Equilibrium

The solubility product ( $K_%7Bsp%7D$ ) of $%5Cmathrm%7BBa(OH)_2%7D$ can be calculated using its solubility ( $S$ ).

Step 1: Dissociation of $%5Cmathrm%7BBa(OH)_2%7D$ .}
$%5Cmathrm%7BBa(OH)_2%7D$ dissociates in water as: $%5Cmathrm%7BBa(OH)_2%20%5C%2C%20%5Clongrightarrow%20%5C%2C%20Ba%5E%7B2%2B%7D%20%2B%202OH%5E-%7D.$

Let the solubility of $%5Cmathrm%7BBa(OH)_2%7D$ be $S$ . Then, the concentrations of the ions at equilibrium are: \begin{itemize} $%5B%5Cmathrm%7BBa%5E%7B2%2B%7D%7D%5D%20%3D%20S$ , $%5B%5Cmathrm%7BOH%5E-%7D%5D%20%3D%202S$ . \end{itemize}

Step 2: Solubility product expression.
The solubility product ( $K_%7Bsp%7D$ ) is given by: $K_%7Bsp%7D%20%3D%20%5B%5Cmathrm%7BBa%5E%7B2%2B%7D%7D%5D%20%5Ccdot%20%5B%5Cmathrm%7BOH%5E-%7D%5D%5E2.$ Substitute the concentrations: $K_%7Bsp%7D%20%3D%20S%20%5Ccdot%20(2S)%5E2.$ Simplify: $K_%7Bsp%7D%20%3D%20S%20%5Ccdot%204S%5E2%20%3D%204S%5E3.$

Step 3: Calculate $K_%7Bsp%7D$ .
Substitute $S%20%3D%201.73%20%5Ctimes%2010%5E%7B-14%7D$ : $K_%7Bsp%7D%20%3D%204%20%5Ccdot%20(1.73%20%5Ctimes%2010%5E%7B-14%7D)%5E3.$
Calculate $(1.73)%5E3$ : $1.73%5E3%20%5Capprox%205.17.$ Calculate $(10%5E%7B-14%7D)%5E3$ : $(10%5E%7B-14%7D)%5E3%20%3D%2010%5E%7B-42%7D.$

Substitute into the equation: $K_%7Bsp%7D%20%3D%204%20%5Ccdot%205.17%20%5Ccdot%2010%5E%7B-42%7D.$

Simplify: $K_%7Bsp%7D%20%3D%2020.7%20%5Ctimes%2010%5E%7B-42%7D.$ Thus, the solubility product is: $K_%7Bsp%7D%20%3D%2020.7%20%5Ctimes%2010%5E%7B-42%7D.$

High-Yield Trend

Chapter Questions
1 MCQs

Official Solution

Equilibrium

High-Yield Trend

Chapter Questions 1 MCQs

Official Solution

Chapter Questions
1 MCQs