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Question: pH of a saturated solution of \( \text{Ba(OH}{{)}_{2}} \) is \( 12 \) . The value of \( {{K}_{sp}} \...

pH of a saturated solution of Ba(OH)2\text{Ba(OH}{{)}_{2}} is 1212 . The value of Ksp{{K}_{sp}} is:
(A) 3⋅3×10−7M3\cdot 3\times {{10}^{-7}}M
(B) 5×10−7M5\times {{10}^{-7}}M
(C) 4×10−6M4\times {{10}^{-6}}M
(D) 5×10−6M5\times {{10}^{-6}}M

Explanation

Solution

Ksp{{\text{K}}_{\text{sp}}} is known as the solubility product constant because it gives the product of the solubility of the ions in mole per liter. First we will write the ionization reaction of barium hydroxide into its constituent ions then we will find pOH\text{pOH} for the solution as barium hydroxide is a base. Once we found the pOH\text{pOH} , the concentration of hydroxyl ions can be easily obtained by using the logarithmic relation of pOH\text{pOH} and the concentration. Using the concentrations we will find the solubility of the ions and hence the solubility product.

Formula Used: pOH=14−pH\text{pOH}=14-\text{pH} , pOH=−log⁡[OH−]\text{pOH}=-\log \left[ \text{O}{{\text{H}}^{-}} \right] , Ksp=[Ba2+][OH−]2{{\text{K}}_{\text{sp}}}=\left[ \text{B}{{\text{a}}^{2+}} \right]{{\left[ \text{O}{{\text{H}}^{-}} \right]}^{2}}

Complete Step By Step Solution
pH for barium hydroxide is 1212 .
So,
pOH=14−pH\text{pOH}=14-\text{pH}
pOH=14−12\text{pOH}=14-12
pOH=2\text{pOH}=2
Using the formula,
pOH=14−12=−log⁡[OH−]\text{pOH}=14-12=-\log \left[ \text{O}{{\text{H}}^{-}} \right]
2=−log⁡[OH−] [OH−]=10−2 \begin{aligned} & 2=-\log \left[ \text{O}{{\text{H}}^{-}} \right] \\\ & \left[ \text{O}{{\text{H}}^{-}} \right]={{10}^{-2}} \\\ \end{aligned}
The dissociation reaction of barium hydroxide is: Ba(OH)2⇌Ba2++2OH−\text{Ba}{{(\text{OH})}_{2}}\rightleftharpoons \text{B}{{\text{a}}^{2+}}+2\text{O}{{\text{H}}^{-}} as seen from this reaction the concentration of barium ions is half of the concentration of hydroxyl ions. So, if concentration of hydroxyl ions is 10−2M{{10}^{-2}}M then concentration of barium ions is: 12×10−2=0⋅5×10−2M\dfrac{1}{2}\times {{10}^{-2}}=0\cdot 5\times {{10}^{-2}}\text{M}
Therefore finding the solubility product,
We have, Ksp=[Ba2+][OH−]2{{\text{K}}_{\text{sp}}}=\left[ \text{B}{{\text{a}}^{2+}} \right]{{\left[ \text{O}{{\text{H}}^{-}} \right]}^{2}}
Ksp=(0⋅5×10−2)×(10−2)2{{\text{K}}_{\text{sp}}}=(0\cdot 5\times {{10}^{-2}})\times {{({{10}^{-2}})}^{2}}
Ksp=0⋅5×10−2×10−4{{\text{K}}_{\text{sp}}}=0\cdot 5\times {{10}^{-2}}\times {{10}^{-4}}
Ksp=0⋅5×10−6{{\text{K}}_{\text{sp}}}=0\cdot 5\times {{10}^{-6}}
Ksp=5×10−7M{{\text{K}}_{\text{sp}}}=5\times {{10}^{-7}}\text{M}
Therefore, the correct choice is (B).

Additional Information
Solubility products represent the level at which a solute dissolves in the solution. The higher value of solubility product represents the higher solubility of the solute. To find the Ksp{{\text{K}}_{\text{sp}}} value we require the concentration of the ion products. If there are coefficients in front of the products they need to be raised to the power of the concentrations of the products. Solubility is affected by temperature, pressure and molecular size. Solubility increases with temperature due to an increase in kinetic energy which allows the solvent molecules to effectively break the solute molecules that are held together by intermolecular forces of attraction. Solute dissolves in solvents mainly of the same polarity.

Note
The coefficient must be raised to the power of the respective concentrations as there were two hydroxyl ions so the power two was raised to its concentration. Similarly one was the coefficient for barium ions so its power was one only. Remember to take the product of the concentrations to find the solubility product constant. One should know the dissociation of various compounds into their respective ions and the charge on them.