pH of 0.1 M HCl vs 0.1 M Acetic Acid — Strong vs Weak Acid

hard CBSE JEE-MAIN NEET • NCERT Class 11 Chapter 7 • 4 min read
Tags Equilibrium

Question

Compare the pH of 0.1 M HCl and 0.1 M CH₃COOH (acetic acid). Calculate the pH of each and explain why they differ despite having the same concentration.

Given: KaK_a of acetic acid = 1.8×10−51.8 \times 10^{-5}

Solution — Step by Step

HCl is a strong acid — it dissociates 100% in water. No equilibrium, no calculation needed.

HCl→H++Cl−\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-

So [H+]=0.1 M=10−1 M[\text{H}^+] = 0.1 \text{ M} = 10^{-1} \text{ M}, which gives pH=−log⁡(10−1)=1\text{pH} = -\log(10^{-1}) = \mathbf{1}.

Acetic acid is a weak acid — it partially dissociates. We must use the KaK_a expression.

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

Let xx = concentration of H+\text{H}^+ formed at equilibrium. Initial concentration C=0.1C = 0.1 M.

SpeciesInitialChangeEquilibrium
CH₃COOH0.1−x0.1 − x
H⁺0+xx
CH₃COO⁻0+xx
 Ka=[H+][CH3COO−][CH3COOH]=x20.1−xK_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} = \frac{x^2}{0.1 - x}

Since Ka=1.8×10−5K_a = 1.8 \times 10^{-5} is very small compared to C=0.1C = 0.1, we can safely 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}

Check: x/C=(1.34×10−3)/0.1=1.34%x/C = (1.34 \times 10^{-3})/0.1 = 1.34\%. Since this is well below 5%, the approximation is valid.

pH=−log⁡(1.34×10−3)=3−log⁡(1.34)≈3−0.127=2.87\text{pH} = -\log(1.34 \times 10^{-3}) = 3 - \log(1.34) \approx 3 - 0.127 = \mathbf{2.87}
AcidConcentration[H⁺]pH
HCl (strong)0.1 M0.1 M1.00
CH₃COOH (weak)0.1 M1.34 × 10⁻³ M2.87

Same concentration, but the weak acid has almost 75 times fewer H⁺ ions in solution.

Why This Works

The core idea is that “concentration” tells us how much acid we dissolved — not how much actually dissociated. A strong acid hands over all its protons immediately. A weak acid holds most of them back, releasing only a small fraction governed by the equilibrium constant KaK_a.

The KaK_a value is the acid’s “generosity” with protons. For acetic acid, Ka=1.8×10−5K_a = 1.8 \times 10^{-5} means at equilibrium, the product side is highly disfavored. Only about 1.3% of the molecules dissociate at this concentration.

This distinction is critical for NEET and JEE: concentration and strength are completely independent properties. A 10 M solution of acetic acid is concentrated but still weak. A 10−410^{-4} M HCl is dilute but still strong.

Alternative Method — Using Degree of Dissociation

For weak acid problems, you can directly use the formula for degree of dissociation α\alpha:

α=KaC\alpha = \sqrt{\frac{K_a}{C}} α=1.8×10−50.1=1.8×10−4=0.01342\alpha = \sqrt{\frac{1.8 \times 10^{-5}}{0.1}} = \sqrt{1.8 \times 10^{-4}} = 0.01342

So α≈1.34%\alpha \approx 1.34\%, meaning [H+]=Cα=0.1×0.01342=1.342×10−3[\text{H}^+] = C\alpha = 0.1 \times 0.01342 = 1.342 \times 10^{-3} M — same answer, faster.

This α=Ka/C\alpha = \sqrt{K_a/C} shortcut works whenever the approximation x≪Cx \ll C holds (i.e., α<5%\alpha < 5\%). Memorise it — it saves 2 minutes in JEE Main’s timed format.

The validity condition for this approximation: C/Ka>400C/K_a > 400. Here, 0.1/1.8×10−5≈55550.1 / 1.8 \times 10^{-5} \approx 5555, well above 400, so we’re safe.

Common Mistake

The most common error: treating 0.1 M acetic acid the same way as 0.1 M HCl and writing [H+]=0.1[\text{H}^+] = 0.1 M, giving pH = 1. This ignores the partial dissociation completely.

The second mistake is forgetting to verify the approximation. If α\alpha comes out above 5% (which happens at very low concentrations or with stronger weak acids), you must solve the full quadratic. Writing x=Ka⋅Cx = \sqrt{K_a \cdot C} blindly without checking is a guaranteed error in CBSE board theory questions.

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