Question

The molar solubility of CaF$_2$ (K$_{sp}$ = 5.3 × 10$^{-11}$) in 0.1 M solution of NaF will be

• A. 5.3 × 10$^{-11}$ mol L$^{-1}$
• B. 5.3 × 10$^{-8}$ mol L$^{-1}$
• C. 5.3 × 10$^{-9}$ mol L$^{-1}$
• D. 5.3 × 10$^{-10}$ mol L$^{-1}$

Answer 1

The Correct Option is C

 The molar solubility of calcium fluoride (\(CaF_2\)) in a solution of sodium fluoride (\(NaF\)) is influenced by the common ion effect due to the presence of fluoride ions \((F^-)\) from \(NaF\). Let's solve this problem step by step:

1. Understanding the Solubility Equilibrium: 
   The solubility product expression for \(CaF_2\) is given by its dissociation: \[ \text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2 \text{F}^-(aq) \] 
   The solubility product (\( K_{sp} \)) expression is: \[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \]
2. Identifying Ion Concentrations: 
   Let the molar solubility of \(CaF_2\) in the presence of \(NaF\) be \(s\).
   In this solution, the concentration of \(Ca^{2+}\) ions is \(s\), and the concentration of \(-\) ions is \(0.1 + 2s \). Sin\)
3. Setting Up the Equation: 
   Substitute the ion concentrations into the \( K_{sp} \) expression: \[ 5.3 \times 10^{-11} = s \times (0.1)^2 \] \[ 5.3 \times 10^{-11} = s \times 0.01 \]
4. Solving for Molar Solubility \(s\): 
   Solve the equation for \(s\): \[ s = \frac{5.3 \times 10^{-11}}{0.01} = 5.3 \times 10^{-9} \]

Hence, the molar solubility of \(CaF_2\) in a 0.1 M solution of \(NaF\) is 5.3 × 10\(^{-9}\) mol L\(^{-1}\).