Only 0.0640 grams of PbF2 will dissolve in 100.0 mL of solution. From this data, calculate the Ksp by two methods.
Which method do you know?
Ksp=produts molarity/reatants molarity
Then you dont know either.
mols PbF2 dissolved - 0.064/molar mass PbF2 = Estimated 0.0004
............PbF2(s) ==> Pb^2+ + 2F^-
I..........solid.................0............0
C.........solid............0.0004.....0.0008
E.........solid............0.0004......0.0008
Ksp = (Pb^2+)(F^-)^2
Ksp = (0.0004)(0.0008)^2 = ?
Another method.
............PbF2 ==> Pb^2+ + 2F^-
I..........solid...........0..............0
C.........solid...........x..............2x
E.........solid...........x..............2x
Ksp = (Pb^2+)(F^-)^2
Ksp = (x)(2x)^2 = 4x^3
You know x = estimated 0.0004 so
Ksp = 4*0.0004)^3 approximately.
Don't forget to get more accurate numbers
To calculate the solubility product constant (Ksp) of PbF2, we can use two methods: the molar solubility method and the concentration method. Let's go through each of them step by step:
Method 1: Molar Solubility Method
Step 1: Determine the molar solubility (s):
Molar solubility (s) is the number of moles of PbF2 that dissolve per liter of solution.
We are given that 0.0640 grams of PbF2 dissolve in 100.0 mL of solution.
First, convert grams to moles:
molar mass of PbF2 = (207.2 g/mol for Pb) + (2 * 19.0 g/mol for F) = 223.2 g/mol
moles of PbF2 = mass / molar mass = 0.0640 g / 223.2 g/mol = 0.000287 mol
Next, convert mL to L:
100.0 mL = 0.100 L
Therefore, the molar solubility (s) is:
s = moles of PbF2 / volume of solution in liters = 0.000287 mol / 0.100 L = 0.00287 M
Step 2: Write the chemical equation:
PbF2(s) ⇌ Pb2+(aq) + 2F-(aq)
Step 3: Write the expression for Ksp:
Ksp = [Pb2+][F-]^2
Since the stoichiometric coefficients in the balanced equation are 1:2:1 for Pb2+, F-, and PbF2, respectively, the concentration of Pb2+ and F- is equal to the molar solubility (s).
Therefore, Ksp = (s)(s^2) = s^3
Plugging in the value of s we found earlier:
Ksp = (0.00287 M)^3 = 2.43 × 10^-8
Method 2: Concentration Method
Step 1: Calculate the concentration of Pb2+ and F- ions in the solution:
Since we know the molar solubility (s) from before, we can calculate the concentration of Pb2+ and F- ions.
[Pb2+] = s = 0.00287 M
[F-] = 2s = 2 * 0.00287 M = 0.00574 M
Step 2: Write the expression for Ksp:
Ksp = [Pb2+][F-]^2
Plugging in the concentrations we found:
Ksp = (0.00287 M)(0.00574 M)^2 = 4.84 × 10^-8
Therefore, the solubility product constant (Ksp) for PbF2, calculated using both methods, is approximately 2.43 × 10^-8 or 4.84 × 10^-8, depending on the method used.