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.

thanks

is it 6.429x10^-9?

I agree.

To calculate the solubility-product constant (Ksp) for lead(II) iodide (PbI2) at 25°C, we need to use the given information about the solution's saturation level.

The balanced equation for the dissociation of PbI2 is:

PbI2 ⇌ Pb2+ + 2I-

Since the concentration of Pb2+ is equal to the concentration of I- due to the stoichiometry of the balanced equation, we can refer to the concentration of Pb2+ as x.

Using the molar mass of PbI2 (461 g/mol), we can convert the mass of PbI2 (0.54 g) into moles:

moles of PbI2 = (0.54 g) / (461 g/mol) = 0.001172 mol

Since the concentration of Pb2+ is equal to x, we can write the solubility-product expression for PbI2:

Ksp = [Pb2+][I-]^2

Given that the initial volume of the solution is 1.00 L, the concentration of Pb2+ can be expressed as:

[Pb2+] = x / 1.00 L = x M

Since the concentration of I- is twice that of Pb2+, it can be expressed as:

[I-] = 2x M

Substituting these expressions back into the solubility-product expression, we get:

Ksp = (x)(2x)^2 = 4x^3

We can substitute the known value of x into the equation. From the given information, we know that 0.54 g of PbI2 dissolved in 1.00 L solution. Therefore, the concentration of Pb2+ (and I-) can be calculated as follows:

[Pb2+] = (0.54 g) / (461 g/mol) / (1.00 L) = 0.001172 M

Substituting this concentration into the solubility-product expression, we get:

Ksp = 4(0.001172)^3 ≈ 7.75 x 10^(-9)

Therefore, the solubility-product constant (Ksp) for lead(II) iodide (PbI2) at 25°C is approximately 7.75 x 10^(-9).