calculate the molar solubility of barium fluoride in water at 25 degrees C. The solubility product constant for BaF2 at this temperature is 1.0x10^-6
See the post above.
6.3x10^-3
To calculate the molar solubility of Barium Fluoride (BaF2) in water at 25 degrees Celsius, we need to use the solubility product constant (Ksp) for BaF2 at this temperature, which is given as 1.0x10^-6.
The solubility product constant (Ksp) expression for BaF2 is as follows:
BaF2(s) ⇌ Ba2+(aq) + 2F-(aq)
Where:
BaF2(s) represents the solid barium fluoride
Ba2+(aq) represents the barium ion in aqueous solution
2F-(aq) represents two fluoride ions in aqueous solution
The molar solubility (s) of BaF2 is the number of moles of BaF2 that can dissolve in 1 L of water.
Using the Ksp expression, we can set up the following equation:
Ksp = [Ba2+(aq)][F-(aq)]^2
Since the stoichiometry of BaF2 is 1:2 (1 Ba2+ ion and 2 F- ions), we can substitute ([F-(aq)])^2 with (2s)^2.
Therefore, the equation becomes:
Ksp = [Ba2+(aq)](2s)^2
Now, let's substitute the Ksp value provided (1.0x10^-6) and solve for s:
1.0x10^-6 = [Ba2+(aq)](2s)^2
To simplify further, let's assume x = [Ba2+(aq)], which represents the molar concentration of Ba2+ ions in aqueous solution.
1.0x10^-6 = x(2s)^2
Now, we need to take the square root of both sides of the equation to solve for s:
√(1.0x10^-6) = √[x(2s)^2]
1.0x10^-3 = 2s√x
To find the molar solubility, we need to isolate s:
s = (1.0x10^-3) / (2√x)
Since x represents the molar concentration of Ba2+(aq), it is also equal to the molar solubility (s), so we can substitute s for x:
s = (1.0x10^-3) / (2√s)
To solve this equation, we'll square both sides:
s^2 = [(1.0x10^-3) / (2√s)]^2
s^2 = (1.0x10^-3)^2 / (4s)
s^2 = 1.0x10^-6 / (4s)
Now, let's cross-multiply and rearrange the equation:
s^3 = 1.0x10^-6 / 4
s = (√(1.0x10^-6 / 4))^⅓
Calculating the value, we can determine the molar solubility of BaF2 in water at 25 degrees Celsius.
To calculate the molar solubility of barium fluoride (BaF2) in water at 25 degrees Celsius, we need to use the solubility product constant (Ksp). The Ksp value represents the equilibrium constant for the dissociation of the compound into its ions in a saturated solution.
For the dissociation of BaF2, the balanced equation is:
BaF2(s) ⇌ Ba2+(aq) + 2F-(aq)
According to the equation, for every BaF2 molecule that dissolves, one Ba2+ ion and two F- ions are produced. Therefore, the solubility of BaF2 in terms of moles per liter (mol/L) is equal to the molar solubility of Ba2+(aq) in terms of mol/L, as the concentration of F- ions is twice that of Ba2+ ions.
Let's assume the molar solubility of BaF2 in water at 25 degrees Celsius is 's' mol/L. Using stoichiometry, we find that the concentrations of Ba2+ and F- ions are 's' mol/L and '2s' mol/L, respectively.
Now, we can write the expression for the solubility product constant (Ksp) using the concentrations of the ions:
Ksp = [Ba2+][F-]^2
Substituting the concentrations, we have:
Ksp = s * (2s)^2
Ksp = 4s^3
We know the value of Ksp at 25 degrees Celsius is 1.0x10^-6. Substituting this value, we get:
1.0x10^-6 = 4s^3
To solve for 's', we can take the cube root of both sides:
∛(1.0x10^-6) = ∛(4s^3)
1.0x10^-6 = s
Therefore, the molar solubility of barium fluoride (BaF2) in water at 25 degrees Celsius is 1.0x10^-6 mol/L.