Assuming that Ba(OH)$$_2$$ is completely ionised in aqueous solution under the given conditions the concentration of H$$_3$$O$$^+$$ ions in 0.005M aqueous solution of Ba(OH)$$_2$$ at 298 K is _________ $$\times 10^{-12}$$ mol L$$^{-1}$$. (Nearest integer)

First, we recall that barium hydroxide dissociates completely in water. The balanced ionic equation is

$$\text{Ba(OH)}_2 \;\longrightarrow\; \text{Ba}^{2+} + 2\,\text{OH}^-$$

We are told that the formal concentration of the Ba(OH)$$_2$$ solution is $$0.005\ \text{mol L}^{-1}$$. Because one formula unit of Ba(OH)$$_2$$ furnishes two hydroxide ions, the concentration of hydroxide ions produced is obtained by simple stoichiometry:

$$[\text{OH}^-] = 2 \times 0.005\ \text{mol L}^{-1} = 0.010\ \text{mol L}^{-1}$$

Next, we invoke the ionic product of water. At $$298\ \text{K}$$ the experimentally measured constant is

$$K_w = [\text{H}_3\text{O}^+]\, [\text{OH}^-] = 1.0 \times 10^{-14}$$

Solving this relation for the hydronium-ion concentration gives

$$[\text{H}_3\text{O}^+] = \frac{K_w}{[\text{OH}^-]}$$

Now we substitute the numerical values:

$$[\text{H}_3\text{O}^+] = \frac{1.0 \times 10^{-14}}{0.010} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-2}}$$

Dividing the powers of ten, we obtain

$$[\text{H}_3\text{O}^+] = 1.0 \times 10^{-12}\ \text{mol L}^{-1}$$

The question asks for the numerical factor that multiplies $$10^{-12}$$. We have found that factor to be exactly $$1.0$$, which rounds to the nearest integer as $$1$$.

So, the answer is $$1$$.