Calculating pH of Strong Acids and Strong Bases
Strong acids and bases completely dissociate in water. This makes their pH calculations straightforward. Let’s learn how to solve these problems step by step.
Strong Acids
Definition: Strong acids completely dissociate in aqueous solution to produce H⁺ ions.
Common Strong Acids:
- HCl, HBr, HI (halogen acids)
- HNO₃ (nitric acid)
- H₂SO₄ (sulfuric acid)
- HClO₄ (perchloric acid)
pH Calculation for Strong Acids
Formula: For a strong monoprotic acid:
pH = -log[H⁺] = -log(concentration of acid)
Worked Examples – Strong Acids
Example 1: 0.1M HCl
HCl completely dissociates:
HCl(aq) → H⁺(aq) + Cl⁻(aq) 0.1M 0.1M 0.1M
[H⁺] = 0.1 M = 1×10⁻¹ mol/dm³
pH = -log(10⁻¹) = 1
Example 2: 0.001M H₂SO₄
H₂SO₄ is diprotic (produces 2 H⁺ ions):
H₂SO₄(aq) → 2H⁺(aq) + SO₄²⁻(aq) 0.001M 0.002M
[H⁺] = 2 × 10⁻³ = 0.002 mol/dm³
pH = -log(0.002) = -log(2×10⁻³) = 2.70
Example 3: Calculate [H⁺] for pH = 3.4
pH = -log[H⁺]
3.4 = -log[H⁺]
log[H⁺] = -3.4
[H⁺] = 10⁻³·⁴ = 3.98 × 10⁻⁴ mol/dm³
Strong Bases
Definition: Strong bases completely dissociate in aqueous solution to produce OH⁻ ions.
Common Strong Bases:
- Group I hydroxides: LiOH, NaOH, KOH
- Group II hydroxides: Ca(OH)₂, Ba(OH)₂
pH Calculation for Strong Bases
Method 1 – Using pOH:
pOH = -log[OH⁻] pH = 14 - pOH
Method 2 – Direct Calculation:
First, calculate pOH, then find pH using pH + pOH = 14
Worked Examples – Strong Bases
Example 1: 0.025M NaOH
NaOH completely dissociates:
NaOH(aq) → Na⁺(aq) + OH⁻(aq) 0.025M 0.025M
[OH⁻] = 0.025 M
pOH = -log(0.025) = 1.60
pH = 14 – pOH = 14 – 1.60 = 12.4
Example 2: 0.01M Ca(OH)₂
Ca(OH)₂ produces 2 OH⁻ ions:
Ca(OH)₂(aq) → Ca²⁺(aq) + 2OH⁻(aq) 0.01M 0.02M
[OH⁻] = 2 × 0.01 = 0.02 M = 2×10⁻² mol/dm³
pOH = -log(2×10⁻²) = 1.70
pH = 14 – 1.70 = 12.30
Mixed Acid-Base Problems
Example: pH of Mixture
25 cm³ of 0.16M NaOH is added to 50 cm³ of 0.1M HCl. What is the pH of the mixture?
Solution:
Number of moles of NaOH = 0.16 × 25/1000 = 0.004 moles
Number of moles of HCl = 0.1 × 50/1000 = 0.005 moles
Excess moles of HCl = 0.005 – 0.004 = 0.001 moles
Total volume = 25 + 50 = 75 cm³ = 0.075 dm³
[H⁺] = 0.001/0.075 = 0.0133 M
pH = -log(0.0133) = 1.88
Summary Table
| Type | Example | Dissociation | pH Calculation |
|---|---|---|---|
| Strong Monoprotic Acid | HCl, HNO₃ | Complete, produces 1 H⁺ | pH = -log[acid] |
| Strong Diprotic Acid | H₂SO₄ | Complete, produces 2 H⁺ | pH = -log(2×[acid]) |
| Strong Base (1 OH⁻) | NaOH, KOH | Complete, produces 1 OH⁻ | pOH = -log[base], pH = 14-pOH |
| Strong Base (2 OH⁻) | Ca(OH)₂ | Complete, produces 2 OH⁻ | pOH = -log(2×[base]) |
Key Points to Remember
- Strong acids and bases completely dissociate in water
- For diprotic acids/bases, multiply concentration by the number of H⁺ or OH⁻ produced
- Always use pH + pOH = 14 at 25°C
- pH is inversely related to hydrogen ion concentration
