Multiple choice

How much water must be added to $300$ mL of $0.2$ M solution of ${ CH }{ 3 }COOH{ (K) }{ a }={ 1.8\times 10 }^{ -5 }$ for the degree of dissociation of the acid to double?

  1. 600 mL

  2. 900 mL

  3. 1200 mL

  4. 1500 mL
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Degree of dissociation alpha = sqrt(Ka/C). If alpha doubles, C must become C/4. Initial C = 0.2 M. Final C = 0.05 M. Initial volume = 300 mL. C1V1 = C2V2 => 0.2 * 300 = 0.05 * V2. V2 = 1200 mL. Water to be added = 1200 - 300 = 900 mL.

AI explanation

According to Ostwald's dilution law, the degree of dissociation of a weak acid is inversely proportional to the square root of its molarity. To double the degree of dissociation, you must decrease the concentration by a factor of four, so the new volume must be four times the initial 300 mL, which is 1200 mL. Subtracting the original 300 mL from 1200 mL means you must add 900 mL of water.

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