Explanation:

To determine the suitable indicator for the titration of X- Na (0.1 M, 10 ml) with 0.1 M HCl, we need to consider the pH range at which the indicator changes color. The suitable indicator should change its color within the pH range of the equivalence point of the reaction.

1. Equivalence point:
The reaction between X- Na and HCl can be represented as follows:
X- Na + HCl → XH + NaCl

At the equivalence point, all the X- ions will react with an equal number of H+ ions from HCl. Therefore, the concentration of X- ions will be completely neutralized by the concentration of H+ ions.

2. pH at the equivalence point:
To find the pH at the equivalence point, we need to calculate the concentration of H+ ions.

Given:
[X-] = 0.1 M (concentration of X- Na)
[HCl] = 0.1 M (concentration of HCl)

Since the reaction is 1:1, the concentration of H+ ions is also 0.1 M.

Using the equation for the ionization of water:
Kw = [H+][OH-]
1.0 x 10^-14 = (0.1)(OH-)
OH- = 1.0 x 10^-13 M

Since the solution is basic (due to the presence of X- ions), the concentration of OH- ions is higher than the concentration of H+ ions. Therefore, the pH at the equivalence point is greater than 7.

3. Suitable indicator:
The suitable indicator for the titration should have a color change within the pH range of the equivalence point.

Given:
kb(X-) = 1.0 x 10^-6

The pOH at the equivalence point can be calculated as follows:
pOH = -log(OH-)
pOH = -log(1.0 x 10^-13)
pOH = 13

Since pOH + pH = 14, the pH at the equivalence point is 14 - 13 = 1.

Therefore, the suitable indicator should change its color within the pH range of 1-14.

Conclusion:
The range of the most suitable indicator for the titration of X- Na with 0.1 M HCl is 3-5, as it falls within the pH range of 1-14. Hence, the correct answer is option B.