The solubility product constant Ksp is an equilibrium constant that represents the maximum amount of a solute that can dissolve in a solution to form a saturated solution. For sodium chloride NaCl , the dissolution process can be represented by the following equation:NaCl s Na+ aq + Cl- aq The Ksp expression for this reaction is:Ksp = [Na+][Cl-]Given that the Ksp of NaCl is 36.8 at 25C, we can determine the solubility of NaCl in water. Let's assume that the solubility of NaCl is 's' mol/L. When NaCl dissolves, it dissociates into equal concentrations of Na+ and Cl- ions. Therefore, the concentrations of Na+ and Cl- in the saturated solution are both 's' mol/L. Now, we can write the Ksp expression in terms of 's':Ksp = s s = s^2Substituting the given Ksp value:36.8 = s^2To find the solubility 's', we take the square root of both sides:s = 36.8 6.07 mol/LSo, the solubility of NaCl in water at 25C is approximately 6.07 mol/L.Now, let's consider the effect of adding other ionic solutes to the solution. According to the common ion effect, the solubility of a sparingly soluble salt decreases in the presence of a common ion. Let's assume we add another soluble salt, MCl where M is a different cation , to the NaCl solution. MCl will also dissociate into M+ and Cl- ions. The presence of additional Cl- ions from MCl will shift the equilibrium of NaCl dissolution to the left, according to Le Chatelier's principle, resulting in a decrease in NaCl solubility.Let's assume that the concentration of Cl- ions from MCl is 'c' mol/L. Now, the concentration of Cl- ions in the solution is s + c mol/L. The Ksp expression for NaCl becomes:Ksp = [Na+][ s + c ]Since Ksp remains constant, we can write:36.8 = s s + c We know that the solubility of NaCl without the presence of MCl is 6.07 mol/L. Therefore, in the presence of MCl, the solubility of NaCl will be less than 6.07 mol/L. This demonstrates that the addition of other ionic solutes to the solution decreases the solubility of NaCl.