The volume of water (in mL) required to be added to a 100 mL solution (aq. 0.1 M) of a weak acid (HA) at 25 °C to double its degree of dissociation is _____ mL.

For a weak acid:

$$\alpha=\sqrt{\frac{K_a}{C}}$$

If the degree of dissociation is doubled:

$$\alpha_2=2\alpha_1$$

Therefore,

$$\sqrt{\frac{K_a}{C_2}}=2\sqrt{\frac{K_a}{C_1}}$$

Squaring both sides:

$$\frac{1}{C_2}=\frac{4}{C_1}$$

$$C_2=\frac{C_1}{4}$$

Initial concentration:

$$C_1=0.1\,\mathrm{M}$$

Therefore,

$$C_2=\frac{0.1}{4}=0.025\,\mathrm{M}$$

Using the dilution equation:

$$C_1V_1=C_2V_2$$

$$0.1\times100=0.025\times V_2$$

$$V_2=400\,\mathrm{mL}$$

Volume of water added:

$$V_{\mathrm{water}}=V_2-V_1$$

$$V_{\mathrm{water}}=400-100$$

$${V_{\mathrm{water}}=300\,\mathrm{mL}}$$

Correct answer: 300 mL