Calculate the solubility of AgCl in 0.10 M aqueous NH3. K_sp(AgCl) = 1.8x 10^-10 and K_f of [Ag(NH3)2]+ = 1.6x 10^7. The complex [Ag(NH3)2]+ is the only silver species in solution, and [NH3] is the free ammonia concentration after complexation, i.e. take [NH3] = 0.10 - 2s where s is the solubility (do not neglect 2s).

JEE Main levelSingle correct optionDifficulty 2 of 3ia151

Calculate the solubility of AgCl in $0.10 M$ aqueous $NH X_{3}$ . $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ and $K_{f}$ of $[Ag (NH X_{3}) X_{2}] X^{+} = 1.6 \times 1 0^{7}$ . The complex $[Ag (NH X_{3}) X_{2}] X^{+}$ is the only silver species in solution, and $[NH X_{3}]$ is the free ammonia concentration after complexation, i.e. take $[NH X_{3}] = 0.10 - 2 s$ where $s$ is the solubility (do not neglect $2 s$ ).

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

K = K_{s p} K_{f} = 2.9 \times 1 0^{- 3}

. With

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

and

[NH X_{3}] = 0.10 - 2 s

s / (0.10 - 2 s) = K = 0.0537

, so

s = 0.0537 \times 0.10/ (1 + 2 \times 0.0537) = 4.85 \times 1 0^{- 3} M