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Question: The \(pH\) of a saturated aqueous solution of \(Ba{{(OH)}_{2}}\) is 10. If the \({{K}_{sp}}\) of \(B...

The pHpH of a saturated aqueous solution of Ba(OH)2Ba{{(OH)}_{2}} is 10. If the Ksp{{K}_{sp}} of Ba(OH)2Ba{{(OH)}_{2}} is 5×10−135\times {{10}^{-13}}. The concentration of Ba2+B{{a}^{2+}} ions in the solution is:
A. 1×10−21\times {{10}^{-2}}
B. 1×10−31\times {{10}^{-3}}
C. 5×10−55\times {{10}^{-5}}
D. 1×10−51\times {{10}^{-5}}

Explanation

Solution

Think about the formula for the solubility product constant (Ksp{{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 (Ksp{{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:

& {{K}_{sp}} = {{[catio{{n}^{+}}]}^{n}}{{[anio{{n}^{-}}]}^{m}} \\\ & \text{For }{{C}_{n}}{{A}_{m}} \\\ \end{aligned}$$ Thus, for barium hydroxide, it will be: $${{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{aligned} & pH + pOH = 14 \\\ & \therefore pOH = 14-pH \\\ \end{aligned}$$ The given $pH$ of $Ba{{(OH)}_{2}}$ is 10. Thus, $pOH = 4$ We know the formula: $$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{aligned} & [B{{a}^{2+}}] = \frac{5\times {{10}^{-13}}}{{{10}^{-8}}} \\\ & [B{{a}^{2+}}] = 5\times {{10}^{-5}} \\\ \end{aligned}$$ 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.