Question

Commercial silver plating operations frequently use a solution containing the complex |Ag(CN)_2] ion. Because the formation constant K_f is quite large, this procedure ensures that the free Ag+ concentration in solution is low to promote uniform electrodeposition. In one process, a chemist added 9.0 L of 1.1M NaCN to 90.0 L of 0.12 M AgN0_3. Calculate the concentration of free Ag+ ions at equilibrium. K_f for this reaction is 1.0 times 10^21. Enter your answer in scientific notation.

Answer 1

no. of moles of NaCN = molarity*vol.in lt = 7.7*9.0 = 69.3 moles

no.of moles of AgNO3 = molarity*vol.in lt = 0.12*90.0 = 10.8 moles

...... .......... Ag^+ + 2CN- .......> [Ag(CN)2]^-

initial           10.8mol....69.3 mol..............0
amounts reacting 10.8 mol...2*10.8................0

amounts at eq.    0 ........(69.3-21.6 = 47.7).....10.8 mol

Equilibrium calculation

...... .......... Ag^+ + 2CN- .......> [Ag(CN)2]^-
initial............0......47.7 mol.......10.8 mol
change.............+x......+2X............ - x
Equilibrium........ +x.....(47.7+2x)......(10.8-x)

      Kf = [Ag(CN)2]^- /[Ag+][CN-]^2
   
     1.0*10^21 = (10.8-x)/(x)(47.7+2x)^2
      2x value is negligible compare to 47.7
       x value is negligible compare to10.8
   2275.29x*10^21 = 10.8
      x = 10.8/2275.29*10^21
      x = 4.75*10^-18 M
[Ag+] = 4.75*10^-18 M