the solubility of BaF2 is 6.29x10^-3M and the solubility of barium fluoride in 0.15M NaF is 4.4x10^-5M. Compare the solubility and explain?
BaF2 ==> Ba^+2 + 2F^-
NaF ==> Na^+ + F^-
Solubility BaF2 alone = the above number
Solubility mixture BaF2 and 0.15 M NaF = a much smaller number. Why? Because of the common ion effect (the fluoride ion from NaF) shifts the BaF2 equilibrium to the left which results in a lower solubility of BaF2 in the mixture.
To compare the solubility of BaF2 in pure water and in a solution of NaF, we'll first determine the common ion present in both cases.
BaF2 dissociates into Ba^2+ ions and F^- ions. When BaF2 dissolves in water, it can be represented by the equation:
BaF2(s) ⇌ Ba^2+(aq) + 2F^-(aq)
In this case, the only source of F^- ions is the dissolution of BaF2 itself.
Now, let's consider the solubility of BaF2 in a 0.15M NaF solution. NaF also dissociates into Na^+ ions and F^- ions. The dissociation equation for NaF is:
NaF(s) ⇌ Na^+(aq) + F^-(aq)
Here, there is an additional source of F^- ions provided by NaF. This means that there is an increased concentration of F^- ions in the solution, which can affect the solubility of BaF2.
The solubility product constant (Ksp) expression for BaF2 is given as:
Ksp = [Ba^2+][F^-]^2
Since the solubility is given in terms of concentration (M), we can rewrite Ksp as:
Ksp = [Ba^2+][F^-]^2 = (s)(2s)^2 = 4s^3
In pure water, the solubility (s) of BaF2 is given as 6.29x10^-3M. So, the Ksp expression becomes:
Ksp_water = (6.29x10^-3)^3 = 2.5x10^-8
In the presence of the 0.15M NaF solution, let the solubility of BaF2 be x M. The additional F^- ions from NaF will increase the concentration of F^- ions to (4.4x10^-5 + x) M. Therefore, the Ksp expression becomes:
Ksp_NaF = (x)((4.4x10^-5 + x))^2
Since both Ksp_water and Ksp_NaF represent the same compound (BaF2), they should be equal. Hence:
2.5x10^-8 = (x)((4.4x10^-5 + x))^2
Now, we can solve this equation to find the solubility of BaF2 in the presence of the 0.15M NaF solution.
However, please note that this is a complex equation to solve algebraically. Therefore, solving it directly might not be feasible without using numerical methods or software.
To summarize, the solubility of BaF2 in a 0.15M NaF solution is given by solving the equation 2.5x10^-8 = (x)((4.4x10^-5 + x))^2.