The molar solubility of PbBr2 at 250C is 1.0x10^-2. Calculate the Ksp.
To calculate the solubility product constant (Ksp) for a compound, we need to know the molar solubility and the balanced chemical equation representing the dissolution of the compound.
The balanced chemical equation for the dissolution of lead(II) bromide (PbBr2) is:
PbBr2(s) ⇌ Pb2+(aq) + 2Br-(aq)
According to the equation, one mole of PbBr2 dissociates to produce one mole of Pb2+ ions and two moles of Br- ions.
Given that the molar solubility of PbBr2 is 1.0x10^-2 M, we know that for every mole of PbBr2 that dissolves, we will get one mole of Pb2+ ions in solution. Therefore, the concentration of Pb2+ ions is also 1.0x10^-2 M.
Now, we can write the expression for Ksp using the concentrations of the ions:
Ksp = [Pb2+][Br-]^2
Since the concentration of Pb2+ is 1.0x10^-2 M and the concentration of Br- is also 1.0x10^-2 M (twice the concentration due to the stoichiometry of the balanced equation), we can substitute these values into the Ksp expression:
Ksp = (1.0x10^-2)(1.0x10^-2)^2
= 1.0x10^-6
Therefore, the solubility product constant (Ksp) for PbBr2 at 25°C is 1.0x10^-6.