The Ksp of PbI2 is 8.73 × 10-9. Find the solubility of PbI2 in M
Please guide me , the I C E table still confuses me, please help me!!
The ice table is the easiest way to do this. It's done this way.
...........PbI2 ==> Pb^2+ + 2I^-
I...........solid.........0..............0
C.........solid..........x...............2x
E..........solid..........x..............2x
Note: I is initially. Initially you have solid and before anything happens you have zero Pb ion and zero I ion. because initially nothing has dissolved.
Change. For every mole of PbI2 that dissociates, you get 1 mol Pb ion an 2 mols I ion.
Equilibrium line you add I line + C line to get E line.
Ksp = (Pb^2+)(I^-)^2
Plug the E line into Ksp expression and solve for x = (Pb^2+) = PbI2
To find the solubility of PbI2 in M using the given Ksp value, you can use an ICE table to determine the concentrations of Pb2+ and I- ions at equilibrium.
1. Start by writing the balanced equation for the dissolution of PbI2:
PbI2 (s) ⇌ Pb2+ (aq) + 2I- (aq)
2. Define the initial concentration (I), the change in concentration (C), and the equilibrium concentration (E) for each species:
For PbI2:
Initial concentration (I): 0 (since it is a solid)
Change in concentration (C): +x (the quantity that dissolves)
Equilibrium concentration (E): x
For Pb2+:
Initial concentration (I): 0
Change in concentration (C): +x
Equilibrium concentration (E): x
For I-:
Initial concentration (I): 0
Change in concentration (C): +2x (since 2 moles of I- are produced for every 1 mole of PbI2 that dissolves)
Equilibrium concentration (E): 2x
3. Write the expression for the solubility product constant (Ksp):
Ksp = [Pb2+][I-]^2
Since Pb2+ and I- both have a concentration of x at equilibrium, the expression becomes:
Ksp = x * (2x)^2
4. Substitute the given Ksp value into the equation and solve for x:
8.73 × 10^-9 = x * (2x)^2
8.73 × 10^-9 = 4x^3
5. Solve for x by taking the cube root of both sides:
x = (8.73 × 10^-9)^(1/3)
x ≈ 6.31 × 10^-3
Therefore, the solubility of PbI2 is approximately 6.31 × 10^-3 M.
Sure, I can help you with that! The solubility of PbI2 refers to the amount of PbI2 that can dissolve in water to form a saturated solution at a certain temperature.
To find the solubility of PbI2 in Molarity (M), we can use the concept of the Ksp (solubility product constant). The Ksp is an equilibrium constant that represents the equilibrium between solid PbI2 dissolving into its ions Pb2+ and 2I- in a saturated solution.
The balanced equation for the dissociation of PbI2 is:
PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)
The stoichiometry of the balanced equation tells us that when 1 molecule of PbI2 dissolves, it generates 1 mol Pb2+ and 2 mol I- ions.
Now, let's use an I C E (Initial, Change, Equilibrium) table to solve this problem step by step:
1. Start with the initial concentration (I) - This is the concentration of solute (PbI2) before any reaction has occurred. In this case, we assume that the solubility of PbI2 is "x" M.
PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)
Initial: x M 0 M 0 M
2. Determine the change in concentration (C) - This is the change in concentration of the species when the reaction reaches equilibrium. Since the solubility of PbI2 is "x" M, the change in concentration of Pb2+ and I- will also be "x" M.
PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)
Change: - x M + x M + 2x M
3. Find the equilibrium concentration (E) - This is the concentration of the species at equilibrium. The solubility product constant (Ksp) is defined as the product of the equilibrium concentrations of the ions raised to the power of their stoichiometric coefficients.
Ksp = [Pb2+][I-]^2 = (x)(2x)^2 = 4x^3
4. Set up the equation for Ksp and solve for x:
8.73 × 10^-9 = 4x^3
Now, you can solve this equation for x using algebraic methods such as cube root and rearranging the equation to isolate x. The resulting value of x will give you the solubility of PbI2 in Molarity (M).
I hope this helps clarify the process! Let me know if you have any further questions.