Aqueous solubility and pH?
For which salt should the aqueous solubility be most sensitive to pH? CaCl2, CaI2, CaF2, CaBr2. I know that the answer is CaI2. Could you, please, explane to me why? Thank you.
Answer:
I believe the answer is CaF2.
Let's say in all these cases, the equation for the dissolving of the calcium salt is:
CaX2 (s) = Ca2+(aq) + 2X-(aq)
Varying the pH will vary the [H+].
Let's we have 4 containers, one with CaCl2, one with CaF2, one with CaF2, and one with CaBr2 dissoved in water. Now pretend we add the same amount of acid, H+ ions, to each beaker.
The effect of the H+ on the solubility will depend on how the H+ interacts with the X- ions.
If the H+ has an affinity for the X- ion, it will remove X- from the solution, causing the solubilty equilibrium to shift to the RIGHT (more CaX2 dissolves).
If the H+ has little affinity for X- ion, nothing will happen, and the solubility will remain constant.
The questions is now, "Which halogen ion will have the most affinity for the H+ ion?"
We can solve this by looking at the relative strengths of acids, whic is: (see source below for pKa's)
HI > HBr > HCl > HF
This means that the equilibrium:
HI = H+ + I-
lies more to the right than in the other acids. Or, in other words, the reverse reaction (H+ and I- getting together to form HI) is highly unlikely to occur.
Since H+ and I- are least likely to come together, the [I-] will not be affected by changing the pH, making CaI2 the LEAST sensitive to pH. Since HF is the weakest of the acids, this means the H+ and F- have a greater chance of becoming HF, removing F- ions, and altering the solubiity. This makes CaF2 the most sensitive to changes in pH.
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Answer:
I believe the answer is CaF2.
Let's say in all these cases, the equation for the dissolving of the calcium salt is:
CaX2 (s) = Ca2+(aq) + 2X-(aq)
Varying the pH will vary the [H+].
Let's we have 4 containers, one with CaCl2, one with CaF2, one with CaF2, and one with CaBr2 dissoved in water. Now pretend we add the same amount of acid, H+ ions, to each beaker.
The effect of the H+ on the solubility will depend on how the H+ interacts with the X- ions.
If the H+ has an affinity for the X- ion, it will remove X- from the solution, causing the solubilty equilibrium to shift to the RIGHT (more CaX2 dissolves).
If the H+ has little affinity for X- ion, nothing will happen, and the solubility will remain constant.
The questions is now, "Which halogen ion will have the most affinity for the H+ ion?"
We can solve this by looking at the relative strengths of acids, whic is: (see source below for pKa's)
HI > HBr > HCl > HF
This means that the equilibrium:
HI = H+ + I-
lies more to the right than in the other acids. Or, in other words, the reverse reaction (H+ and I- getting together to form HI) is highly unlikely to occur.
Since H+ and I- are least likely to come together, the [I-] will not be affected by changing the pH, making CaI2 the LEAST sensitive to pH. Since HF is the weakest of the acids, this means the H+ and F- have a greater chance of becoming HF, removing F- ions, and altering the solubiity. This makes CaF2 the most sensitive to changes in pH.
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