Question

K_sp of BaSO₄ is 8 × 10^–11. If the solubility in presence of 0.1 M CaSO₄ is?

Answer 1

We are given that the solubility product (\( K_{sp} \)) of \( \text{BaSO}_4 \) is \( 8 \times 10^{-11} \), and the concentration of \( \text{CaSO}_4 \) in the solution is 0.1 M. We need to find the solubility of \( \text{BaSO}_4 \) in the presence of \( \text{CaSO}_4 \). 

Step 1: Understanding the effect of common ion 
The presence of \( \text{CaSO}_4 \) in the solution introduces a common ion, \( \text{SO}_4^{2-} \), which will reduce the solubility of \( \text{BaSO}_4 \) due to the common ion effect. The solubility equilibrium for \( \text{BaSO}_4 \) is: \[ \text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq) \] Let the solubility of \( \text{BaSO}_4 \) in the presence of \( 0.1 \, \text{M} \) \( \text{CaSO}_4 \) be \( s \). This means that the concentration of \( \text{Ba}^{2+} \) in the solution is \( s \), and the concentration of \( \text{SO}_4^{2-} \) is \( 0.1 + s \), because \( \text{CaSO}_4 \) contributes \( 0.1 \, \text{M} \) \( \text{SO}_4^{2-} \). 

Step 2: Write the expression for the solubility product 
The solubility product (\( K_{sp} \)) is given by: \[ K_{sp} = [\text{Ba}^{2+}] [\text{SO}_4^{2-}]. \] Substituting the concentrations of the ions: \[ 8 \times 10^{-11} = s(0.1 + s). \] 

Step 3: Solve the quadratic equation 
Since \( s \) is very small compared to \( 0.1 \, \text{M} \), we can approximate \( 0.1 + s \approx 0.1 \). Therefore, the equation simplifies to: \[ 8 \times 10^{-11} = s \times 0.1, \] \[ s = \frac{8 \times 10^{-11}}{0.1} = 8 \times 10^{-10} \, \text{M}. \] 

Final Answer: The solubility of \( \text{BaSO}_4 \) in the presence of \( 0.1 \, \text{M} \) \( \text{CaSO}_4 \) is \( \boxed{8 \times 10^{-10}} \, \text{M} \), or \( 8 \times 10^{-10} \, \text{mol/L} \).