What is the molar solubility of AgCl in 0.10 M NaCN if the colorless complex ion Ag(CN)₂^- forms? K_sp for AgCl is 1.8 × 10^-10 and K_f for Ag(CN)₂^- is 1.0 × 10²¹. A)0.050 M B)0.10 M C)0.20 M D)0.40 M

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The formation of the complex ion Ag(CN)₂^- implies that some of the Ag+ and CN- ions are consumed in forming the complex. Let x be the amount (in mol/L) of AgCl which dissolves in the presence of NaCN. Since AgCl dissociates into Ag+ and Cl- ions, the concentration of Ag+ is also x. As a result, the concentration of CN- ion in solution will increase by x mol/L. Then, the equilibrium concentrations of the ions will be: [Ag+]= x [CN-]= 0.10 + x [Cl-]= 0.10 - x Using the solubility product expression for AgCl, we can write: Ksp= [Ag+][Cl-]= (x)(0.10 - x) The equilibrium concentration of [Ag(CN)₂^-] complex is not given, but we know its formation constant, so we can write: Kf= [Ag+][CN-]/[Ag(CN)₂^-]= x(0.10 + x)/[Ag(CN)₂^-] Now, we can solve for x by equating Ksp and Kf: (x)(0.10 - x)= 1.8 × 10^-10 x(0.10 + x)/[Ag(CN)₂^-]= 1.0 × 10²¹ Solving these two equations simultaneously, we get x= 0.050 M. Therefore, the molar solubility of AgCl in 0.10 M NaCN is 0.050 M.

What Is the Molar Solubility of AgCl in 0