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JEE Main 2021 · Chemistry · Ionic Equilibrium · hard

Silver chloride dissolves in aqueous ammonia as AgCl+2NH3rightleftharpoons[Ag(NH3)2]^++Cl^-. Given Ksp(AgCl)=1.8×10^-10 and…

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Silver chloride dissolves in aqueous ammonia as AgCl+2NH3⇌[Ag(NH3)2]++Cl−\mathrm{AgCl}+2\mathrm{NH_3}\rightleftharpoons[\mathrm{Ag(NH_3)_2}]^++\mathrm{Cl^-}. Given Ksp(AgCl)=1.8×10−10K_{sp}(\mathrm{AgCl})=1.8\times10^{-10} and Kf([Ag(NH3)2]+)=1.6×107K_f([\mathrm{Ag(NH_3)_2}]^+)=1.6\times10^{7}, the solubility of AgCl\mathrm{AgCl} in 0.10 M0.10\ \mathrm{M} NH3\mathrm{NH_3} (taking the free NH3\mathrm{NH_3} concentration as 0.10−2s0.10-2s) is
  1. A. 5.4×10−3 M5.4\times10^{-3}\ \mathrm{M}
  2. B. 4.8×10−3 M4.8\times10^{-3}\ \mathrm{M}
  3. C. 1.3×10−5 M1.3\times10^{-5}\ \mathrm{M}
  4. D. 2.4×10−3 M2.4\times10^{-3}\ \mathrm{M}
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Answer: (B) 4.8×10−3 M4.8\times10^{-3}\ \mathrm{M}

The overall constant is K=KspKf=1.8×10−10×1.6×107=2.88×10−3K=K_{sp}K_f=1.8\times10^{-10}\times1.6\times10^7=2.88\times10^{-3}. With solubility ss: K=s⋅s(0.10−2s)2K=\dfrac{s\cdot s}{(0.10-2s)^2}, so s0.10−2s=K=0.0537\dfrac{s}{0.10-2s}=\sqrt{K}=0.0537. Solving, s=0.0537(0.10−2s)s=0.0537(0.10-2s), s(1+0.1073)=5.37×10−3s(1+0.1073)=5.37\times10^{-3}, giving s=4.85×10−3 Ms=4.85\times10^{-3}\ \mathrm{M}. Ignoring 2s2s gives the tempting 5.4×10−35.4\times10^{-3}, and 1.3×10−51.3\times10^{-5} is the solubility in pure water, Ksp\sqrt{K_{sp}}.
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