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NCERT Exemplar · Q8

Q.The ionisation constant of an acid, Ka, is the measure of strength of an acid. The Ka values of acetic acid, hypochlorous acid and formic acid are 1.74 × 10^-5, 3.0 × 10^-8 and 1.8 × 10^-4 respectively. Which of the following orders of pH of 0.1 mol dm^-3 solutions of these acids is correct?

(i) acetic acid > hypochlorous acid > formic acid
(ii) hypochlorous acid > acetic acid > formic acid
(iii) formic acid > hypochlorous acid > acetic acid
(iv) formic acid > acetic acid > hypochlorous acid
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For weak acids, pH is inversely related to KaK_a: the smaller the KaK_a, the higher the pH. Since KaK_a values are 1.8×10−41.8 \times 10^{-4} (formic), 1.74×10−51.74 \times 10^{-5} (acetic), 3.0×10−83.0 \times 10^{-8} (hypochlorous), the pH order is hypochlorous acid > acetic acid > formic acid, which is option (ii).

The key idea is simple: a weak acid’s strength is measured by its ionization constant KaK_a. A larger KaK_a means more dissociation, more H+H^+ in solution, and therefore a lower pH. For a given concentration, the pH of a weak acid solution depends directly on KaK_a.

Let’s see why this works.

  1. The relationship between KaK_a and [H+][H^+] For a weak acid HAHA of initial concentration cc, the equilibrium is:

HA⇌H++A−HA \rightleftharpoons H^+ + A^-

If the degree of ionization is α\alpha, then [H+]=cα[H^+] = c\alpha and [A−]=cα[A^-] = c\alpha, while [HA]=c(1−α)[HA] = c(1-\alpha). The ionization constant is:

Ka=[H+][A−][HA]=(cα)2c(1−α)K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(c\alpha)^2}{c(1-\alpha)}

For a weak acid, α\alpha is small, so 1−α≈11-\alpha \approx 1. This gives the approximation:

Ka≈cα2⇒α≈KacK_a \approx c\alpha^2 \quad \Rightarrow \quad \alpha \approx \sqrt{\frac{K_a}{c}}

Hence:

[H+]=cα≈Kac[H^+] = c\alpha \approx \sqrt{K_a c}

And pH is:

pH=−log⁡[H+]≈−log⁡Kac=12(−log⁡Ka−log⁡c)\text{pH} = -\log[H^+] \approx -\log\sqrt{K_a c} = \frac{1}{2}(-\log K_a - \log c)

For a weak acid of concentration cc, pH≈12(pKa−log⁡c)\text{pH} \approx \frac{1}{2}(\text{p}K_a - \log c), where pKa=−log⁡Ka\text{p}K_a = -\log K_a.

Since cc is the same for all three acids (0.1 mol dm−3^{-3}), the pH depends only on KaK_a: larger KaK_a → smaller pH.

  1. Compare the KaK_a values

    • Formic acid: Ka=1.8×10−4K_a = 1.8 \times 10^{-4} (largest)
    • Acetic acid: Ka=1.74×10−5K_a = 1.74 \times 10^{-5} (middle)
    • Hypochlorous acid: Ka=3.0×10−8K_a = 3.0 \times 10^{-8} (smallest)

    So formic acid is the strongest among the three, and hypochlorous acid is the weakest.

  2. Translate to pH order

    Stronger acid → lower pH. Therefore:

    • Formic acid has the lowest pH. …

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