Using a value of Ksp = 1.8 x 10-2 for the reaction PbCl2(s) = Pb+2(aq) + 2Cl-(aq).
If a solution has a current value of Keq of 1.2 x 10-2, which statement is true?
Finish the question.
Im not sure
Too much solid has dissolved, additional precipitate is forming, the solution is unsaturated, or the ions are now combining to reduce their concentrations.
To determine which statement is true, we need to compare the given equilibrium constant, Keq, to the solubility product constant, Ksp, for the reaction.
The equation for the reaction is PbCl2(s) ⇌ Pb+2(aq) + 2Cl-(aq)
The solubility product constant, Ksp, is the equilibrium constant for the dissociation of the solid salt into its constituent ions in a saturated solution at a specific temperature. It can be calculated by multiplying the concentrations of the ions raised to the power of their stoichiometric coefficients:
Ksp = [Pb+2][Cl-]^2
Given that Ksp = 1.8 x 10^-2, we know that the product of the concentrations of Pb+2 and Cl- squared is equal to this value.
The equilibrium constant, Keq, on the other hand, is the ratio of the equilibrium concentrations of products to reactants, each raised to the power of their stoichiometric coefficients. In this case, it is given that Keq = 1.2 x 10^-2.
To relate Keq and Ksp, we need to consider the stoichiometry of the reaction. According to the balanced chemical equation, the stoichiometric coefficients of Pb+2 and Cl- are 1 and 2, respectively. Therefore, the relationship between Keq and Ksp is:
Keq = [Pb+2][Cl-]^2
Comparing this equation with the Ksp expression, we can conclude that if Keq > Ksp, then the statement is true, and vice versa.
Since 1.2 x 10^-2 > 1.8 x 10^-2, the correct statement is: Keq is greater than Ksp.