When solid BaF2 is added to H2O the following equilibrium is established. BaF2(s) ⇌ Ba2+(aq) + 2 F–( — Equilibrium / Solubility Chemistry Question
Problem Context
When solid BaF2 is added to H2O the following equilibrium is established.
BaF2(s) ⇌ Ba2+(aq) + 2 F–(aq) Ksp = 1.5×10–6 at 25 °C
Calculate the molar solubility of barium fluoride at 25 °C.
Model Answer
If S = molar solubility of BaF2 (s), then [Ba2+] = S, [F–] = 2S
Ksp = [Ba2+][F–]2 = (S)(2S)2 = 4S3 = 1.5 × 10–6
S = 0.00721 mol/L
Explain how adding each of the following substances affects the solubility of BaF2 in water.
i. 0.10 M Ba(NO3)2
Model Answer
Adding Ba2+ ion will decrease the molar solubility of BaF2 due to the common ion effect.
ii. 0.10 M HNO3
Model Answer
Adding H+ will increase the molar solubility of BaF2 as the F– ion will react with H+ to form HF, thereby causing more BaF2 to dissolve by Le Chatelier’s Principle.
In an experiment to determine the Ksp of PbF2 a student starts with 0.10 M Pb(NO3)2 and 0.10 M KF and uses the method of serial dilutions to find the lowest [Pb2+] and [F–] that form a precipitate when mixed. If the student uses the concentration of the ions in the combined solution to determine Ksp, will the value of Ksp calculated be too large, too small or just right? Explain.
Ksp for PbF2 = 4.0 × 10–8
Model Answer
The calculated Ksp will be too large because the student is relying on seeing the formation of a precipitate at the moment that Q exceeds Ksp. The student will miss the exact moment that happens, so the calculated value of Ksp will be too large.
Other possible issues: Protolysis will decrease the concentration of fluoride, so more fluoride will need to be added to cause precipitation; therefore measured Ksp will be too large. Likewise, some complex ions such as PbF+ or PbF2 (aq) may form, again leading to an experimental value that is too large.
i. In a solution of 0.010 M barium nitrate and 0.010 M lead(II) nitrate, which will precipitate first, BaF2 or PbF2, as NaF(s) is added? Assume volume changes are negligible. Explain (support your answer with calculations).
Model Answer
As both BaF2 and PbF2 are 1:2 compounds, and the concentrations of the metal ions are both 0.010 M, you can tell that PbF2 will precipitate first, because it has the lower Ksp. For calculations to support this:
For PbF2, 4.0 x 10–8 = (0.01)[F–]2 [F–]2 = 4.0 x 10–6 [F–] = 2.0 x 10–3 M
For BaF2, 1.5 x 10–6 = (0.01)[F–]2 [F–]2 = 1.5 x 10–4 [F–] = 1.2 x 10–2 M
The PbF2 will precipitate first because a lower value for the concentration of fluoride is needed.
ii. When the more soluble fluoride begins to precipitate, what is the concentration of the cation for the less soluble fluoride that remains in solution?
Model Answer
From part (i) we know that the BaF2 precipitates second, when the [F–] reaches 1.2 x 10–2 M
Since PbF2 (s) is present, then [Pb2+][F–]2 = Ksp = 4.0 x 10–8
[Pb2+](1.2 x 10–2)2 = 4.0 × 10–8
[Pb2+] = 2.8 × 10–4 M