Question

Calculate the molar solubility of CaF2 in a solution buffered to a pH of 8.250

Answer 1

1) The chemical equation:

CaF₂ ⇌ Ca²⁺ + 2F¯

2) The K_sp expression:

K_sp = [Ca²⁺] [F¯]²

3) Use the pH to get the pOH:

14.00 - 8.250 = 5.75

4) Use the pOH to get the [OH¯]:

[OH¯] = 10¯^pOH = 10¯^5.75 = 1.7782 x 10¯⁶ M

5) From the chemical equation, we note that the [Ca²⁺] is half the value of the [F¯], therefore:

[Ca²⁺] = 1.7782 x 10¯⁶ M divided by 2 = 8.8913 x 10¯⁷ M

6) We now have the necessary values to put into the K_sp expression:

K_sp = (8.8913 x 10¯⁷) (1.7782 x 10¯⁶)²

K_sp =(8.8913 x 10¯⁷) (3.1619 x 10¯¹²)

K_sp =28.1142 x 10¯¹⁹

CaF2 = Ca++ + 2 F-

Let "X" equal the molar solubility of CaF2. Then:

[X][2X]^2 = 28.1142 x 10¯¹⁹

4 X^3 = 28.1142 x 10¯¹⁹

X^3 = 7.02X 10^-19

X = (7.02X 10^-19)^-3

X= 8.8874 x 10^-7 is the the molar solubility of CaF2.

X = 2.04 X10^-4

So the molar solubility of CaF2 = 2.04 X 10^4 M