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For a weak acid HA\text{HA}, the percentage of dissociation is nearly 1%1\% at equilibrium. If the concentration of acid is 0.1 mol L−10.1\text{ mol L}^{-1}, then the correct option for its KaK_a at the same temperature will be:

A

1×10−41 \times 10^{-4}

B

1×10−61 \times 10^{-6}

C

1×10−51 \times 10^{-5}

D

1×10−31 \times 10^{-3}

Step-by-Step Solution

For a weak acid HA, the ionization constant (KaK_a) is related to its concentration (cc) and degree of dissociation (α\alpha) by the formula: Ka=cα21−αK_a = \frac{c\alpha^2}{1-\alpha}

Given the values: Concentration of acid, c=0.1 mol L−1=10−1 Mc = 0.1\text{ mol L}^{-1} = 10^{-1}\text{ M} Percentage of dissociation = 1%1\% Degree of dissociation, α=1100=0.01=10−2\alpha = \frac{1}{100} = 0.01 = 10^{-2}

Since α\alpha (0.010.01) is very small compared to 11, the term (1−α)(1 - \alpha) in the denominator can be approximated as 11 . The formula simplifies to: Ka≈cα2K_a \approx c\alpha^2

Substituting the given values: Ka=(10−1)×(10−2)2K_a = (10^{-1}) \times (10^{-2})^2 Ka=10−1×10−4K_a = 10^{-1} \times 10^{-4} Ka=10−5K_a = 10^{-5}

Thus, the ionization constant KaK_a is 1×10−51 \times 10^{-5}.

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