Determine the solubility of BaF2 in g/L if Ksp of BaF2 = 1.7 × 10-6.
Well, I can certainly help you with that. But first, let me put on my lab coat and my funny wig. Now, let's get down to business!
So, the solubility of BaF2 can be determined using its solubility product constant, Ksp, which is given as 1.7 × 10-6.
Now, the solubility of BaF2 can be determined by taking the square root of Ksp, since the stoichiometry of the reaction is 1:1.
Taking the square root of 1.7 × 10-6 gives us approximately 1.30 × 10-3.
So, the solubility of BaF2 in grams per liter (g/L) is approximately 1.30 × 10-3 g/L.
Remember, this is just an approximation. If you need more precise values, feel free to consult your friendly neighborhood chemist!
By the way, why did the molecule go to school? Because it had a lot to molecule over!
To determine the solubility of BaF2 in g/L, we need to use the given Ksp value and a few equations. The Ksp expression for the reaction is:
Ksp = [Ba2+][F-]^2
BaF2(s) ⇌ Ba2+(aq) + 2F-(aq)
Since BaF2 dissociates into one Ba2+ ion and two F- ions, the concentration of Ba2+ is equal to the solubility of BaF2, while the concentration of F- is double the solubility.
Let's denote the solubility of BaF2 as "S". Therefore, the concentration of Ba2+ is S, and the concentration of F- is 2S.
Plugging these values into the Ksp expression:
1.7 × 10^-6 = (S)(2S)^2
Now, solve for S:
1.7 × 10^-6 = 4S^3
Simplify:
4S^3 = 1.7 × 10^-6
Divide both sides by 4:
S^3 = (1.7 × 10^-6) / 4
Simplify further:
S^3 = 4.25 × 10^-7
Take the cube root of both sides:
S = ∛(4.25 × 10^-7)
Using a calculator, we find:
S ≈ 0.00797 g/L
Therefore, the solubility of BaF2 is approximately 0.00797 g/L.
To determine the solubility of BaF2 in g/L, we need to first understand what the Ksp value represents. Ksp is the solubility product constant, which measures the extent to which a solid compound dissociates or dissolves in water.
The Ksp expression for BaF2 can be written as follows:
BaF2 ⇌ Ba2+ + 2F-
The Ksp expression is the product of the concentrations or activities of the dissociated ions raised to the power of their stoichiometric coefficients.
To calculate the solubility of BaF2 in g/L, we need to determine the molar solubility, which is the number of moles of BaF2 that dissolve per liter of solution.
Let's assume 'x' is the molar solubility of BaF2 in mol/L. Since BaF2 dissociates to form one Ba2+ ion and two F- ions, the concentration of Ba2+ ions in the solution will be 'x', and the concentration of F- ions in the solution will be '2x'.
Using the balanced equation and the molar solubility values, we can write the expression for the solubility product constant (Ksp) as:
Ksp = [Ba2+][F-]^2
Substituting the concentrations with our values, we get:
1.7 × 10^-6 = x * (2x)^2 = 4x^3
Now, we can solve for 'x', the molar solubility of BaF2:
4x^3 = 1.7 × 10^-6
To solve for 'x', we take the cube root of both sides:
x = (∛(1.7 × 10^-6) / ∛4)
Calculating this expression, we find that the molar solubility of BaF2 is approximately 0.00123 mol/L.
To determine the solubility in g/L, we need to consider the molar mass of BaF2. Barium (Ba) has a molar mass of approximately 137.33 g/mol, and fluorine (F) has a molar mass of approximately 18.998 g/mol. Thus, the molar mass of BaF2 is:
137.33 g/mol (Ba) + 2 * 18.998 g/mol (F) = 175.326 g/mol
Now, we can calculate the solubility of BaF2 in g/L:
Solubility (g/L) = molar solubility (mol/L) * molar mass (g/mol)
Solubility = 0.00123 mol/L * 175.326 g/mol
The solubility of BaF2 in g/L is approximately 0.215 g/L.
See this problem above.
https://www.jiskha.com/questions/1834093/calculate-the-molar-solubility-of-ag2so4-when-dissolved-in-water-ksp-1-8-x-10-5