Final answer:
The molar solubility of CaF₂ in a 0.100-M solution of HF can be calculated using the concept of common ion effect and stoichiometry. The molar solubility is approximately 6.0 x 10⁻³ M.
Step-by-step explanation:
The molar solubility of CaF2 in a 0.100-M solution of HF can be determined using the concept of common ion effect. When a common ion is introduced into a solution, it reduces the solubility of a slightly soluble salt. In this case, HF is a common ion because it contains the fluoride ion (F¯), which is also present in CaF2. Therefore, the solubility of CaF2 in the solution of HF will decrease due to the common ion effect.
The molar solubility of CaF2 in a 0.100-M solution of HF can be calculated by considering the equilibrium between the dissolution of CaF2 and the ionization of HF:
CaF2 (s) ⇌ Ca²+ (aq) + 2F¯ (aq)
Let's assume the molar solubility of CaF2 in the solution is represented by 's'. As per the stoichiometry of the dissolution equation, the molar concentration of fluoride ions (F¯) is twice that of the calcium ions (Ca²+). Therefore, [F¯] = 2s.
Now, we can write the expression for the equilibrium constant (Ksp) of CaF2:
Ksp = [Ca²+][F¯]2
Substituting the values, we get:
Ksp = (2s)(2s)2 = 4s3
Since the value of Ksp is given as 3.45 × 10-11, we can set up the equation:
4s3 = 3.45 × 10-11
Solving for 's', we find the molar solubility of CaF2 in the 0.100-M solution of HF to be approximately 6.0 x 10-3 M.