What is the solubility of lead bromide at 25C? (Ksp = 4.6 x E^-6)
You have used another name here but work the problem the same way I showed you for AgCl. The only difference is that
PbBr2(s) ==> Pb^+2 + 2Br^-
Post your work if you get stuck.
382.5 x 10 -24
To determine the solubility of lead bromide (PbBr2) at 25°C, we need to use the solubility product constant (Ksp) and the stoichiometry of the dissociation equation.
The balanced dissociation equation for lead bromide is:
PbBr2(s) ⇌ Pb^2+(aq) + 2Br^-(aq)
The Ksp expression for lead bromide is:
Ksp = [Pb^2+][Br^-]^2
We are given the value of Ksp, which is 4.6 x 10^(-6).
To find the solubility of lead bromide (x), we need to make an assumption that the solubility is much less than the initial concentration of PbBr2. This allows us to approximate the change in concentration of Pb^2+ and Br^- as x, neglecting the change in concentration of PbBr2.
Since the stoichiometry of the dissociation equation shows that 1 PbBr2 produces 1 Pb^2+ and 2 Br^-, the concentrations of Pb^2+ and Br^- can be written as x and 2x, respectively.
Substituting these values into the Ksp expression:
Ksp = x * (2x)^2
4.6 x 10^(-6) = 4x^3
Rearranging the equation:
x^3 = (4.6 x 10^(-6)) / 4
x^3 = 1.15 x 10^(-6)
Now we can solve for x by taking the cube root of both sides:
x = (1.15 x 10^(-6))^(1/3)
x ≈ 0.0103
Therefore, the solubility of lead bromide at 25°C is approximately 0.0103 M.