What mass of NaF must be added to 1.0 L of 0.0036 M Pb2+ to initiate precipitation of PbF2(s)? (Ksp of PbF2 is 3.3 × 10-8; assume no volume change on addition of NaF.)
can you please explain
..................PbF2 ==> Pb^2+ + 2F^-
I...................solid...........0............0
C..................solid..........x.............2x
E...................solid..........x............2x
Ksp = (Pb^+)(F^-)^2
You know Ksp. You know (Pb^2+) from the problem. You want to know (F^-). Solve for that (in mols/L). Convert (F^-) to (NaF) in mols/L and change that to grams/L. The idea here is that PbF2 will ppt when Ksp is exceeded. So you're calculating the amount of NaF to add to the solution so (Pb^+)(F^-)^2 is greater than Ksp for PbF2. Post your work if you get stuck.
To determine the mass of NaF needed to initiate precipitation of PbF2(s), we can use the solubility product constant (Ksp) of PbF2.
The balanced equation for the precipitation reaction is:
Pb2+ (aq) + 2F- (aq) ⟶ PbF2 (s)
The Ksp expression for PbF2 is:
Ksp = [Pb2+][F-]^2
Given that the concentration of Pb2+ is 0.0036 M, we need to find the concentration of F- ions required to reach the Ksp.
Since the stoichiometric ratio between Pb2+ and F- is 1:2, the concentration of F- ions will be twice the concentration of Pb2+, i.e. 2 * 0.0036 M = 0.0072 M.
Now, let's set up the Ksp expression using the concentrations of Pb2+ and F- ions:
Ksp = (0.0036)(0.0072)^2
Since Ksp is given as 3.3 × 10^-8, we can set up the equation:
3.3 × 10^-8 = (0.0036)(0.0072)^2
To solve for the mass of NaF, we need to calculate the moles of F- ions required to reach the Ksp. The molar mass of NaF is 41 g/mol.
First, find the moles of F-:
moles of F- = (0.0072 M) x (1.0 L) = 0.0072 mol
Next, find the mass of NaF:
mass of NaF = moles of F- x molar mass of NaF
mass of NaF = 0.0072 mol x 41 g/mol = 0.2952 g
Therefore, approximately 0.2952 grams of NaF must be added to 1.0 L of 0.0036 M Pb2+ to initiate precipitation of PbF2(s).
To solve this problem, we need to use the concept of solubility product constant (Ksp) and stoichiometry.
First, let's write the balanced equation for the precipitation reaction:
Pb2+ + 2F- -> PbF2(s)
Based on the equation, we can see that one mole of Pb2+ reacts with two moles of F- to form one mole of PbF2. Therefore, the stoichiometric ratio of Pb2+ to F- is 1:2.
From the given information, we know that the concentration of Pb2+ is 0.0036 M, and the Ksp of PbF2 is 3.3 × 10-8.
To find the mass of NaF required to initiate precipitation, we need to determine the concentration of F- ions at which the reaction will start.
Let's denote the concentration of NaF as x M. Since NaF completely dissociates in water, the concentration of F- ions provided by NaF will also be x M.
According to stoichiometry, the concentration of F- provided by NaF is twice the concentration of Pb2+. Therefore, the concentration of F- ions derived from Pb2+ is (2 * 0.0036) M = 0.0072 M.
To calculate the concentration of F- ions from NaF, we can add the two concentrations:
Total concentration of F- = concentration of F- from Pb2+ + concentration of F- from NaF
Total concentration of F- = 0.0072 M + x M
Now, we have the concentration of F- ions at equilibrium. We can use the Ksp expression to set up the equation:
Ksp = [Pb2+][F-]^2
Substituting the known values, we get:
3.3 × 10-8 = (0.0036 M) * [(0.0072 M + x M)^2]
The concentration of NaF, represented by x, can be assumed to be small compared to 0.0072 M. Thus, we can approximate (0.0072 M + x M)^2 as approximately (0.0072 M)^2.
Now let's solve the equation:
3.3 × 10-8 = (0.0036 M) * (0.0072 M)^2
Rearranging and solving for x, we get:
x^2 + 0.0144 M * x - 3.3 × 10-8 / 0.0036 M = 0
Using the quadratic formula, we can solve for x. The value of x will represent the concentration of NaF required to initiate the precipitation.
Once we have the value of x, we can calculate the mass of NaF required using the formula:
Mass = concentration (mol/L) * volume (L) * molar mass (g/mol)
Given that the volume is 1.0 L, we can substitute the values and find the mass of NaF required to initiate the precipitation of PbF2(s).