Ionic EquilibriumHard
Question
How much water must be added to 300 ml of 0.2 M solution of CH3COOH for the degree of dissociation of the acid to double? The value of Ka for acetic acid = 1.8 × 10–5.
Options
A.1200 ml
B.300 ml
C.600 ml
D.900 ml
Solution
$\alpha_{1} = 2 \times \alpha_{1} \Rightarrow \sqrt{\frac{K_{a}}{n} \times V_{1}} = 2 \times \sqrt{\frac{K_{a}}{n} \times V_{1}}$
$\Rightarrow V_{2} = 4 \times V_{1} = 4 \times 300 = 1200\text{ ml} $$$\therefore V_{\text{water added}} = 1200 - 300 = 900\text{ ml}$$
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