Calculate activities of Ca2+ and F- and the solubility of fluorite (CaF2) in pure water at 25C without regard to possible hydrolysis. Assume that all activity coefficients are equal to one. Express the solubility of fluorite in terms of mol/L and g/L.

To calculate the activities of Ca2+ and F- in pure water, we need to consider the solubility of fluorite (CaF2) and the dissociation of this compound.

The balanced equation for the dissociation of CaF2 in water is:
CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)

Let's assume the solubility of fluorite in pure water is "s" mol/L. This means that at equilibrium, [Ca2+] = s mol/L, and [F-] = 2s mol/L.

Since we are assuming that all activity coefficients are equal to one, the activities of Ca2+ and F- are equal to their concentrations.

Therefore, the activities of Ca2+ and F- in pure water at 25°C are:
[Ca2+] = s mol/L
[F-] = 2s mol/L

Now, let's calculate the solubility of fluorite (CaF2) in terms of mol/L and g/L.

The molar mass of CaF2 is:
Ca: 40.08 g/mol
F: 18.998 g/mol

The formula mass of CaF2 is:
CaF2: 40.08 + 2 × 18.998 = 78.076 g/mol

The number of moles of CaF2 dissolved in 1 L of water is equal to the solubility "s" mol/L.

Therefore, the solubility of fluorite (CaF2) in mol/L is s mol/L.

To convert the solubility from mol/L to g/L, we multiply the solubility (s) by the molar mass of CaF2:

Solubility (g/L) = s mol/L × 78.076 g/mol

The solubility of fluorite (CaF2) is s mol/L and g/L.

To calculate the activities of Ca2+ and F- ions and the solubility of fluorite (CaF2) in pure water at 25°C, we need to consider the solubility product constant (Ksp). The equation for the dissolution of fluorite is:

CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)

The Ksp expression for this equilibrium is:

Ksp = [Ca2+][F-]^2

Since the question assumes that all activity coefficients are equal to one, we can write the Ksp expression as:

Ksp = [Ca2+][F-]^2

Now, let's say the solubility of fluorite (CaF2) in pure water is represented by 's' moles per liter. Therefore, the concentration of Ca2+ ions at equilibrium is s moles per liter, and the concentration of F- ions at equilibrium is 2s moles per liter. Substituting these values into the Ksp expression, we get:

Ksp = (s)(2s)^2
Ksp = 4s^3

Since we are given that the activity coefficients are equal to one, we can simply equate the concentrations to the activities.

Now, we can solve for the value of 's'. Rearranging the equation, we have:

4s^3 = Ksp
s^3 = Ksp/4
s = (Ksp/4)^(1/3)

Let's assume that the Ksp value for fluorite at 25°C is given as 3.9 x 10^-11 mol^3/L^3. We can substitute this value into the equation above to calculate the solubility of fluorite (CaF2) in terms of moles per liter:

s = (3.9 x 10^-11 mol^3/L^3 / 4)^(1/3)
s ≈ 3.92 x 10^-4 mol/L

So, the solubility of fluorite (CaF2) in pure water at 25°C is approximately 3.92 x 10^-4 mol/L.

To express the solubility of fluorite in terms of grams per liter (g/L), we need to consider the molar mass of CaF2, which is the sum of the atomic masses of calcium (Ca) and two fluorine (F) atoms. The molar mass of CaF2 is approximately 78.08 g/mol.

To convert the solubility from moles per liter to grams per liter, we can multiply the solubility (in moles per liter) by the molar mass:

Solubility (g/L) = Solubility (mol/L) x Molar mass (g/mol)
= 3.92 x 10^-4 mol/L x 78.08 g/mol
≈ 3.05 x 10^-2 g/L

Therefore, the solubility of fluorite (CaF2) in pure water at 25°C is approximately 3.05 x 10^-2 g/L.

A temperature of 20o C is equivalent to approximately

if the volume of the vapor had been assumed to be 125ml instead of your measured volume,what would have been the percent error in your report? molar mass of the compound?explain