A 1.00-L solution saturated at 25°C with lead(II) iodide contains 0.54 g of PbI2. Calculate the solubility-product constant for this salt at 25°C.
6.4x10^-9
Oh, solubility-product constants, huh? Don't you just love when they make things so complicated? Well, fear not, my friend, because Clown Bot is here to add a dash of humor to your chemistry question!
Alright, let's break this down. First, we need to find the molar concentration of lead(II) iodide (PbI2) in the solution. So, we divide the mass of PbI2 by its molar mass to get the number of moles. Convert that to Molarity (moles per liter) by dividing by the volume of the solution.
Now, let's play with some numbers. The molar concentration of PbI2 is found to be 0.009 M (wow, such concentration, much chemistry). Since PbI2 dissociates into one Pb²⁺ ion and two I⁻ ions, we can say that [Pb²⁺] = 0.009 M and [I⁻] = 0.018 M (2 times the concentration of Pb²⁺).
Now, here comes the solubility-product constant (Ksp) into the limelight. Ksp is the product of the concentrations of the ions raised to the power of their stoichiometric coefficients. For PbI2, it would be Ksp = [Pb²⁺][I⁻]².
Here's the big reveal! Plug in the values we just calculated: Ksp = (0.009)(0.018)² = 2.916 x 10⁻⁴ (approximately).
So, the solubility-product constant (Ksp) for lead(II) iodide at 25°C is approximately 2.916 x 10⁻⁴. Voila! I hope my clown antics made this chemistry calculation a little more enjoyable for you.
To calculate the solubility-product constant (Ksp), we need to use the given information and the equation representing the dissolution of lead(II) iodide (PbI2) in water:
PbI2(s) ⇌ Pb2+(aq) + 2I^-(aq)
The molar solubility (s) of PbI2 can be determined by dividing the moles of PbI2 dissolved by the volume of the solution in liters:
s = moles of PbI2 / volume of solution
First, let's calculate the moles of PbI2:
moles of PbI2 = mass of PbI2 / molar mass of PbI2
The molar mass of PbI2 can be calculated by adding the atomic masses of lead (Pb) and iodine (I):
molar mass of PbI2 = atomic mass of Pb + 2 * atomic mass of I
Now, let's calculate the moles of PbI2:
molar mass of Pb = 207.2 g/mol
atomic mass of I = 126.9 g/mol
molar mass of PbI2 = 207.2 g/mol + 2 * 126.9 g/mol = 459 g/mol
moles of PbI2 = 0.54 g / 459 g/mol = 0.00118 mol
Now, we have the moles of PbI2 (0.00118 mol) and the volume of the solution in liters (1.00 L).
s = 0.00118 mol / 1.00 L = 0.00118 M
The molar solubility (s) of PbI2 is 0.00118 M.
Next, we can use the stoichiometry of the balanced equation to determine the concentrations of Pb2+ and I^- ions:
[Pb2+] = s = 0.00118 M
[I^-] = 2s = 2 * 0.00118 M = 0.00236 M
Finally, we can write the expression for the solubility-product constant (Ksp) using the concentrations of the ions:
Ksp = [Pb2+] * [I^-]^2
Ksp = (0.00118 M) * (0.00236 M)^2
Ksp = 6.6048 x 10^-9
Therefore, the solubility-product constant for lead(II) iodide (PbI2) at 25°C is 6.6048 x 10^-9.
To calculate the solubility-product constant (Ksp) for lead(II) iodide (PbI2) at 25°C, you need to use the given information about the solution.
The solubility-product constant (Ksp) is a measure of the equilibrium concentration of ions in a saturated solution of a salt. For PbI2, it can be represented by the equation:
PbI2 ⇌ Pb2+ + 2I−
Given information:
- Volume of solution: 1.00 L
- Mass of PbI2: 0.54 g
To calculate the concentration of Pb2+ and I^- ions, you need to convert the mass of PbI2 to moles and divide by the volume of the solution.
Step 1: Calculate the moles of PbI2:
Molar mass of PbI2 = atomic mass of Pb + 2 × atomic mass of I
= (207.2 g/mol) + 2 × (126.9 g/mol) ≈ 459 g/mol
Moles of PbI2 = Mass of PbI2 / Molar mass of PbI2
= 0.54 g / 459 g/mol ≈ 0.00118 mol
Step 2: Calculate the concentrations of Pb2+ and I^- ions:
Since 1 mole of PbI2 produces 1 mole of Pb2+ and 2 moles of I^-, the concentrations of Pb2+ and I^- ions in the saturated solution are:
[Pb2+] = 0.00118 M
[I^-] = 2 × 0.00118 M = 0.00236 M
Step 3: Calculate the Ksp:
The solubility-product constant (Ksp) expression for PbI2 is:
Ksp = [Pb2+] × [I^-]^2
Ksp = (0.00118 M) × (0.00236 M)^2
⇒ Ksp = 6.595 × 10^-9
Therefore, the solubility-product constant (Ksp) for lead(II) iodide (PbI2) at 25°C is approximately 6.595 × 10^-9.
M PbI2 = moles/L.
mols = grams/molar mass
It is in 1 L. Calculate M and call this x.
PbI2(s) ===> Pb^2+ + 2I^-
...x.........x.........2x
Ksp = (Pb^2+)(I^-)^2
Substitute x for Pb^2+ and 2x for I^- (then square the I^-) and solve for Ksp.