Question

65. If the solubility of AgCl (formula mass=143) in water
at 25°C is 1.43 x104 g/100 mL of solution then
the value of K, will be :-
(1) 1 x 10-5
(2) 2 x 10-5
(3) 1 x 10-10
(4) 2 x 10-10​

Answer 1

The value of k is equal to 1×10^-10 M²

• It is given that 1.43×10^-4 g of AgCl is dissolves per 100 ml of solution.
• So 1 L of the solution will dissolve 14.3 × 10^-4 g of AgCl (i.e. 10^-5 moles)
• Molar solubility(S) of AgCl is 10^-5 M.Since molarity is the number of moles present per litre of the solution.
• k=S² for AgCl since it dissociates Ag+ and Cl- ions with concentration equal to S on dissolution.
• k=(10^-5)²=10^-10.