Question

Find the solubility product of \( \mathrm{Ba(OH)_2} \)?
Where \( S = 1.73 \times 10^{-14} \).

• A. \( 20.7 \times 10^{-42} \)
• B. \( 1.73 \times 10^{-28} \)
• C. \( 3.24 \times 10^{-42} \)
• D. None of these

Hint:

When solving solubility product problems, remember to account for the stoichiometric coefficients in the dissociation equation. Multiply the solubility by the appropriate power for each ion's concentration.

Answer 1

The Correct Option is D

The solubility product (\( K_{sp} \)) of \( \mathrm{Ba(OH)_2} \) can be calculated using its solubility (\( S \)). 

Step 1: Dissociation of \( \mathrm{Ba(OH)_2} \).} 
\( \mathrm{Ba(OH)_2} \) dissociates in water as: \[ \mathrm{Ba(OH)_2 \, \longrightarrow \, Ba^{2+} + 2OH^-}. \] 

Let the solubility of \( \mathrm{Ba(OH)_2} \) be \( S \). Then, the concentrations of the ions at equilibrium are: \begin{itemize} \( [\mathrm{Ba^{2+}}] = S \), \( [\mathrm{OH^-}] = 2S \). \end{itemize} 

Step 2: Solubility product expression. 
The solubility product (\( K_{sp} \)) is given by: \[ K_{sp} = [\mathrm{Ba^{2+}}] \cdot [\mathrm{OH^-}]^2. \] Substitute the concentrations: \[ K_{sp} = S \cdot (2S)^2. \] Simplify: \[ K_{sp} = S \cdot 4S^2 = 4S^3. \] 

Step 3: Calculate \( K_{sp} \). 
Substitute \( S = 1.73 \times 10^{-14} \): \[ K_{sp} = 4 \cdot (1.73 \times 10^{-14})^3. \] 
Calculate \( (1.73)^3 \): \[ 1.73^3 \approx 5.17. \] Calculate \( (10^{-14})^3 \): \[ (10^{-14})^3 = 10^{-42}. \] 

Substitute into the equation: \[ K_{sp} = 4 \cdot 5.17 \cdot 10^{-42}. \] 

Simplify: \[ K_{sp} = 20.7 \times 10^{-42}. \] Thus, the solubility product is: \[ K_{sp} = 20.7 \times 10^{-42}. \]