To find the dissociation constant of acetic acid, we use the Henderson-Hasselbalch equation for the given system and conditions. The equation is as follows:

$ \text{pH} = \text{pK}_\text{a} + \log \left(\frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}\right) $

In the solution mixture:

We have 25 mL of 0.2 M $\text{CH}_3\text{COONa}$ and 25 mL of 0.02 M $\text{CH}_3\text{COOH}$.

The concentration ratio $\left(\frac{[\text{CH}_3\text{COONa}]}{[\text{CH}_3\text{COOH}]}\right)$ becomes $\frac{25 \times 0.2}{25 \times 0.02} = 10$.

Given that the pH of the solution is 5, we substitute into the equation:

$ 5 = \text{pK}_\text{a} + \log 10 $

Since $\log 10 = 1$, we solve for $\text{pK}_\text{a}$:

$ 5 = \text{pK}_\text{a} + 1 \quad \Rightarrow \quad \text{pK}_\text{a} = 4 $

Converting from $\text{pK}_\text{a}$ to $K_\text{a}$, we use:

$ K_\text{a} = 10^{-\text{pK}_\text{a}} = 10^{-4} $

Thus, since the dissociation constant $K_\text{a}$ is given as $x \times 10^{-5}$, compare:

$ 10^{-4} = 10 \times 10^{-5} $

Therefore, $x = 10$.