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**Hint:**Think about the formula for the solubility product constant (${{K}_{sp}}$). Consider how it determines the relationship between the concentration of the ions present and their solubility in the solvent.

**Complete step by step answer:**

We know that the solubility product constant (${{K}_{sp}}$) is defined as the product of the concentrations of the ion species found in the solvent. This constant measures the degree up to which a compound is solvated. The formula for the solubility product constant is:

\[\begin{align}

& {{K}_{sp}} = {{[catio{{n}^{+}}]}^{n}}{{[anio{{n}^{-}}]}^{m}} \\

& \text{For }{{C}_{n}}{{A}_{m}} \\

\end{align}\]

Thus, for barium hydroxide, it will be:

\[{{K}_{sp}}=[B{{a}^{2+}}]{{[O{{H}^{-}}]}^{2}}\]

To find the concentration of the $B{{a}^{2+}}$ ions, we need to first find the concentration of the hydroxide ions. To do this, we can use the $pH$ of the solution that is given.

Since, barium hydroxide is a base, we will consider the $pOH$ to find the concentration of the $O{{H}^{-}}$ ions. We know the formula:

\[\begin{align}

& pH + pOH = 14 \\

& \therefore pOH = 14-pH \\

\end{align}\]

The given $pH$ of $Ba{{(OH)}_{2}}$ is 10.

Thus, $pOH = 4$

We know the formula:

\[pOH=-\log [O{{H}^{-}}]\]

Rearranging the formula and replacing the value of $pOH$, we get:

\[[O{{H}^{-}}]={{10}^{-4}}\]

The value of ${{K}_{sp}}$ of $Ba{{(OH)}_{2}}$ is given as $5\times {{10}^{-13}}$. Now putting these values in the formula for the solubility product constant, we get:

\[5\times {{10}^{-13}}=[B{{a}^{2+}}]\times {{({{10}^{-4}})}^{2}}\]

Now, solving for $[B{{a}^{2+}}]$, we get:

\[\begin{align}

& [B{{a}^{2+}}] = \frac{5\times {{10}^{-13}}}{{{10}^{-8}}} \\

& [B{{a}^{2+}}] = 5\times {{10}^{-5}} \\

\end{align}\]

Thus, the concentration of the barium cations is $5\times {{10}^{-5}}$.

**Hence, the answer to this question is ‘C. $5\times {{10}^{-5}}$’.**

**Note:**Always check whether you are dealing with an acid or a base and the value of the potency of the protons or the hydroxyl ions is given. Always convert the given value into $pH$ or $pOH$ as required. Remember to take the power into consideration while formulating the formula for the solubility product constant.

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