A solution is 0.050 M in Cl- and contains 0.020 M total silver, all of it present as [Ag(NH3)2]+ for practical purposes. What minimum total concentration of NH3 (free plus bound) is needed to just prevent precipitation of AgCl? K_sp(AgCl) = 1.8 x 10^-10, K_f([Ag(NH3)2]+) = 1.6 x 10^7. Two decimals, in mol L^-1.

JEE Advanced levelNumerical answerDifficulty 3 of 3ia089

A solution is $0.050 M$ in $Cl X^{-}$ and contains $0.020 M$ total silver, all of it present as $[Ag (NH X_{3}) X_{2}] X^{+}$ for practical purposes. What minimum total concentration of $NH X_{3}$ (free plus bound) is needed to just prevent precipitation of $AgCl$ ? $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ , $K_{f} ([Ag (NH X_{3}) X_{2}] X^{+}) = 1.6 \times 1 0^{7}$ . Two decimals, in $mol L^{- 1}$ .

Numerical answer type, answer in mol/L, tolerance 0.01.

Largest allowed

[Ag X^{+}] = 1.8 \times 1 0^{- 10} /0.050 = 3.6 \times 1 0^{- 9} M

. From

K_{f} = [Ag (NH X_{3}) X_{2} X^{+}] / ([Ag X^{+}] [NH X_{3}]^{2})

[NH X_{3}]^{2} = 0.020/ (1.6 \times 1 0^{7} \times 3.6 \times 1 0^{- 9}) = 0.3472

, so free

[NH X_{3}] = 0.5893 M

. Add

2 \times 0.020

bound: total = 0.6293, i.e. 0.63.