Question

The $pH$ of a saturated aqueous solution of $Ba{{(OH)}_{2}}$ is 10. If the ${{K}_{sp}}$ of $Ba{{(OH)}_{2}}$ is $5\times {{10}^{-13}}$. The concentration of $B{{a}^{2+}}$ ions in the solution is:
A. $1\times {{10}^{-2}}$
B. $1\times {{10}^{-3}}$
C. $5\times {{10}^{-5}}$
D. $1\times {{10}^{-5}}$

Answer 1

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.