In a group 1 analysis, a student adds hydrochloric acid to the unknown solution to make [Cl-] = 0.15 M. Some PbCl2 precipitates. Calculate the concentration of Pb2+ remaining in solutions.
To calculate the concentration of Pb2+ remaining in solution, we need to use the solubility product constant (Ksp) for PbCl2 and the stoichiometry of the reaction.
The balanced chemical equation for the dissociation of PbCl2 in water is:
PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq)
The solubility product expression for this equilibrium is:
Ksp = [Pb2+][Cl-]^2
Given that the concentration of Cl- is 0.15 M, we can substitute this value into the Ksp expression:
Ksp = [Pb2+](0.15)^2
Now we need to solve for [Pb2+]. Rearranging the equation and solving for [Pb2+], we get:
[Pb2+] = Ksp / (0.15)^2
To calculate [Pb2+], we also need the value of the Ksp for PbCl2. The Ksp for PbCl2 can vary depending on temperature, but at room temperature (25°C), it is approximately 1.17 x 10^-5.
Substituting this value into the equation, we have:
[Pb2+] = (1.17 x 10^-5) / (0.15)^2
Calculating this expression will give us the concentration of Pb2+ remaining in solution.
To calculate the concentration of Pb2+ remaining in solution, we need to use the chemical equation for the reaction between PbCl2 and HCl.
The balanced chemical equation is:
PbCl2 + 2HCl -> PbCl2 + 2Cl-
From the equation, we can see that for every 1 mole of PbCl2, 2 moles of HCl will react to form 2 moles of Cl-.
Given the concentration of Cl- in the solution is 0.15 M, and considering that the ratio between Cl- and PbCl2 is 2:1, the concentration of Pb2+ can be calculated as follows:
[Cl-] = [PbCl2]
0.15 M = [PbCl2]
Now, to convert the concentration of PbCl2 to the concentration of Pb2+, we use the formula
[Pb2+] = [PbCl2] / 2
Substituting the value we found for [PbCl2]:
[Pb2+] = 0.15 M / 2
Simplifying:
[Pb2+] = 0.075 M
Therefore, the concentration of Pb2+ remaining in solution is 0.075 M.
.............PbCl2 ==> Pb^2+ + 2Cl^-
I............solid......0......0.15
C............solid......x........x
E............solid......x......0.15+x
Ksp = (Pb2+)(Cl^-)^2
Ksp = (x)(x+0.15)
Solve for x.