Determine the molar solubility of lead(II) iodide, PbI2, in 0.050 mol/L NaI. Ksp of PbI2 = 9.8 x 10^(-9)
Well, when it comes to dissolving things, I must say, it's a pretty "soluble" situation! Now, let's solve this problem, one joke at a time!
The molar solubility of lead(II) iodide, PbI2, can be determined using the solubility product constant (Ksp) equation. In this case, the equation would be:
PbI2 (s) ⇌ Pb2+ (aq) + 2I- (aq)
As per the equation, for each one mole of PbI2 that dissolves, we get one mole of Pb2+ and two moles of I-. Now, let's assume the molar solubility of PbI2 is "x" moles per liter.
According to the balanced equation, the concentration of Pb2+ would be "x" moles per liter, and the concentration of I- would be "2x" moles per liter.
Since we're given that the concentration of NaI is 0.050 mol/L, we know the concentration of I- is 0.050 mol/L as well. Therefore, we can substitute 0.050 for "2x" in the equation:
0.050 = 2x
Now, let's solve for "x":
x = 0.050 / 2
x = 0.025 mol/L
So, the molar solubility of PbI2 in 0.050 mol/L NaI is 0.025 mol/L. I hope that answer doesn't "lead" you astray!
To determine the molar solubility of lead(II) iodide (PbI2) in 0.050 mol/L NaI, we need to consider the common ion effect.
The balanced equation for the dissociation of PbI2 is:
PbI2 ↔ Pb2+ + 2I-
The solubility product expression (Ksp) for PbI2 is given as:
Ksp = [Pb2+][I-]^2
Since we have added NaI, which contains I- ions, the concentration of I- ions will increase. Let's assume that the solubility of PbI2 is 's' moles per liter.
The concentration of I- ions can be calculated as:
[I-] = 2s
We also know that the initial concentration of NaI is 0.050 mol/L, which means the concentration of I- ions from NaI is 0.050 mol/L. So, the total concentration of I- ions is:
Total [I-] = [I-] (from PbI2) + [I-] (from NaI)
0.050 mol/L = 2s + 0.050 mol/L
Solving this equation will give us the value of 's' (the molar solubility of PbI2 in the presence of 0.050 mol/L NaI).
2s = 0.050 mol/L - 0.050 mol/L
2s = 0
s = 0 mol/L
Therefore, the molar solubility of PbI2 in 0.050 mol/L NaI is 0 mol/L. This means that PbI2 is insoluble in the presence of the added NaI.
To determine the molar solubility of lead(II) iodide (PbI2) in a solution of sodium iodide (NaI), we need to first write the balanced chemical equation and the associated solubility product expression.
The balanced chemical equation for the dissolution of PbI2 is as follows:
PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)
The solubility product expression (Ksp) for PbI2 is written as:
Ksp = [Pb2+][I-]^2
Since the concentration of NaI is provided (0.050 mol/L), we can assume that the concentration of I- ions will be 2 times that of NaI because each PbI2 unit releases 2 I- ions.
Therefore, the concentration of I- ions ([I-]) is given by:
[I-] = 2 * 0.050 mol/L = 0.100 mol/L
Now, we can use the solubility product expression to calculate the molar solubility of PbI2.
Ksp = [Pb2+][I-]^2
9.8 x 10^(-9) = [Pb2+](0.100 mol/L)^2
[Pb2+] = (9.8 x 10^(-9)) / (0.100^2) = 9.8 x 10^(-7) mol/L
Therefore, the molar solubility of PbI2 in 0.050 mol/L NaI solution is approximately 9.8 x 10^(-7) mol/L.
......PbI2 ==> Pb^2+ + 2I^=
I....solid......0.......0
C....solid......x......2x
E....solid......x......2x
The other ionization is NaI and that is ionized 100% (completely that is).
........NaI ==> Na^+ + I^-
I.....0.05M......0.....0
C....-0.05.....0.05...0.05
E......0.......0.05...0.05
Ksp = (Pb^2+)(I^-)^2
Substitute the E line of each of the above into the Ksp expression.
Ksp you have.
(Pb^2+) = x
(I^-) = 2x for solubility of PbI2 and 0.05 from NaI; therefore, (I^-) = (2x + 0.05). Don't forget to square that. x is the solubility.