Answer :
Final answer:
After dissolving 0.400 g KCl in saturated AgCl, the molarity of Ag+ is 1.6×10-10M when equilibrium is reestablished.
Explanation:
The student is tasked with calculating the molarity of Ag+ ions after equilibrium is reestablished when 0.400 g of KCl is dissolved in a 2.00 L solution of saturated AgCl with a solubility product (Ksp) of 1.6×10-10.
Given that a saturated solution of AgCl has a chloride ion concentration of (1.0 + x) M, where x is negligibly small and can be approximated as [Ag+] = x M, the Ksp expression can be simplified to Ksp = [Ag+][Cl-] ~ x(1.0). By solving this equation, x, which represents the concentration of Ag+ ions, is equal to the Ksp value (1.6×10-10M).
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