How much Pb(OH)2(in mg) can be added to 4.0liter of a 0.095 M KOH solution before precipitation starts? Ksp 4.2x10^-5
Ksp Pb(OH)2 = (Pb^2+)(OH^-)^2
Use 0.095M = (OH^-) and solve for Pb^2+.
That gives M Pb^2+ to start pptn.
M Pb^+2 = mols Pb^2+/L = mols Pb(OH)2/L.
M Pb^2+ from above and L (4.00) give mols. Then mols= grams/molar mass. You know molar mass and mols, solve for grams and convert to mg.
Awesome. Thank you so much!!!!!
which molar mass do you use? pb(OH)2 or pb2+?
To determine the maximum amount of Pb(OH)2 that can be added before precipitation starts, we need to use the solubility product constant (Ksp) for Pb(OH)2. The Ksp expression for Pb(OH)2 is:
Ksp = [Pb2+][OH-]^2
Given that the Ksp value for Pb(OH)2 is 4.2 x 10^-5, we can set up the following equation:
4.2 x 10^-5 = [Pb2+][OH-]^2
Since the concentration of KOH is given (0.095 M), we know the concentration of OH- ions is also 0.095 M. Therefore, we can substitute this value into the equation:
4.2 x 10^-5 = [Pb2+](0.095)^2
To find the concentration of Pb2+, rearrange the equation:
[Pb2+] = (4.2 x 10^-5) / (0.095)^2
[Pb2+] = 0.0048 M
Now, we can calculate the moles of Pb2+ that can be present in the 4.0 L solution using the given volume:
moles of Pb2+ = [Pb2+] × volume of solution
moles of Pb2+ = 0.0048 M × 4.0 L
moles of Pb2+ = 0.0192 mol
Finally, to convert moles of Pb2+ to milligrams (mg), we need to multiply by the molar mass of Pb(OH)2. The molar mass of Pb(OH)2 can be calculated as follows:
molar mass of Pb(OH)2 = atomic mass of Pb + (2 × atomic mass of O) + (2 × atomic mass of H)
molar mass of Pb(OH)2 = 207.2 g/mol + (2 × 16.0 g/mol) + (2 × 1.01 g/mol)
molar mass of Pb(OH)2 = 241.22 g/mol
Now, we can calculate the amount of Pb(OH)2 in milligrams:
mass of Pb(OH)2 = moles of Pb2+ × molar mass of Pb(OH)2
mass of Pb(OH)2 = 0.0192 mol × 241.22 g/mol
mass of Pb(OH)2 = 4.633 mg (rounded to three decimal places)
Therefore, approximately 4.633 mg of Pb(OH)2 can be added to 4.0 L of 0.095 M KOH solution before precipitation starts.