QNEET-2024-Ch-113: NEET 2024 Chemistry - Class 11 NEET-2024-Ch-113 Solution: The ratio of solubility of AgCl in 0.1 M KCl solution to the…

To find the ratio of solubility of AgCl in 0.1 M KCl solution to the solubility of AgCl in water, we first need to understand how common ion effect influences solubility. Let's denote the solubility product constant of AgCl as K s p . Given K s p of AgCl = 10 _ 10 . First, we calculate the solubility of AgCl in pure water: In water, the dissociation of AgCl can be represented as: AgCl (s) _ Ag + (aq) + Cl _ (aq) If s is the solubility of AgCl in water, then [ A g + ] = s [ C l _ ] = s Hence, the solubility product K s p can be written as: K s p = [ A g + ] [ C l _ ] K s p = s _ s = s 2 So, s 2 = 10 _ 10 s = 10 _ 5 Now, let's consider the solubility of AgCl in 0.1 M KCl solution. Because of the common ion effect, the presence of C l _ ions from KCl will suppress the solubility of AgCl. Here, [ C l _ ] from KCl is 0.1 M. Let the new solubility of AgCl in this solution be s _ . Then, [ A g + ] = s _ [ C l _ ] = 0.1 + s _ � 0.1 Since s _ is much smaller than 0.1 M, we can approximate: K s p = [ A g + ] [ C l _ ] 10 _ 10 = s _ _ 0.1 s _ = 10 _ 10 0.1 s _ = 10 _ 9 Finally, we find the ratio of solubility in 0.1 M KCl to that in pure water: Ratio = s _ s = 10 _ 9 10 _ 5 = 10 _ 4 Therefore, the correct answer is: Option A 10 _ 4