Find the solubility of AgCl in 1.0 M aqueous NH3 (in mol L^-1). K_sp(AgCl) = 1.8 x 10^-10, K_f([Ag(NH3)2]+) = 1.6 x 10^7. Neglect any other complex.

JEE Main levelSingle correct optionDifficulty 2 of 3ia078

Find the solubility of $AgCl$ in $1.0 M$ aqueous $NH X_{3}$ (in $mol L^{- 1}$ ). $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ , $K_{f} ([Ag (NH X_{3}) X_{2}] X^{+}) = 1.6 \times 1 0^{7}$ . Neglect any other complex.

K = [Ag (NH X_{3}) X_{2} X^{+}] [Cl X^{-}] / [NH X_{3}]^{2} = 2.9 \times 1 0^{- 3}

. With

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

and

[NH X_{3}] = 1 - 2 s

s / (1 - 2 s) = K = 5.37 \times 1 0^{- 2}

, so

s = 5.37 \times 1 0^{- 2} / (1 + 2 \times 5.37 \times 1 0^{- 2}) = 4.8 \times 1 0^{- 2} M