pH of Strong Acid vs Weak Acid — Why Same Concentration Different pH

easy CBSE JEE-MAIN • NCERT Class 11 • 3 min read
Tags Equilibrium

Question

Two solutions: 0.1 M HCl and 0.1 M CH₃COOH (acetic acid). Same concentration, same temperature. Which has lower pH, and why?

Calculate the pH of 0.1 M CH₃COOH given Ka = 1.8 × 10⁻⁵.

Solution — Step by Step

HCl is a strong acid — it dissociates 100% in water. So for 0.1 M HCl:

[H+]=0.1 M[\text{H}^+] = 0.1 \text{ M}

Acetic acid is weak — it reaches an equilibrium, meaning only a fraction of molecules give up their proton.

The dissociation: CH3COOH⇌CH3COO−+H+\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+

Ka=[CH3COO−][H+][CH3COOH]K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]}

Let xx = concentration of H⁺ produced. At equilibrium: numerator = x⋅xx \cdot x, denominator = 0.1−x0.1 - x.

Since Ka is very small (1.8 × 10⁻⁵) compared to 0.1 M, we assume x≪0.1x \ll 0.1, so (0.1−x)≈0.1(0.1 - x) \approx 0.1.

1.8×10−5=x20.11.8 \times 10^{-5} = \frac{x^2}{0.1} x2=1.8×10−6x^2 = 1.8 \times 10^{-6} x=[H+]=1.34×10−3 Mx = [\text{H}^+] = 1.34 \times 10^{-3} \text{ M}

For HCl:

pH=−log⁡(0.1)=1.0\text{pH} = -\log(0.1) = \boxed{1.0}

For CH₃COOH:

pH=−log⁡(1.34×10−3)=2.87\text{pH} = -\log(1.34 \times 10^{-3}) = 2.87

HCl has pH = 1.0; acetic acid has pH = 2.87 at the same 0.1 M concentration.

Degree of dissociation α=xC=1.34×10−30.1=0.0134\alpha = \frac{x}{C} = \frac{1.34 \times 10^{-3}}{0.1} = 0.0134

So only 1.34% of acetic acid molecules actually dissociate. That’s why the H⁺ concentration — and hence acidity — is so much lower than HCl.

Why This Works

The pH difference comes entirely from how completely the acid hands over its proton to water. A strong acid like HCl has such a low pKa (≈ −7) that the reverse reaction essentially doesn’t happen — every molecule dissociates. The equilibrium lies completely to the right.

Weak acids like CH₃COOH have Ka values around 10⁻⁵, meaning the undissociated form is heavily favored. Most molecules stay intact, so the actual H⁺ in solution is far less than the nominal concentration suggests.

This is why “concentration” alone doesn’t tell you the pH of a weak acid — you always need Ka. Two 0.1 M solutions can have H⁺ concentrations differing by a factor of 75, as we see here (0.1 vs 0.00134).

Quick approximation rule: for weak acid with concentration C and dissociation constant Ka, [H+]≈Ka⋅C[\text{H}^+] \approx \sqrt{K_a \cdot C}. Valid when α<5%\alpha < 5\% — always verify this after calculating.

Alternative Method — Using Degree of Dissociation Formula

For weak acids, there’s a direct formula:

α=KaC\alpha = \sqrt{\frac{K_a}{C}} α=1.8×10−50.1=1.8×10−4=0.0134\alpha = \sqrt{\frac{1.8 \times 10^{-5}}{0.1}} = \sqrt{1.8 \times 10^{-4}} = 0.0134

Then [H+]=C⋅α=0.1×0.0134=1.34×10−3[\text{H}^+] = C \cdot \alpha = 0.1 \times 0.0134 = 1.34 \times 10^{-3} M — same answer, faster.

This formula is especially handy in MCQs where you need the degree of dissociation directly without setting up the full ICE table.

Common Mistake

Treating weak acid concentration as [H⁺] directly. Students write pH = −log(0.1) = 1 for acetic acid — the same as HCl. This is wrong. For a weak acid, [H⁺] ≠ initial concentration. You must use Ka to find the actual [H⁺] at equilibrium. This mistake costs marks in both CBSE board papers and JEE Main, where the same-concentration comparison is a favourite MCQ setup.

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