Multiple choice

How much water must be added in $900\ ml$ of $0.1\ M\ CH_{3}COOH$ solution to triple its degree of dissociation (Assume $\alpha < 5\%$ is negligible ) : ($K_{a}=1.8 \times 10^{-5}$)

  1. $7.2\ L$
  2. $3.6\ L$
  3. $5.8\ L$
  4. $2.4\ L$
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A Correct answer
Explanation

Degree of dissociation alpha is proportional to 1/sqrt(C). To triple alpha, the concentration C must decrease by a factor of 9. Initial volume V1 = 900 ml. New volume V2 = 9 * 900 = 8100 ml. Water to add = 8100 - 900 = 7200 ml = 7.2 L.

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