Question

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

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

1) The chemical equation:

CaF2 ⇌ Ca2+ + 2F¯

2) The Ksp expression:

Ksp = [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 Ksp expression:

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

Ksp =(8.8913 x 10¯7) (3.1619 x 10¯12)

Ksp =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

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