The dissociation constant of acetic acid is $$x \times 10^{-5}$$. When 25 mL of 0.2 M $$CH_3COONa$$ solution is mixed with 25 mL of 0.02 M $$CH_3COOH$$ solution, the pH of the resultant solution is found to be equal to 5. The value of x is _____.

When 25 mL of 0.2 M $$CH_3COONa$$ is mixed with 25 mL of 0.02 M $$CH_3COOH$$, the total volume is 50 mL, and the concentrations become $$[CH_3COONa] = \frac{25 \times 0.2}{50} = 0.1$$ M and $$[CH_3COOH] = \frac{25 \times 0.02}{50} = 0.01$$ M.

This is a buffer solution, so using the Henderson-Hasselbalch equation, $$\text{pH} = \text{p}K_a + \log\frac{[\text{Salt}]}{[\text{Acid}]} = \text{p}K_a + \log\frac{0.1}{0.01} = \text{p}K_a + 1$$.

Given pH = 5, it follows that $$\text{p}K_a = 4$$, hence $$K_a = 10^{-4} = 10 \times 10^{-5}$$ and therefore $$x = \boxed{10}$$.