Final answer:
The molar solubility of AgCl in pure water at 25°C is calculated using the Ksp value of 1.77 × 10−10, resulting in a solubility of 1.33 × 10−5 M.
Step-by-step explanation:
Calculating Molar Solubility of AgCl in Pure Water
The question revolves around the solubility of AgCl in pure water at 25°C. To calculate this, we use the solubility product constant (Ksp) for AgCl, which is given as 1.77 × 10−10. When we let x represent the solubility of AgCl, it means at equilibrium, we will have [Ag+] = [Cl]− = x M. Substituting these values into the solubility product expression, we get Ksp = (x)(x) = x².
By solving for x, we find that x (the solubility of AgCl in pure water at 25°C) is 1.33 × 10−5 M. This calculation shows that AgCl has a low molar solubility in pure water.