Lead iodide, PbI2, has a Ksp in water of 7.1e-9 M3 at room temperature.
Calculate the solubility of PbI2 in water. Express your answer in grams per liter.
What is the solubility of PbI2 in 0.1 M NaI (aq)? Express your answer in grams per liter.
This is simple Stoichiometry problem
Write down PbI2 -> Pb2+ + 2I-
Therefore
Ksp = [Pb2+][I-]^2
From the stoich you can derive that
Ksp = Cs(2Cs)^2 = 4Cs^3
Hence Cs = (Ksp/4) ^ 1/3
Then multiply the Molar Weight to get grams / Liter unit that you like
The second problem is common ion effect
First see is the added ion concentration more than or less than your Cs.
0.1 seems to be >>>> than Cs so common ion will happen
Ksp = Cs'(0.1)^2 = 0.01Cs'
Hence
Cs' = Ksp/0.01
Then just multiply by Molar weight again to convert to g/L units .
Note use the Molar Weight of PbI2 !!!!
To calculate the solubility of a compound, we need to use the concept of Ksp (solubility product constant). Ksp is determined using the equilibrium expression for dissolving a compound in water.
The solubility of PbI2 in water can be calculated using the Ksp value and the formula weight of PbI2.
1. Calculate the molar solubility of PbI2 in water:
Let's assume the solubility of PbI2 in water is "s" moles per liter.
The dissociation of PbI2 in water can be written as follows:
PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)
The equilibrium expression is:
Ksp = [Pb2+] [I-]^2
Since the stoichiometry of PbI2 is 1:2 (Pb:I), we can say that [Pb2+] = s, and [I-] = 2s.
Substituting these values into the Ksp expression, we have:
Ksp = s * (2s)^2 = 4s^3
Now we can solve for "s":
7.1e-9 M^3 = 4s^3
s^3 = 7.1e-9 M^3 / 4
s^3 = 1.775e-9 M^3
s ≈ 1.26e-3 M (rounded to 3 significant figures)
2. Convert the molar solubility to grams per liter:
To convert from moles per liter to grams per liter, we need to know the formula weight of PbI2, which can be calculated as follows:
Formula weight of PbI2 = atomic weight of Pb + 2 * atomic weight of I
= (207.2 g/mol) + 2 * (126.9 g/mol)
= 207.2 g/mol + 253.8 g/mol
= 461 g/mol
Now, we can convert the molar solubility to grams per liter:
Solubility = molar solubility * formula weight of PbI2
= 1.26e-3 M * 461 g/mol
= 0.580 g/L (rounded to 3 significant figures)
Therefore, the solubility of PbI2 in water is approximately 0.580 grams per liter.
To calculate the solubility of PbI2 in 0.1 M NaI (aq), we need to consider the common ion effect. The presence of 0.1 M NaI will introduce additional I- ions into the solution, which will reduce the solubility of PbI2. To calculate the solubility, we need to consider the solubility product expression and the new concentration of I- ions.
The concentration of I- ions in 0.1 M NaI is given as 0.1 M.
1. Calculate the new concentration of I- ions after considering the common ion effect:
The new concentration of I- ions can be calculated by adding the concentration of I- ions from NaI to the original concentration of I- ions from the dissociation of PbI2 in water.
[I-] = 2s + 0.1 M
2. Substitute the new concentration of I- ions into the solubility product expression:
Ksp = [Pb2+][I-]^2
7.1e-9 M^3 = s * ([2s + 0.1 M])^2
3. Solve the equation iteratively to find the value of "s":
Start with an initial guess for "s" (e.g., 1 mM or 0.001 M).
Substitute the value of "s" into the equation and solve for a new value of "s".
Repeat this process until the value of "s" converges to a consistent result.
Using this iterative approach, the solubility of PbI2 in 0.1 M NaI can be calculated.