AgCl dissolves in ammonia: AgCl + 2NH3 [Ag(NH3)2]+ + Cl-. Take K_sp(AgCl) = 1.8x10^-10 and K_f([Ag(NH3)2]+) = 1.6x10^7. The equilibrium [NH3] needed to just dissolve 0.010 mol AgCl in 1 L is closest to

JEE Main levelSingle correct optionDifficulty 2 of 3qa209

$AgCl$ dissolves in ammonia: $AgCl + 2 NH X_{3} [Ag (NH X_{3}) X_{2}] X^{+} + Cl X^{-}$ . Take $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ and $K_{f} ([Ag (NH X_{3}) X_{2}] X^{+}) = 1.6 \times 1 0^{7}$ . The equilibrium $[NH X_{3}]$ needed to just dissolve 0.010 mol $AgCl$ in 1 L is closest to

The overall constant is

K = K_{s p} K_{f} = 2.88 \times 1 0^{- 3}

. With

[[Ag (NH X_{3}) X_{2}] X^{+}] = [Cl X^{-}] = 0.010

K = (0.010)^{2} / [NH X_{3}]^{2}

[NH X_{3}] = 0.010/ K = 0.19