Question

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
[Given: K_a of HA at 25 °C=1.8 x10⁻⁵ ]

• A. 100
• B. 200
• C. 300
• D. 400

Answer 1

The Correct Option is C

To solve this problem, we need to determine how much additional water should be added to a weak acid solution to double its degree of dissociation. 

Given:

• Initial concentration of HA, \([HA]_0 = 0.1 \, \text{M}\)
• Degree of dissociation initially, \(\alpha_1\)
• Dissociation constant, \(K_a = 1.8 \times 10^{-5}\)

The degree of dissociation, \(\alpha\), for a weak acid is given by:

\(\alpha = \sqrt{\frac{K_a}{[HA]_0}}\)

Initial degree of dissociation is:

\(\alpha_1 = \sqrt{\frac{1.8 \times 10^{-5}}{0.1}}\)

\(\alpha_1 = \sqrt{1.8 \times 10^{-4}} = 0.0134\)

We want to double the degree of dissociation:

\(\alpha_2 = 2\alpha_1 = 2 \times 0.0134 = 0.0268\)

The final concentration of the acid after dilution should make the new degree of dissociation equal to \(\alpha_2\):

\(\alpha_2 = \sqrt{\frac{K_a}{[HA]_2}}\)

\(0.0268 = \sqrt{\frac{1.8 \times 10^{-5}}{[HA]_2}}\)

Squaring both sides gives:

\(0.00071824 = \frac{1.8 \times 10^{-5}}{[HA]_2}\)

\([HA]_2 = \frac{1.8 \times 10^{-5}}{0.00071824} = 0.0251 \, \text{M}\)

The concentration changes with dilution. Using the dilution formula:

\(C_1 V_1 = C_2 V_2\)

\(0.1 \times 100 = 0.0251 \times V_2\)

\(V_2 = \frac{10}{0.0251} = 398.41 \, \text{mL}\)

The volume of water to be added is:

\(V_2 - 100 \, \text{mL} = 398.41 - 100 = 298.41 \, \text{mL}\)

Since we need to choose the closest option: 300 mL is the correct volume to add.

Therefore, the correct answer is 300 mL.