Calculate the concentration of IO3– in a 4.73 mM Pb(NO3)2 solution saturated with Pb(IO3)2. The Ksp of Pb(IO3)2 is 2.5 × 10-13. Assume that Pb(IO3)2 is a negligible source of Pb2 compared to Pb(NO3)2.
This is a Ksp problem with a common ion of Pb from the Pb(NO3)2.
..........Pb(IO3)2 ==> Pb^2+ + 2IO3^-
I.........solid.........0........0
C.........solid.........x........2x
E.........solid.........x........2x
Pb(NO3)2 is 100% dissociated.
........Pb(NO3)2 ==> Pb^2+ + 2NO3^-
I.......4.73 mM........0........0
C......-4.73 mM......4.73mM..2*4.73mM
E........0...........4.73mM..9.46 mM
Ksp Pb(IO3)2 = (Pb^2+)(IO3^-)^2
Ksp = 2.5E-13
(Pb^2+) = x from Pb(IO3)2 which the problem tells you to ignore + 4.73 mM from the Pb(NO3). Convert this to 0.00473M
(IO3^-) is 2x
Substitute into the Ksp expression as follows:
2.5E-13 = (0.00473)(2x)^2 and solve for x = (IO3). The unit will be M or mols/L.
To calculate the concentration of IO3– in the solution, we need to use the concept of solubility product constant (Ksp) and the assumption that the concentration of Pb(IO3)2 is negligible compared to Pb(NO3)2.
The Ksp expression for Pb(IO3)2 is:
Ksp = [Pb2+][IO3–]^2
Since the concentration of Pb(IO3)2 is assumed to be negligible, the concentration of Pb2+ can be considered to come from Pb(NO3)2 entirely.
To determine the concentration of Pb2+ from Pb(NO3)2, we need to consider the dissociation of Pb(NO3)2:
Pb(NO3)2 → Pb2+ + 2NO3–
Since the concentration of Pb(NO3)2 is given as 4.73 mM, the concentration of Pb2+ is also 4.73 mM.
Now, let's use the Ksp expression to calculate the concentration of IO3–:
Ksp = [Pb2+][IO3–]^2
Substituting the known values:
2.5 × 10^-13 = (4.73 mM)([IO3–]^2)
Solving for [IO3–]:
[IO3–]^2 = (2.5 × 10^-13) / (4.73 mM)
[IO3–]^2 = 5.2919 × 10^-14
Taking the square root of both sides:
[IO3–] = √(5.2919 × 10^-14)
[IO3–] = 7.273 × 10^-8 M
So, the concentration of IO3– in the solution is approximately 7.273 × 10^-8 M.