What is the maximum mass(g) of KCL that can be added to 1.00L of a 0.0100 M lead(ii) chloride solution without causing any precipitation of PbCl2? Assume the additional KCl does not affect the volume of the solution. For, PbCl2, Ksp=1.6x10^-5
Personally I'm at a loss here. I cant figure out where to start since Cl is in both compounds. I imagine the K ends up on it's own and that it doesn't really effect the equation but how can the Qsp of PbCl2 be raised by adding more Cl? Is there an excess of Pb that I am not aware of? Or will it force more Cl to bind with the Pb? I might be over thinking this.....
I think you are trying to make this too hard. To start the 0.01 PbCl2 is in solution.
......PbCl2 ==> Pb^2+ + 2Cl^-
I.....0.01......0.01....0.02
You can show this is in solution.
Qsp = (0.01)(0.02)^2 = 4E-6 which is < Ksp so it hasn't pptd and all is in solution.
What must (Cl^-) be to cause pptn?
Ksp = (Pb^2+)(Cl^-)^2
1.6E-5 = (0.01)(Cl^-)^2
(Cl^-) = 0.04 but you should confirm that. You already have 0.02 there. So you must add 0.02 M KCl. Convert that to grams in the 1L. Right?
Well, well, well, looks like we’ve stumbled upon a chemistry question! Let me clown around a bit and try to help you out.
So, we have a lead(ii) chloride solution, PbCl2, and we want to find out the maximum mass of KCl we can add without causing any precipitation. It's like playing a balancing act with chemicals!
You're right, both PbCl2 and KCl contain Cl, so let's focus on the lead(ii) chloride equation: PbCl2 ⇌ Pb2+ + 2Cl-
The equilibrium expression for the solubility product (Ksp) can be written as: Ksp = [Pb2+][Cl-]^2
Now, we need to figure out the maximum concentration of Cl- ions we can add before exceeding the solubility product.
Since the Ksp of PbCl2 is given as 1.6x10^-5, we know that [Pb2+][Cl-]^2 should never exceed this value.
Considering that the initial concentration of Cl- is from the lead(ii) chloride solution, which is 0.0100 M, we can calculate the max [Cl-] as sqrt(Ksp/[Pb2+]). Plug in the values and do the math!
Once you have the maximum concentration of Cl-, you can calculate the moles of Cl- ions in the solution. Remember, moles = concentration x volume.
Finally, multiply the moles of Cl- by the molar mass of KCl to get the maximum mass of KCl that can be added.
Phew, chemistry can be a real clown sometimes! I hope this explains it for you.
To determine the maximum mass of KCl that can be added without causing any precipitation of PbCl2, we need to compare the value of Qsp (the reaction quotient) to the solubility product constant, Ksp.
The solubility product constant for PbCl2 is given as Ksp = 1.6x10^-5. The formula for the dissolving of PbCl2 in water is:
PbCl2(s) ⇌ Pb2+(aq) + 2Cl^-(aq)
From this equation, we can see that the concentration of Pb2+ ions in solution is equal to the molar solubility of PbCl2, and the concentration of Cl^- ions is twice the molar solubility.
Considering the given concentration of the lead(ii) chloride solution as 0.0100 M, we can calculate the initial concentrations of Pb2+ and Cl^- ions. Since one mole of PbCl2 gives one mole of Pb2+, the concentration of Pb2+ ions is also 0.0100 M.
The concentration of Cl^- ions is twice the molar solubility of PbCl2, denoted as s, which is unknown at the moment. Therefore, the concentration of Cl^- ions is 2s.
Now, we can calculate the value of Qsp as:
Qsp = [Pb2+][Cl^-]^2 = (0.0100)(2s)^2 = 4s^2
To prevent any precipitation of PbCl2, the value of Qsp should be less than or equal to the value of Ksp.
Therefore, we have the inequality:
4s^2 ≤ Ksp
Plugging in the given value for Ksp (1.6x10^-5), we can solve for s.
4s^2 ≤ 1.6x10^-5
Simplifying:
s^2 ≤ (1.6x10^-5)/4
s^2 ≤ 4x10^-6
Taking the square root of both sides, we get:
s ≤ √(4x10^-6)
s ≤ 2x10^-3
Since the concentration of Cl^- ions is twice the molar solubility of PbCl2, the molar solubility of PbCl2 (s) is equal to the maximum concentration of Cl^- ions that can be added without precipitation.
Now, let's calculate the maximum mass of KCl that can be added without causing precipitation of PbCl2.
1. Convert the volume of the solution from liters to milliliters:
1.00 L = 1000 mL
2. Calculate the amount of PbCl2 that is already present in the lead(ii) chloride solution:
Amount of PbCl2 = (0.0100 M PbCl2) x (1000 mL) = 10.0 millimoles
3. The molar mass of PbCl2 = 207.2 g/mol + (2 x 35.45 g/mol) = 278.1 g/mol
4. Calculate the maximum amount of PbCl2 that can potentially precipitate:
Maximum amount of PbCl2 = (molar solubility of PbCl2) x (1000 mL) = (2x10^-3 M) x (1000 mL) = 2 millimoles
5. Calculate the maximum mass of PbCl2 that can potentially precipitate:
Maximum mass of PbCl2 = (maximum amount of PbCl2) x (molar mass of PbCl2) = (2 millimoles) x (278.1 g/mol) = 556.2 milligrams
Therefore, the maximum mass of KCl that can be added without causing any precipitation of PbCl2 is approximately 556.2 milligrams.
To determine the maximum mass of KCl that can be added without causing any precipitation of PbCl2, we need to consider the solubility product constant (Ksp) and the common ion effect.
The solubility product constant (Ksp) is a measure of the extent to which a sparingly soluble salt dissolves in water. For the compound PbCl2, the Ksp value is given as 1.6x10^-5. This means that when PbCl2 dissolves in water, it will dissociate to some extent and form lead ions (Pb2+) and chloride ions (Cl-). The Ksp expression for PbCl2 can be written as [Pb2+][Cl-]^2 = 1.6x10^-5.
Now, let's consider the addition of KCl to the lead(ii) chloride solution. When KCl dissolves in water, it will dissociate to form potassium ions (K+) and chloride ions (Cl-). The chloride ions from KCl will compete with the chloride ions from PbCl2 for binding with lead ions (Pb2+). As more chloride ions are added from the KCl, the concentration of chloride ions (Cl-) will increase, which can potentially exceed the solubility product constant (Ksp) for PbCl2 and cause precipitation.
To determine the maximum mass of KCl that can be added without causing precipitation, we need to calculate the concentration of chloride ions (Cl-) in the solution after adding KCl. This can be done using the formula:
Cl- concentration = initial concentration of Cl- + concentration of Cl- from KCl
Since we are assuming that the volume of the solution remains constant, we should use the equation:
initial concentration of Cl- = concentration of Cl- from the PbCl2 solution = 2 * (concentration of PbCl2)
In this case, the initial concentration of Cl- would be 2 times the concentration of PbCl2, because the stoichiometric ratio between PbCl2 and the chloride ions (Cl-) is 1:2.
Once we have the initial concentration of Cl-, we can determine the maximum concentration of chloride ions (Cl-) that can be added from the KCl solution without causing precipitation. From there, we can calculate the maximum mass of KCl using its molar mass.
To summarize the steps:
1. Calculate the initial concentration of Cl- based on the given concentration of PbCl2.
2. Determine the maximum concentration of Cl- that can be added from the KCl without exceeding the solubility product constant (Ksp) for PbCl2.
3. Convert the maximum concentration of Cl- in moles to grams using the molar mass of KCl to find the maximum mass of KCl that can be added.
I hope this explanation helps you understand how to approach and solve the problem.