Zn + PbCl2 --> ZnCl2 + Pb
The initial mass of a zinc strip is 2.684. After being left overnight in a lead (II) chloride solution, the mass of zinc is 2.581g. Calculate the Ksp of lead (II) chloride.
So far I have:
PbCl2 --> Pb(aq)2+ + 2Cl(aq)-
Ksp = [Pb(aq)2+] [Cl(aq)-]^2
Using my ICE table, I get Ksp = 4x^3.
What do I do now?
grams Zn = 2.684-2.581 g = ??
mols Zn = ??/65.38 = xx mols (check that atomic mass).
mols Zn = mols Pb and mols Pb = what you have called x.
To calculate the Ksp of lead (II) chloride, you need to first determine the moles of zinc that reacted with lead (II) chloride. Here's how you can do that:
1. Calculate the mass of zinc that reacted:
grams of zinc = initial mass - final mass = 2.684 g - 2.581 g = 0.103 g
2. Convert the mass of zinc to moles:
moles of zinc = grams of zinc / molar mass of zinc
The molar mass of zinc (Zn) is 65.38 g/mol.
moles of zinc = 0.103 g / 65.38 g/mol = 0.001574 mol
3. Since Zn and Pb have a 1:1 stoichiometric ratio in the reaction, the moles of lead (II) chloride (PbCl2) that reacted is also 0.001574 mol.
Now that you have the moles of lead (II) chloride that reacted, you can proceed with calculating the Ksp:
Ksp = [Pb(aq)2+] [Cl(aq)-]^2
Since the stoichiometric ratio between PbCl2 and Pb(aq)2+ is 1:1, the concentration of Pb(aq)2+ is equal to the moles of PbCl2 that reacted:
[Pb(aq)2+] = 0.001574 mol/L
And since the stoichiometric ratio between PbCl2 and Cl(aq)- is 2:1, the concentration of Cl(aq)- is twice the moles of PbCl2 that reacted:
[Cl(aq)-] = 2 * 0.001574 mol/L = 0.003148 mol/L
Substituting these values into the equation for Ksp:
Ksp = (0.001574 mol/L)(0.003148 mol/L)^2
Now you can calculate Ksp.