Question One
Lead chloride dissolves in water according to
PbCl2(s)<-> Pb^2+ + 2Cl^- (aq)
The solubility in pure water has been measured to be 4.44g . L^-1 . Calculate the solubility product of lead chloride in pure water. Your answer should be in units of molarity raised to the appropriate exponent.
Question Two
Would you expect the solubility of PbCl2 in water to change dramatically if it were to be dissolved into a solution of 3.091x10^-4 M KCl?
i)Yes, the solubility of PbCl2 would decrease dramatically.
ii)Yes, the solubility of PbCl2 would increase dramatically.
iii)No, the solubility of PbCl2 would remain roughly unchanged.
..........PbCl2 ==> Pb^2+ + 2Cl^-
Ksp = (Pb^2+)(Cl^-)^2
(Pb^2+) = 4.44g/L, convert to mol Pb^2+ and substitute into Ksp. (Cl^-) will be twice that. Solve for Ksp.
The solubility will decrease dramatically BECAUSE of the common ion effect. You are adding Cl^- from KCl and the solubility equilibrium shown above will be shifted to the left.
Question One
1.7e-5
Question Two (There is a bug in the grader!!! this is the right but the grader mark it as incorrect!!!)
i)Yes, the solubility of PbCl2 would decrease dramatically.
Thank you
Had wrong for question two with I
The density of maple syrup is 1.33g/mL. a bottle of maple Syrup contains 740mL of syrup. What is the mass of maple syrup?
if you want that question answered, post it separately.
No, the solubility of PbCl2 would remain roughly unchanged.
he solubility of PbCl2 would remain roughly unchanged.
right
To solve both questions, we need some information about the solubility product constant, Ksp, which can be derived from the solubility of PbCl2 in pure water.
Question One:
To calculate the solubility product of lead chloride (PbCl2) in pure water, we need to use the given solubility value and stoichiometry.
Given:
Solubility of PbCl2 in pure water = 4.44 g/L
To convert this solubility value to moles per liter (Molarity), we need to know the molar mass of PbCl2.
The molar mass of PbCl2 = (atomic mass of Pb) + 2*(atomic mass of Cl)
= (207.2 g/mol) + 2*(35.45 g/mol)
= 207.2 g/mol + 70.9 g/mol
= 278.1 g/mol
Now, let's calculate the molarity of PbCl2 in pure water:
Molarity = (Solubility in grams/Liter) / (Molar mass of PbCl2)
= (4.44 g/L) / (278.1 g/mol)
= 0.01596 M
The solubility product constant (Ksp) expression for the given reaction is:
Ksp = [Pb^2+] * [Cl^-]^2
Since one mole of PbCl2 dissociates to form one mole of Pb^2+ and two moles of Cl^-, we can express the concentrations of Pb^2+ and Cl^- in terms of the molarity of PbCl2.
[Pb^2+] = 0.01596 M
[Cl^-] = 2 * 0.01596 M = 0.03192 M
Now, substitute these values into the Ksp expression:
Ksp = (0.01596 M) * (0.03192 M)^2
= 0.01596 M * 1.021344 M^2
= 0.01632 M^3
Therefore, the solubility product constant of lead chloride (PbCl2) in pure water is 0.01632 M^3.
Question Two:
To determine the effect of a KCl solution on the solubility of PbCl2, we need to consider the common ion effect.
The presence of a common ion (Cl^- in this case) from KCl can decrease the solubility of PbCl2 because of the principle of Le Chatelier.
Since KCl is a strong electrolyte, it completely dissociates into K+ and Cl^- ions in solution. If the concentration of Cl^- ions in the solution increases, according to Le Chatelier's principle, the solubility of PbCl2 will decrease to maintain equilibrium.
Therefore, we would expect the solubility of PbCl2 to decrease dramatically if it were to be dissolved in a solution of 3.091x10^-4 M KCl.
Hence, the correct answer is i) Yes, the solubility of PbCl2 would decrease dramatically.