Final answer:
To determine the equilibrium concentration of Ag+ ions, calculate the initial moles of Ag+ and IO3- ions, take into account the total volume after mixing, utilize the Ksp for AgIO3, and consider the excess ions. The resultant equilibrium concentration of Ag+ is found to be 6.0 × 10−6 M.
Step-by-step explanation:
To calculate the equilibrium concentration of Ag+ ions after mixing a 60.0-mL sample of 0.00200 M AgNO3 with 60.0 mL of 0.0100 M NaIO3, we need to use the Ksp value for AgIO3.
First, we find the initial moles of Ag+ and IO3− in the solution:
- Ag+: 0.00200 M * 0.060 L = 0.00012 moles
- IO3−: 0.0100 M * 0.060 L = 0.0006 moles
When the two solutions are mixed, the total volume is 120 mL (0.120 L), and the reaction to form AgIO3 will occur:
Ag+(aq) + IO3−(aq) → AgIO3(s)
Since the AgIO3 will precipitate out of the solution until the maximum amount of solid is formed, we can assume the resulting equilibrium concentrations of Ag+ and IO3− will be determined by the Ksp value. The Ksp expression is:
Ksp = [Ag+][IO3−] = 3.0 × 10−8
Because we have excess IO3− ions, we can assume that the final concentration of Ag+ at equilibrium will be approximately equal to the Ksp divided by the initial concentration of IO3−, which after dilution is (0.0006 moles / 0.120 L = 0.00500 M):
[Ag+] = Ksp / [IO3−] = (3.0 × 10−8) / 0.00500 M = 6.0 × 10−6 M
The equilibrium concentration of Ag+ in the solution is 6.0 × 10−6 M.