**2. Answer the following in one sentence :**

**Given:**

- Dissociation constant (\(K_a\)) = \(1.8 \times 10^{-5}\),
- Concentration (\(c\)) = 0.01 M

**To find:**

- Percent dissociation

**Formulae:**

- \(K_a = \alpha^2c\)
- Percent dissociation = \(\alpha \times 100\)

**Calculation:**

\(c = 0.01\) M

= \(1 \times 10^{-2}\) M

Using formula (i),

Therefore, \(\alpha = \sqrt{\frac{K_a}{c}}\)

= \(\sqrt{\frac{1.8 \times 10^{-5}}{1 \times 10^{-2}}}\)

= \(\sqrt{1.8 \times 10^{-3}}\)

= \(\sqrt{18 \times 10^{-4}}\)

= \(4.242 \times 10^{-2}\)

Using formula (ii),

Percent dissociation = \(\alpha \times 100\)

= \(4.242 \times 10^{-2} \times 100\)

= \(4.242\%\)

Percent dissociation of 0.01 M acetic acid solution is 4.242%.