1. What is the value of Ksp if a saturated solution of PbCO3 is 2.72x10-7M?
a. 5.2x10-4
b. 7.4x10-14 ***
c. 1.5x10-33
d. 2.7x10-7
(2.72 * 10^-7)(2.72 * 10^-7)= 7.4 * 10^-14
Looks good to me
To find the value of Ksp (solubility product constant) for a saturated solution of PbCO3, you need to use the equation:
Ksp = [Pb2+][CO32-]
Given that the concentration of the saturated solution of PbCO3 is 2.72x10^-7 M, we assume that the concentration of Pb2+ and CO32- ions in the solution is also 2.72x10^-7 M.
Substituting the values into the equation, we have:
Ksp = (2.72x10^-7)^2 = 7.4x10^-14
Therefore, the value of Ksp for the saturated solution of PbCO3 is 7.4x10^-14.
Hence, the correct answer is b. 7.4x10^-14.
To find the value of the solubility product constant (Ksp) for a saturated solution of PbCO3, you need to use the given concentration of the solute (PbCO3) and apply the stoichiometry of the balanced equation for the dissolution reaction.
The solubility product constant (Ksp) is the equilibrium constant for the dissociation of an ionic compound into its constituent ions in a saturated solution at a given temperature.
The balanced equation for the dissociation of PbCO3 is:
PbCO3(s) ⇌ Pb2+(aq) + CO32-(aq)
In this equation, PbCO3 dissociates to form one Pb2+ ion and one CO32- ion.
Given that the concentration of the saturated solution of PbCO3 is 2.72x10^-7 M, this concentration represents the concentration of both Pb2+ and CO32- ions in the solution.
Since the stoichiometry of the dissociation reaction is 1:1, the concentration of both Pb2+ and CO32- ions will also be 2.72x10^-7 M.
To calculate the Ksp value, you need to square the concentration of either Pb2+ or CO32- ions (since the stoichiometry is 1:1). In this case, you can square the concentration of Pb2+ or CO32-, both will give you the same Ksp value since they have the same concentration.
(2.72x10^-7) * (2.72x10^-7) = 7.4x10^-14
Therefore, the value of Ksp for the saturated solution of PbCO3 is 7.4x10^-14. Therefore, option b (7.4x10^-14) is the correct choice.