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JEE Main 2024 · Chemistry · Ionic Equilibrium · hard

Solid AgCl is shaken with 1.00 M aqueous NH3 at 298 K until equilibrium, forming [Ag(NH3)2]^+. Given Ksp(AgCl)=1.8×10^-10 and Kf=1.5×10^7 for…

Numerical answer

Solid AgCl is shaken with 1.00 M1.00\ \mathrm{M} aqueous NH3\mathrm{NH_3} at 298 K298\ \mathrm{K} until equilibrium, forming [Ag(NH3)2]+[\mathrm{Ag(NH_3)_2}]^+. Given Ksp(AgCl)=1.8×10−10K_{sp}(\mathrm{AgCl})=1.8\times10^{-10} and Kf=1.5×107K_f=1.5\times10^{7} for [Ag(NH3)2]+[\mathrm{Ag(NH_3)_2}]^+, the solubility of AgCl, in millimolar (mM), rounded to the nearest integer, is (neglect the concentration of free Ag+\mathrm{Ag^+})
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Answer: 47

AgCl(s)+2NH3⇌[Ag(NH3)2]++Cl−\mathrm{AgCl(s)+2NH_3\rightleftharpoons[Ag(NH_3)_2]^++Cl^-} with K=KspKf=2.7×10−3K=K_{sp}K_f=2.7\times10^{-3}. If the solubility is SS, then [[Ag(NH3)2]+]=[Cl−]=S[\mathrm{[Ag(NH_3)_2]^+}]=[\mathrm{Cl^-}]=S and free [NH3]=1.00−2S[\mathrm{NH_3}]=1.00-2S. So S2(1−2S)2=2.7×10−3⇒S1−2S=0.0520\dfrac{S^2}{(1-2S)^2}=2.7\times10^{-3}\Rightarrow\dfrac{S}{1-2S}=0.0520, giving S=0.05201+0.104=0.0471 M≈47 mMS=\dfrac{0.0520}{1+0.104}=0.0471\ \mathrm{M}\approx47\ \mathrm{mM}.
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