Find the solubility of AgCl in a solution that is 0.50 M in NH3 (before dissolving) and also contains 0.10 M NaCl. K_sp = 1.8x 10^-10, K_f([Ag(NH3)2]+) = 1.6x 10^7. Dissolved silver is almost entirely the complex; account for the Cl- released and the NH3 consumed.

JEE Advanced levelSingle correct optionDifficulty 3 of 3ia159

Find the solubility of AgCl in a solution that is $0.50 M$ in $NH X_{3}$ (before dissolving) and also contains $0.10 M$ $NaCl$ . $K_{s p} = 1.8 \times 1 0^{- 10}$ , $K_{f} ([Ag (NH X_{3}) X_{2}] X^{+}) = 1.6 \times 1 0^{7}$ . Dissolved silver is almost entirely the complex; account for the $Cl X^{-}$ released and the $NH X_{3}$ consumed.

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

[Ag X^{+}] = K_{s p} / (0.10 + s)

[NH X_{3}] = 0.50 - 2 s

, and

s = K_{f} [Ag X^{+}] [NH X_{3}]^{2}

gives

s (0.10 + s) = K_{f} K_{s p} (0.50 - 2 s)^{2}

. Solving the quadratic-type equation:

s = 6.42 \times 1 0^{- 3} M

. Neglecting the extra chloride gives

2.4 \times 1 0^{- 2} M

; neglecting

s

in both places gives

7.2 \times 1 0^{- 3} M

(too large).