Calculation of Solubility of Caf in 0.0005 M Solution of Baf2

Given: Kps of Caf = 3.4 × 10^-11

Step 1: Write the equilibrium reaction

Caf(s) ⇌ Ca2+(aq) + F-(aq)

Step 2: Write the expression for Ksp of Caf

Ksp = [Ca2+][F-]

Since Caf is sparingly soluble, the concentration of Ca2+ and F- in the solution will be equal to the solubility of Caf, denoted by S.

Ksp = S × S = S^2

Step 3: Write the expression for the reaction between Baf2 and CaF2

Baf2(s) + CaF2(s) ⇌ 2CaF(s) + Ba2+(aq)

Since Baf2 is present in excess, its concentration will remain constant throughout the reaction. Therefore, we can write the expression for the equilibrium constant of this reaction as follows:

K = [CaF]2/[Ba2+]

Step 4: Write the expression for the concentration of CaF in terms of solubility of Caf

Since CaF is formed from the reaction between Ca2+ and F-, its concentration can be expressed in terms of the solubility of Caf, denoted by S.

CaF = S

Therefore, the expression for the equilibrium constant can be written as:

K = S^2/[Ba2+]

Step 5: Calculate the value of K from the given information

K = 0.0005 M

Step 6: Calculate the solubility of Caf

Substitute the value of K and Ksp in the expression for K to obtain:

0.0005 = S^2/3.4 × 10^-11

S^2 = 1.7 × 10^-14

S = 1.3 × 10^-7 M

Therefore, the correct answer is option 4) 1.36 × 10^-7.

Explanation

The solubility of Caf in a 0.0005 M solution of Baf2 is determined by the equilibrium constant of the reaction between Baf2 and CaF2. The concentration of CaF2 can be expressed in terms of the solubility of Caf, and the equilibrium constant can be written as the ratio of the concentration of CaF2 to that of Ba2+. By substituting the given information into the expression for the equilibrium constant, the solubility of Caf can be calculated. The correct answer is 1.36 × 10^-7 M.