Consider the two equilibria:
BaF2=Ba2+ + 2F- Ksp=1.7x10^(-6)
F- + H20=HF + OH- Kb=2.9x10^(-11)
Determine the solubility of BaF2 at pH=5?
What I'm mostly confused about is how to determine if F- is dominant or if you have to take HF into account while doing your calculations.
You must take HF into consideration during the calculation.
The whole point of these problems is to show you that BaF2 has a certain solubility in neutral solution. As the solution becomes more and more acid the formation of HF becomes more (because HF is a weak acid); that shifts the solubility equilibrium to the right and makes BaF2 more soluble in acid solutions.
To determine the solubility of BaF2 at pH=5, you need to consider the effect of pH on the equilibria involving F- and HF.
At pH=5, you have an acidic solution, so some of the F- ions will react with water to form HF and OH-. Therefore, you need to take HF into account when calculating the solubility.
First, let's write the reaction for the dissociation of BaF2:
BaF2 ⇌ Ba2+ + 2F-
The solubility product constant (Ksp) of BaF2 is given as 1.7x10^(-6), which represents the equilibrium expression:
Ksp = [Ba2+][F-]^2
Now, let's consider the second equilibrium involving F- and HF:
F- + H2O ⇌ HF + OH-
Kb = [HF][OH-]/[F-]
To determine whether F- or HF is dominant in the solution, we need to compare the concentration of F- to the concentration of HF. Since we are provided with the Kb value, we can use it as a ratio to calculate the concentration of HF.
Using the equation for Kb, rearrange it to solve for [HF]:
[HF] = Kb[F-]/[OH-]
Now, we need to determine the concentration of OH- in the solution. At pH 5, we can assume that [H+] is 10^(-5) M (based on the definition of pH). Since water is a neutral solvent, [H+] equals [OH-], so [OH-] = 10^(-5) M.
Now, substitute the values into the equation to calculate [HF]:
[HF] = (2.9x10^(-11) M)([F-])/(10^(-5) M)
Once you have the concentration of [HF], you can substitute this value into the Ksp expression for BaF2:
Ksp = [Ba2+][F-]^2
Now comes the tricky part. Since HF is a weak acid, it will partially dissociate into F- and H+. So, we need to consider the common ion effect. The concentration of F- will be the sum of the concentration of F- from the BaF2 dissociation and the concentration of F- from the ionization of HF.
Let's assume the solubility of BaF2 is denoted as 's'. The concentration of F- due to the dissociation of BaF2 will be '2s', and the concentration of F- due to the ionization of HF will be [HF]. Therefore, the total concentration of F- will be '2s + [HF]'.
Substitute the concentration of F- into the Ksp expression:
1.7x10^(-6) = [Ba2+][(2s + [HF])]^2
Now, substitute the calculated [HF] value:
1.7x10^(-6) = [Ba2+][(2s + (2.9x10^(-11) M)([F-])/(10^(-5) M))^2
Since we are trying to solve for the solubility of BaF2, we can set up a quadratic equation and solve for 's'. However, solving this equation involves several assumptions and approximations that are beyond the scope of this explanation.
In summary, to determine the solubility of BaF2 at pH=5, you need to consider the effect of pH on the equilibria involving F- and HF. You need to calculate the concentration of HF using the Kb value, taking into account the concentration of OH- in the solution. Then, you use this concentration of HF in the Ksp expression for BaF2, considering the common ion effect of F- resulting from BaF2 dissociation and the ionization of HF. Finally, you solve the resulting equation for the solubility 's' of BaF2 using various assumptions and approximations.