What [PB2+] should be maintained in Pb(NO3) to produce solubility of 1.0*10^-4 mol PbI2/L when PbI2 (s) is added? Ksp for PbI2 is 7.1*10^-9
I can't figure it out...
.........PbI2(s) ==> Pb^2+ + 2I^-
You want PbI2 = 1E-4; therefore, you want Pb^2+ to be 1E-4M
To find the concentration of [PB2+] that should be maintained in Pb(NO3) to produce a solubility of 1.0*10^-4 mol PbI2/L, you need to use the Ksp (solubility product constant) for PbI2. The Ksp expression for PbI2 is:
Ksp = [Pb2+][I-]^2
Where [Pb2+] represents the concentration of Pb2+ ions and [I-] represents the concentration of I- ions.
In this case, since you want to find the concentration of [Pb2+], you need to rearrange the equation:
[Pb2+] = Ksp / [I-]^2
However, the concentration of [I-] can be calculated using the stoichiometry of the reaction. Each PbI2 molecule forms one Pb2+ ion and two I- ions. So, when 1.0*10^-4 mol PbI2 dissolves, it will produce 1.0*10^-4 mol of Pb2+ and 2.0*10^-4 mol of I-.
Now you can substitute these values into the equation:
[Pb2+] = Ksp / ([I-]^2)
[Pb2+] = (7.1*10^-9) / ((2.0*10^-4)^2)
Simplifying the equation gives you:
[Pb2+] = 1.775 * 10^-1 mol/L
Therefore, to produce a solubility of 1.0*10^-4 mol PbI2/L, a concentration of approximately 1.775 * 10^-1 mol Pb2+/L should be maintained in Pb(NO3).