Final answer:
To find the Ksp for copper(I) chloride, convert the solubility from 3.91 mg/100.0 mL to molarity, get the molar concentration of each ion, and then use the formula Ksp = [Cu+][Cl-] to calculate it. The Ksp for copper(I) chloride is 1.54×10⁻·.
Step-by-step explanation:
To calculate the solubility product constant (Ksp) for copper(I) chloride, we must first convert the given solubility from mg per 100.0 mL to moles per liter (M). The molar mass of copper(I) chloride (CuCl) is roughly 99.45 g/mol. Thus, 3.91 mg is equivalent to 3.91×10⁻³ g, which translates to 3.91×10⁻³ g / 99.45 g/mol = 3.93×10⁻µ moles. This is for 100 mL, so for a liter (1000 mL), it would be 3.93×10⁻´ moles/L. Copper(I) chloride dissociates into one copper(I) ion (Cu⁺) and one chloride ion (Cl⁻). Therefore, the molar concentration of each ion in solution will be the same as the molarity of copper(I) chloride, which is 3.93×10⁻´ M.
The expression for the Ksp of copper(I) chloride is given by:
Ksp = [Cu⁺][Cl⁻]
Substitute the values we calculated:
Ksp = (3.93×10⁻´) × (3.93×10⁻´)
Ksp = 1.54×10⁻·
Therefore, the Ksp for copper(I) chloride is 1.54×10⁻·.