If the maximum concentration of PbCl2 in water is 0.01M at 298K, it's ...
Calculation of Maximum Concentration of PbCl2 in 0.1 M NaCl
The solubility of PbCl2 in water is given as 0.01 M at 298 K. We need to find out the maximum concentration of PbCl2 in 0.1 M NaCl. To calculate this, we need to use the common ion effect.
What is Common Ion Effect?
Common ion effect is a phenomenon in which the solubility of a salt in a solution is reduced when a common ion is added to the solution. In this case, the common ion is Cl- ion, which is present in both PbCl2 and NaCl.
How to Calculate the Maximum Concentration of PbCl2?
To calculate the maximum concentration of PbCl2 in 0.1 M NaCl, we need to first calculate the concentration of Cl- ions in the solution. This can be done using the formula:
[Cl-] = [NaCl] + [PbCl2]
where [NaCl] is the concentration of NaCl and [PbCl2] is the concentration of PbCl2 in the solution.
Now, we know that the concentration of NaCl is 0.1 M. Let's assume that the concentration of PbCl2 in the solution is x M. Therefore, the concentration of Cl- ions in the solution would be:
[Cl-] = 0.1 + x
Now, we need to use the solubility product expression for PbCl2 to find out the maximum concentration of PbCl2 in the solution. The solubility product expression for PbCl2 is given as:
Ksp = [Pb2+][Cl-]^2
where Ksp is the solubility product constant for PbCl2, [Pb2+] is the concentration of Pb2+ ions in the solution and [Cl-] is the concentration of Cl- ions in the solution.
Substituting the values of Ksp and [Cl-] in the above equation, we get:
Ksp = [Pb2+](0.1 + x)^2
Now, we can solve this equation for x, which is the concentration of PbCl2 in the solution:
x = (sqrt(Ksp/[Pb2+])) - 0.1
Substituting the given values of Ksp and [Pb2+], we get:
x = (sqrt(1.17 * 10^-5/1)) - 0.1
x = 0.00395 M
Therefore, the maximum concentration of PbCl2 in 0.1 M NaCl is 0.00395 M or 3.95 * 10^-3 M.
Answer: A) 4*10^-3M