AgCl(s) Ag+ (aq) + Cl (aq) KC= 1.8 x 10-10 Shown above is

Last updated: 7/9/2022

AgCl(s) Ag+ (aq) + Cl (aq) KC= 1.8 x 10-10 Shown above is information about the dissolution of AgCl(s) in water at 298 K. 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 [CI] = 0.010 M) instead of distilled water and the student did not account for the Cl in the tap water? The value obtained for K, would be too small because Cl(aq) ions would be attracted to the Ag+ ions in the AgCl crystals, thus preventing water molecules from reaching the crystals. The value obtained for K 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. The value obtained for K would be too large because more AgCl(s) would dissolve because of the attractions between Ag+ ions in the AgCl crystals and the Cl(aq) ions in the water. The results of the experiment would not be altered because 0.010 M is such a small concentration of Cl(aq) ions and thus has no effect on the dissolution of AgCl(s).