The molar solubility of CaF2 is 2.6 × 10-5 M in pure water at a certain temperature. Calculate the Ksp for CaF2 at this temperature.
The molar solubility of CaF2 in pure water is given as 2.6 × 10-5 M. This means that at equilibrium, the concentration of Ca2+ ions and F- ions from the dissociation of CaF2 is 2.6 × 10-5 M.
The balanced equation for the dissociation of CaF2 is:
CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)
From the balanced equation, we can see that the molar solubility represents the concentration of Ca2+ ions, so [Ca2+] = 2.6 × 10-5 M.
Using the solubility product expression, Ksp = [Ca2+][F-]^2, we can plug in the values to calculate Ksp:
Ksp = (2.6 × 10-5)(2.6 × 10-5)^2
= 2.6 × 10-5 × 6.76 × 10-10
= 1.756 × 10-14
Therefore, the Ksp for CaF2 at this temperature is 1.756 × 10-14.
Bot, you forgot to multiply by 2 to obtain F^-.
CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)
solid..........2.6E-5.......2*2.6E-5
Ksp = (Ca^2+)(F^-)^2 = (2.6E-5)(5.2E-5)^2
Ksp = 7.03E-14
You are correct, I apologize for the error. The correct calculation is as follows:
The molar solubility of CaF2 in pure water is 2.6 × 10-5 M, which represents the concentration of Ca2+ ions. Since the balanced equation shows that 2 moles of F- ions are produced for every mole of CaF2 that dissolves, the concentration of F- ions is 2 * (2.6 × 10-5) = 5.2 × 10-5 M.
Using the solubility product expression:
Ksp = (Ca2+)(F-)2 = (2.6 × 10-5)(5.2 × 10-5)2
= 7.03 × 10-14
Therefore, the correct Ksp for CaF2 at this temperature is 7.03 × 10-14.
To calculate the Ksp (solubility product constant) for CaF2 at a certain temperature, you can use the molar solubility (s) of CaF2 in pure water. The general equation for the dissolution of CaF2 is:
CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)
The Ksp expression for this dissolution is written as:
Ksp = [Ca2+][F-]^2
Since the molar solubility of CaF2 is given as 2.6 × 10^-5 M, the concentration of Ca2+ and F- ions can be considered to be equal to 2.6 × 10^-5 M.
Substituting these values into the Ksp expression, we get:
Ksp = (2.6 × 10^-5)(2.6 × 10^-5)^2
Ksp = (2.6 × 10^-5)(6.76 × 10^-10)
Ksp = 1.756 × 10^-14
Therefore, the Ksp for CaF2 at this temperature is 1.756 × 10^-14.
To calculate the solubility product constant (Ksp) for CaF2 at a given temperature, we need to use the molar solubility value. The equilibrium expression for the dissociation of CaF2 is as follows:
CaF2 (s) ⇌ Ca2+ (aq) + 2F- (aq)
The molar solubility of CaF2 indicates the number of moles of CaF2 that dissolve to form 1 liter of saturated solution.
Given that the molar solubility of CaF2 is 2.6 × 10-5 M, we can write the equilibrium constant expression (Ksp) as follows:
Ksp = [Ca2+][F-]^2
To calculate Ksp, we need to determine the concentrations of Ca2+ and F- ions in the saturated solution. In this case, since 1 mole of CaF2 dissociates to form 1 mole of Ca2+ ions and 2 moles of F- ions, the concentrations of Ca2+ and F- ions will be twice the molar solubility:
[Ca2+] = 2 × (2.6 × 10-5) M = 5.2 × 10-5 M
[F-] = 2 × 2 × (2.6 × 10-5) M = 2.6 × 10-4 M
Now we can substitute these concentrations into the equilibrium expression for Ksp:
Ksp = (5.2 × 10-5)(2.6 × 10-4)^2
Ksp = 3.52 × 10-11
Therefore, the solubility product constant (Ksp) for CaF2 at this temperature is 3.52 × 10-11.