For a weak acid HA, the percentage of dissociation is nearly 1 % at equilibrium. If the concentration of acid is 0.1 mol L^-1, then the correct option for its K_a at the same temperature is :

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$1 \times 10^{- 4}$

$1 \times 10^{- 6}$

$1 \times 10^{- 5}$

$1 \times 10^{- 3}$

answer is C.

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Detailed Solution

Given:

For a weak acid:

HA ↔ H⁺ + A^-

Initial concentration of HA = 0.1 M

Dissociation is 1%, so:

[H⁺] = 1% of 0.1 M = (1/100) × 0.1 = 0.001 M

Remaining undissociated acid:

[HA] = 0.1 - 0.001 = 0.099 ≈ 0.1 M

The acid dissociation constant is given by:

K_a = [H⁺][A^-] / [HA]

Since dissociation is small:

K_a = (0.001 × 0.001) / 0.1
= 0.000001 / 0.1
= 1 × 10^-5

Final Answer

The correct option is: 1 × 10^-5

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