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 (\( K_{sp} \)) for \( \mathrm{Ba(OH)_2} \) is determined from its solubility (\( S \)).

Step 1: Dissociation of \( \mathrm{Ba(OH)_2} \). 
\( \mathrm{Ba(OH)_2} \) dissolves in water according to the equation: \[ \mathrm{Ba(OH)_2 \, \longrightarrow \, Ba^{2+} + 2OH^-}. \] 

If the solubility of \( \mathrm{Ba(OH)_2} \) is \( S \), the equilibrium ion concentrations are:

• \( [\mathrm{Ba^{2+}}] = S \)
• \( [\mathrm{OH^-}] = 2S \)

Step 2: Solubility product expression. 
The solubility product (\( K_{sp} \)) is defined as: \[ K_{sp} = [\mathrm{Ba^{2+}}] \cdot [\mathrm{OH^-}]^2. \] Substituting the ion concentrations yields: \[ K_{sp} = S \cdot (2S)^2. \] Simplifying this expression gives: \[ K_{sp} = S \cdot 4S^2 = 4S^3. \] 

Step 3: Calculation of \( K_{sp} \). 
Using \( S = 1.73 \times 10^{-14} \), the \( K_{sp} \) is calculated as: \[ K_{sp} = 4 \cdot (1.73 \times 10^{-14})^3. \] 
The cube of \( 1.73 \) is approximately: \[ 1.73^3 \approx 5.17. \] The cube of \( 10^{-14} \) is: \[ (10^{-14})^3 = 10^{-42}. \] 

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

The final simplified result is: \[ K_{sp} = 20.7 \times 10^{-42}. \] Therefore, the solubility product is: \[ K_{sp} = 20.7 \times 10^{-42}. \]