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Answer the following in brief · Q6

Q.vi. Acetic acid is 5% ionised in its decimolar solution. Calculate the dissociation constant of acid (Ans : 2.63×10−42.63 \times 10^{-4})

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Step 1. 'Decimolar' means c=0.1c=0.1 M; '5% ionised' means α=0.05\alpha=0.05.

Step 2. Since α=0.05\alpha=0.05 is not negligibly small, use the exact Ostwald's-law relation (Eq. 3.6), Ka=α2c1−αK_a=\dfrac{\alpha^2c}{1-\alpha}, rather than the simplified Ka≈α2cK_a\approx\alpha^2c.

Step 3. Substitute: Ka=(0.05)2×0.11−0.05=0.0025×0.10.95=0.000250.95K_a=\dfrac{(0.05)^2\times0.1}{1-0.05}=\dfrac{0.0025\times0.1}{0.95}=\dfrac{0.00025}{0.95}. …

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