A solution of PbI2 has [Pb2+] = 4.5 x 10-5 and [I-] = 6.5 x 10-4. PbI2 has Ksp = 8.7 x 10-9. Write down the reaction that is taking place. Calculate Q. Is Q larger or smaller than the Ksp? Is the solution unsaturated, saturated, or supersaturated? Will a precipitate form in this solution?
The reaction taking place is the dissociation of PbI2:
PbI2 ⇌ Pb2+ + 2I-
To calculate Q, we need to write the expression for Q, which is the reaction quotient. The expression for Q is similar to the expression for Ksp, but instead of using the equilibrium concentrations, Q uses the initial concentrations:
Q = [Pb2+][I-]^2
Given the initial concentrations [Pb2+] = 4.5 x 10^-5 and [I-] = 6.5 x 10^-4, we can substitute these values into the expression to calculate Q:
Q = (4.5 x 10^-5)(6.5 x 10^-4)^2
Q ≈ 1.819 x 10^-12
To determine if Q is larger or smaller than Ksp, we compare their values. Given that Ksp = 8.7 x 10^-9 and Q = 1.819 x 10^-12, we see that Q is smaller than Ksp.
If Q < Ksp, it means that the reaction has not yet reached equilibrium and the solution is unsaturated. This means there is still the potential for more PbI2 to dissolve, and no precipitate will form.
In summary:
- Q is smaller than Ksp.
- The solution is unsaturated.
- No precipitate will form in this solution.