Assuming that the degree of hydrolysis is small, the pH of 0.1M solution of sodium acetate (K$$_a$$ = $$1.0 \times 10^{-5}$$) will be:

Sodium acetate (CH3COONa) is a salt formed from a weak acid (acetic acid, CH3COOH) and a strong base (NaOH). When dissolved in water, it undergoes hydrolysis, producing a basic solution. The hydrolysis reaction is:

$$CH_{3}COO- + H_{2}O <=> CH_{3}COOH + OH-$$

Given that the acid dissociation constant (Ka) for acetic acid is $$1.0 \times 10^{-5}$$, we can find the hydrolysis constant (Kb) using the relation:

$$K_w = K_a \times K_b$$

where Kw is the ion product of water, equal to $$1.0 \times 10^{-14}$$ at 25°C. Solving for Kb:

$$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-5}} = 1.0 \times 10^{-9}$$

The initial concentration of sodium acetate is 0.1 M, so the initial concentration of acetate ions (CH3COO-) is 0.1 M. Let the degree of hydrolysis be h. At equilibrium:

$$[CH_{3}COOH] = h \times 0.1$$

$$[OH-] = h \times 0.1$$

$$[CH_{3}COO-] = 0.1 - h \times 0.1 \approx 0.1$$

The approximation assumes h is small, so we neglect the change in [CH3COO-]. The Kb expression is:

$$K_b = \frac{[CH_{3}COOH][OH-]}{[CH_{3}COO-]}$$

Substituting the equilibrium concentrations:

$$1.0 \times 10^{-9} = \frac{(h \times 0.1) \times (h \times 0.1)}{0.1}$$

Simplifying the numerator and denominator:

$$1.0 \times 10^{-9} = \frac{(0.1h) \times (0.1h)}{0.1} = \frac{0.01h^2}{0.1}$$

Dividing 0.01 by 0.1:

$$1.0 \times 10^{-9} = 0.1h^2$$

Solving for h2:

$$h^2 = \frac{1.0 \times 10^{-9}}{0.1} = \frac{1.0 \times 10^{-9}}{1.0 \times 10^{-1}} = 1.0 \times 10^{-8}$$

Taking the square root:

$$h = \sqrt{1.0 \times 10^{-8}} = 1.0 \times 10^{-4}$$

The hydroxide ion concentration is:

$$[OH-] = h \times 0.1 = (1.0 \times 10^{-4}) \times 0.1 = 1.0 \times 10^{-5}$$

Now, calculate pOH:

$$\text{pOH} = -\log(1.0 \times 10^{-5}) = 5.0$$

Then, pH is:

$$\text{pH} = 14 - \text{pOH} = 14 - 5.0 = 9.0$$

The degree of hydrolysis (h = 10-4) is small compared to 1, validating the approximation. Therefore, the pH of the 0.1 M sodium acetate solution is 9.0.

Hence, the correct answer is Option D.