Calculate the minimum number of moles of Pb(NO3)2 required to start precipitation in 50.0mL of 0.15M ZnI2
I got 4.72x10^-9
PbI2 ==> Pb^2+ + 2I^-
Ksp = (Pb^2+)(I^-)^2
(I^-) = 0.15M*2 = 0.30M
Substitute and solve for (Pb^2+); the unit is mols/L. You want mols in 50 mL (0.050L) and mols = M x L = ?
Well, I'm not a mathematician, but let me take a guess at this. If we assume that the reaction between Pb(NO3)2 and ZnI2 forms a precipitate, then we can use stoichiometry to figure out the minimum amount of Pb(NO3)2 needed.
The balanced formula for the reaction is:
Pb(NO3)2 + ZnI2 -> PbI2 + Zn(NO3)2
From this balanced equation, we can see that one mole of Pb(NO3)2 reacts with one mole of ZnI2 to form one mole of PbI2.
So if we have 50.0 mL of a 0.15 M ZnI2 solution, we can calculate the number of moles of ZnI2 using the formula:
moles = concentration (M) x volume (in liters)
moles = 0.15 M x 0.05 L
moles = 0.0075 mol
Since the stoichiometry of the reaction is 1 mole of Pb(NO3)2 to 1 mole of ZnI2, we would need at least 0.0075 mol of Pb(NO3)2 to start precipitation.
But wait! I'm just a clown bot, not a chemistry professor. So take my answer with a pinch of salt, or maybe some clown makeup.
To calculate the minimum number of moles of Pb(NO3)2 required to start precipitation in 50.0mL of 0.15M ZnI2, we need to use the concept of stoichiometry and the balanced chemical equation of the reaction.
The balanced chemical equation for the reaction between Pb(NO3)2 and ZnI2 is:
Pb(NO3)2 + ZnI2 -> PbI2 + Zn(NO3)2
From the balanced chemical equation, we can see that the mole ratio between Pb(NO3)2 and ZnI2 is 1:1. This means that one mole of Pb(NO3)2 reacts with one mole of ZnI2 to form one mole of PbI2 and one mole of Zn(NO3)2.
Given that the concentration of ZnI2 is 0.15M in 50.0mL, we can calculate the number of moles of ZnI2 using the formula:
Number of moles = concentration (M) * volume (L)
First, convert the volume from mL to L:
Volume = 50.0 mL = 50.0 mL * (1 L / 1000 mL) = 0.050 L
Number of moles of ZnI2 = 0.15 M * 0.050 L = 0.0075 moles
Since the mole ratio between Pb(NO3)2 and ZnI2 is 1:1, the minimum number of moles of Pb(NO3)2 required to react with 0.0075 moles of ZnI2 is also 0.0075 moles.
Therefore, the minimum number of moles of Pb(NO3)2 required to start precipitation in 50.0mL of 0.15M ZnI2 is 0.0075 moles.