Final answer:
The molar solubility of barium fluoride in a 0.11 M Ba(NO3)2 solution is calculated by considering the common ion effect and using the Ksp expression Ksp = [Ba2+][F-]^2, rearranging it to solve for solubility 's'.
Step-by-step explanation:
To calculate the molar solubility of barium fluoride in the presence of 0.11 M Ba(NO3)2, we have to consider the common ion effect. The solubility product constant (Ksp) for barium fluoride is 2.45 × 10−5. For barium fluoride, the dissolution equation is BaF2 → Ba2+ + 2 F−. The Ksp expression is then Ksp = [Ba2+][F−]^2.
Since the student already has a source of Ba2+ from Ba(NO3)2, which is 0.11 M, this will affect the solubility of BaF2. Let's assume the molar solubility of BaF2 in this solution is 's', which will give us 's' moles of Ba2+ (which we can ignore because the concentration from Ba(NO3)2 is much higher) and 2s moles of F−. The Ksp expression can be rearranged to solve for s: Ksp = (0.11)(2s)^2, which gives us s = √(Ksp / (0.11 × 4)).
Plugging the Ksp value into this equation and solving for 's' will yield the molar solubility of barium fluoride in a 0.11 M Ba(NO3)2 solution. It's crucial to remember that the presence of a common ion (in this case Ba2+) will decrease the solubility of the precipitate.