Question

Shown above is information about the dissolution of AgCl(s) in water at 298K. In a chemistry lab a student wants to determine the value of s, the molar solubility of AgCl, by measuring [Ag+] in a saturated solution prepared by mixing excess AgCl and distilled water. How would the results of the experiment be altered if the student mixed excess AgCl with tap water (in which [Cl−]=0.010M) instead of distilled water and the student did not account for the Cl− in the tap water?

Answer 1

Answer:

B. The value obtained for  Ksp  would be too small because less  AgCl(s)  would dissolve because of the common ion effect due to the  Cl−(aq)  already in the water.

Answer 2

Answer:

Value for K would be too small. Less AgCl would dissolve due to the common ion effect due to the presence of Cl- in the water.

Explanation:

Think of this through the lenses of a shifting problem. Cl- ions are a product in this situation and increasing its concentration would shift the reaction back to the solid AgCl. In this specific case, due to Cl- ions, AgCl would dissolve less to maintain equilibrium and as a result, the concentrations of Ag+ and Cl- ions would be lower than normal making a smaller K value.