Answer:
The molar solubility of lead bromide at 298K is 0.010 mol/L.
Explanation:
In order to solve this problem, we need to use the Nernst Equaiton:
E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.
At equilibrium, E = 0, therefore:
[red] = 0.010 M
The reduction will happen in the anode, therefore, the concentration of the reduced specie is equivalent to the molar solubility of lead bromide.
this is known as condensation when cold and warm temperatures meet
1.40E+5 (140000)
I believe this is right