Question:
Solubility product of BaCl\(_2\) is \( 4 \times 10^{-9} \). Its solubility in mol/L would be:
Solubility product of BaCl\(_2\) is \( 4 \times 10^{-9} \). Its solubility in mol/L would be:
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To calculate solubility from the solubility product, remember to account for the number of ions produced.
Updated On: Apr 22, 2026
- \( 1 \times 10^{-3} \)
- \( 1 \times 10^{-9} \)
- \( 4 \times 10^{-27} \)
- \( 1 \times 10^{-27} \)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the solubility product.
The solubility product (\( K_{sp} \)) of BaCl\(_2\) is given by: \[ K_{sp} = [Ba^{2+}][Cl^-]^2 \] Let the solubility of BaCl\(_2\) be \( s \). The concentration of \( Ba^{2+} \) will be \( s \), and the concentration of \( Cl^- \) will be \( 2s \) (because there are two chloride ions for each BaCl\(_2\)).
Step 2: Setting up the equation.
Substitute the concentrations into the solubility product expression: \[ K_{sp} = s(2s)^2 = 4s^3 \] We are given \( K_{sp} = 4 \times 10^{-9} \). So: \[ 4s^3 = 4 \times 10^{-9} \] \[ s^3 = 10^{-9} \] Taking the cube root: \[ s = 1 \times 10^{-3} \]
Step 3: Conclusion.
The solubility of BaCl\(_2\) is \( 1 \times 10^{-3} \) mol/L.
The solubility product (\( K_{sp} \)) of BaCl\(_2\) is given by: \[ K_{sp} = [Ba^{2+}][Cl^-]^2 \] Let the solubility of BaCl\(_2\) be \( s \). The concentration of \( Ba^{2+} \) will be \( s \), and the concentration of \( Cl^- \) will be \( 2s \) (because there are two chloride ions for each BaCl\(_2\)).
Step 2: Setting up the equation.
Substitute the concentrations into the solubility product expression: \[ K_{sp} = s(2s)^2 = 4s^3 \] We are given \( K_{sp} = 4 \times 10^{-9} \). So: \[ 4s^3 = 4 \times 10^{-9} \] \[ s^3 = 10^{-9} \] Taking the cube root: \[ s = 1 \times 10^{-3} \]
Step 3: Conclusion.
The solubility of BaCl\(_2\) is \( 1 \times 10^{-3} \) mol/L.
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