 mol. n(Ba(OH)⊂2;) = 0.00720 / 2 = 0.00360 mol.

C(Ba(OH)⊂2;) = 0.00360 / 0.01500 = 0.240 mol/L

b) Use phenolphthalein. Weak acid + strong base → equivalence point is basic (pH > 7, around pH 8–9). Phenolphthalein changes from colourless to pink in the range 8.2–10.0, which straddles the equivalence point. Methyl orange changes at pH 3.1–4.4, well before the equivalence point — it would give a wrong endpoint.

Acids, Bases & pH

Conjugate Acid-Base Pairs

Calculating pH of a Strong Acid

Ka and Kb for Weak Acids & Bases

Indicators