pK$$_{a}$$ of a weak acid HA and pK$$_{b}$$ of a weak base BOH are 3.2 and 3.4 respectively. The pH of their salt AB solution at 25°C is:

We are given the salt $$AB$$, which is obtained by neutralising a weak acid $$HA$$ with a weak base $$BOH$$. Such a salt undergoes hydrolysis in water, and the pH of its solution at 25 °C is conveniently calculated by a standard relation.

Formula to be used. For a salt formed from a weak acid and a weak base we have the expression

$$\text{pH}=7+\dfrac{1}{2}\left(\text{p}K_a-\text{p}K_b\right).$$

This result comes from setting up the hydrolysis equilibrium, writing the hydrolysis constant $$K_h=\dfrac{K_w}{K_aK_b}$$, solving for $$\left[H^+\right]$$, and finally taking the negative logarithm. Because both the acid and the base are weak, the derivation simplifies to the compact form stated above.

Now we substitute the numerical values provided in the question. We have

$$\text{p}K_a = 3.2,\qquad \text{p}K_b = 3.4.$$

First find the difference that appears in the formula:

$$\text{p}K_a - \text{p}K_b = 3.2 - 3.4 = -0.2.$$

Next, multiply this difference by the factor $$\dfrac{1}{2}$$ that is present in the formula:

$$\dfrac{1}{2}\left(\text{p}K_a - \text{p}K_b\right) = \dfrac{1}{2}\left(-0.2\right) = -0.1.$$

Finally, add the result to $$7$$, the neutral pH at 25 °C:

$$\text{pH} = 7 + (-0.1) = 7 - 0.1 = 6.9.$$

So, according to the calculation, the solution of the salt $$AB$$ is slightly acidic with a pH of $$6.9$$.

Hence, the correct answer is Option A.