The effect of increasing the concentration of NaCl in a solution on the solubility of a sparingly soluble salt, such as AgCl, can be explained using the common ion effect. The common ion effect states that the solubility of a sparingly soluble salt decreases in the presence of a common ion.AgCl is a sparingly soluble salt that dissociates into its ions as follows:AgCl s Ag aq + Cl aq The solubility product constant Ksp for AgCl is 1.8 x 10.Ksp = [Ag][Cl]When NaCl is added to the solution, it dissociates completely into its ions:NaCl s Na aq + Cl aq Since Cl is a common ion in both reactions, the increase in Cl concentration due to the addition of NaCl will shift the equilibrium of the AgCl dissociation reaction to the left, according to Le Chatelier's principle. This results in a decrease in the solubility of AgCl.To show this effect mathematically, let's assume that the initial concentration of Cl ions due to the addition of NaCl is [Cl]. Let 's' be the solubility of AgCl in the presence of NaCl. Then, the total concentration of Cl ions in the solution will be [Cl] + s. The concentration of Ag ions will be equal to 's'.Now, we can write the Ksp expression for AgCl in the presence of NaCl:Ksp = [Ag][Cl] = s [Cl] + s Since AgCl is sparingly soluble and Ksp is very small, we can assume that s is much smaller than [Cl]. Therefore, [Cl] + s [Cl].Now, the Ksp expression becomes:Ksp = s[Cl]Solving for s:s = Ksp / [Cl]As we can see from this equation, the solubility of AgCl s is inversely proportional to the concentration of Cl ions in the solution. Therefore, increasing the concentration of NaCl, and consequently the concentration of Cl ions, will result in a decrease in the solubility of AgCl.