the solubility of CaF in water is 7*10^-2 gm/l. find its solubility in 0.5m CaCl2 soln
Based on data given and using a formula wt of CaF2 = 78g/mol;
Molar solubility CaF2 = (7E-2/78)M = 8.9E-4M. Ksp = [Ca][F]^2 =(8.9E-4)(8.9E-4)^2 = 7E-10 which does not match published Ksp-values for CaF2. Never the less, In presence of 0.50M CaCl2...
CaF2 <=> Ca+2 + 2F^-
Ceq: --- 0.50M 2x
Ksp = [Ca][F]^2 = (0.50)(2x)^2 = (0.50)(4x^2) = 2x^2 = 7E-10 => x = Solubility in the presence of 0.50M CaCl2 = [(7E-10/2)]^1/2 = 1.9E-5M
Comparing to solubility in water S(HOH)=8.9E-4M > S(Ca)=1.9E-5M. This is consistent with common ion effect.
To find the solubility of CaF in a 0.5M CaCl2 solution, you can use the concept of common ion effect. The common ion effect states that the solubility of a salt is reduced when a common ion is present in the solution.
In this case, the common ion is the Ca2+ ion from CaCl2. Since CaF is a salt containing Ca2+ and F- ions, the solubility of CaF will be affected by the presence of Ca2+ ions.
The solubility of CaF in water is given as 7*10^-2 gm/L. This means that in pure water, 7*10^-2 grams of CaF can dissolve in 1 liter of water.
Now, when a 0.5M CaCl2 solution is added, it means that the concentration of Ca2+ ions is 0.5M. This concentration of Ca2+ ions will reduce the solubility of CaF according to the common ion effect.
To calculate the solubility of CaF in the presence of CaCl2, you can use an ICE (Initial, Change, Equilibrium) table.
Let's assume the solubility of CaF in the presence of CaCl2 is 'x' grams per liter.
The equation for the dissolution of CaF can be written as:
CaF (s) ↔ Ca2+ (aq) + 2F- (aq)
Using the stoichiometry of this equation, the concentration of Ca2+ ions and F- ions will both be 'x' M.
The concentration of Ca2+ ions from CaCl2 is 0.5M, and the concentration of F- ions is 2x M (since there are two F- ions produced for each CaF that dissolves).
Applying the common ion effect, we can write the equation:
Ksp = [Ca2+] * [F-]^2
Where Ksp is the solubility product constant for CaF.
Plugging in the values, we have:
Ksp = (0.5) * (2x)^2 = 4 * x^2
From the given solubility in water, we know that Ksp = 7*10^-2 gm/L.
Setting the two equations equal to each other, we can solve for 'x':
4 * x^2 = 7*10^-2
Dividing both sides by 4:
x^2 = (7*10^-2) / 4
Taking the square root of both sides:
x = sqrt((7*10^-2) / 4)
Now you can calculate the solubility of CaF in the 0.5M CaCl2 solution by plugging in the value of 'x' into the equation.
Please note that you will need to convert the units accordingly, as the given solubility is in gm/L, and the concentration of CaCl2 is in M.