Final answer:
The Ksp of AgCl at 25°C is calculated by converting the given solubility from grams per liter to molarity, which is then squared as per the solubility product constant expression Ksp = [Ag+][Cl-]. The calculated Ksp for AgCl at this temperature is 1.77 × 10-10.
Step-by-step explanation:
The student's question is asking how to calculate the solubility product constant (Ksp) for AgCl at 25°C given the solubility in grams per liter. To find the Ksp, we first need to convert the solubility from grams per liter to moles per liter (molarity). Since AgCl dissociates into Ag+ and Cl- ions in solution, we can denote the molar solubility as x for both Ag+ and Cl-. The Ksp expression for AgCl is thus Ksp = [Ag+][Cl-], which equals to x², as the concentration of Ag+ and Cl- are the same in solution due to the 1:1 stoichiometry of their dissociation from AgCl.
Given the solubility of AgCl in pure water at 25°C as 1.79x10-3 g/L, and the molar mass of AgCl as approximately 143.32 g/mol, we can calculate the molar solubility (x). After finding the molar solubility, we can substitute it into the Ksp expression to find the solubility product constant. For AgCl, solubility product constant at 25°C is Ksp = 1.77 × 10-10.
Therefore, by dissolving 1.79x10-3 g of AgCl in one liter of pure water at 25°C, it will result in a molarity of 1.33 × 10-5 M for both Ag+ and Cl- ions. Squaring this value gives us the Ksp of AgCl, which is 1.77 × 10-10.
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