Lithium fluoride has a Ksp of 1.84 x 10-3. If 1.5 g are dissolved, what will be the percent ionization?
To determine the percent ionization of lithium fluoride (LiF) when 1.5 grams are dissolved, we first need to calculate the concentration of LiF in the solution.
Step 1: Calculate the molar mass of LiF
The molar mass of lithium (Li) is approximately 6.94 g/mol, and the molar mass of fluorine (F) is approximately 19.00 g/mol.
Molar mass of LiF = (1 * 6.94) + (1 * 19.00) = 6.94 + 19.00 = 25.94 g/mol
Step 2: Convert grams of LiF to moles
We can use the molar mass to convert grams of LiF to moles.
moles of LiF = mass of LiF / molar mass of LiF
moles of LiF = 1.5 g / 25.94 g/mol
Step 3: Calculate the concentration of LiF
The concentration of LiF in a solution is given by the formula:
concentration = moles / volume
To find the concentration, we need to know the volume of the solution. Let's assume that the volume is 1 liter (1000 mL) for simplicity.
concentration = (moles of LiF) / (volume of solution in liters)
concentration = (1.5 g / 25.94 g/mol) / 1 L
Step 4: Calculate the ionization of LiF
Now we can calculate the ionization of LiF using its solubility product constant (Ksp).
Ksp = [Li+][F-]
Since LiF dissociates into one Li+ ion and one F- ion:
[L] = [F] = x (assuming they have the same concentration)
Using the concentration calculated in step 3, substitute these values into the Ksp expression:
Ksp = x * x = x^2
Step 5: Solve for x
Now we can solve for x, which represents the concentration of Li+ ions and F- ions in the solution:
x = sqrt(Ksp)
x = sqrt(1.84 x 10^-3)
Step 6: Calculate percent ionization
Finally, we can calculate the percent ionization by comparing the concentration of the ions (x) to the initial concentration (concentration of LiF) and multiplying by 100:
percent ionization = (x / concentration of LiF) * 100
Substitute the values into the formula:
percent ionization = (sqrt(1.84 x 10^-3) / (1.5 g / 25.94 g/mol)) * 100
Calculate the value to find the percent ionization.