Find the solubility of AgCl in a solution containing 0.10 M NaCl and 1.0 M NH3 (assume no volume change and ignore [AgCl2]-). K_sp(AgCl) = 1.8 x 10^-10, K_f([Ag(NH3)2]+) = 1.6 x 10^7.

JEE Advanced levelSingle correct optionDifficulty 3 of 3ia085

Find the solubility of $AgCl$ in a solution containing $0.10 M$ $NaCl$ and $1.0 M$ $NH X_{3}$ (assume no volume change and ignore $[AgCl X_{2}] X^{-}$ ). $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ , $K_{f} ([Ag (NH X_{3}) X_{2}] X^{+}) = 1.6 \times 1 0^{7}$ .

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

with

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

[Cl X^{-}] = 0.10 + s

[NH X_{3}] = 1.0 - 2 s

. So

s (0.10 + s) / (1 - 2 s)^{2} = 2.9 \times 1 0^{- 3}

. Solving this quadratic gives

s = 2.17 \times 1 0^{- 2} M

(the chloride from

NaCl

suppresses the solubility compared with

4.85 \times 1 0^{- 2} M

1.0 M

NH X_{3}

alone).