AgCl dissolves in ammonia: AgCl(s) + 2NH3 [Ag(NH3)2]+ + Cl-. Given K_sp = 1.6x10^-10 and K_f([Ag(NH3)2]^+) = 1.6x10^7, and taking the equilibrium (free) [NH3] as 0.50 M, the solubility of AgCl is

JEE Main levelSingle correct optionDifficulty 2 of 3ia190

$AgCl$ dissolves in ammonia: $AgCl (s) + 2 NH X_{3} [Ag (NH X_{3}) X_{2}] X^{+} + Cl X^{-}$ . Given $K_{s p} = 1.6 \times 1 0^{- 10}$ and $K_{f} ([Ag (NH X_{3}) X_{2}]^{+}) = 1.6 \times 1 0^{7}$ , and taking the equilibrium (free) $[NH X_{3}]$ as $0.50 M$ , the solubility of $AgCl$ is

K = K_{s p} K_{f} = 2.56 \times 1 0^{- 3}

. With

[complex] = [Cl X^{-}] = s

s / [NH X_{3}] = K

, so

s = 0.0506 \times 0.50 = 0.0253 M