Question

Part A A 110.0-mL sample of a solution that is 2.7 x 10- M in AgNO, is mixed with a 230.0-mL sample of a solution that is 0.11 M in NaCN. For Ag(CN)2,Kf = 1.0 x 1021 After the solution reaches equilibrium, what concentration of Ag+ (aq) remains? Express your answer using two significant figures. IVO ACV O O ? [Ag +) =

Answer 1

Given. volume of AgNO₃ = 110ml = 0.112 Concentration / Molarity of Ay NO₂ = 2.78103 M Volume of NaCW = 230ml 0.23L Holarity of Nal = 0.11M. moles of AgNO₃ = 2.7 x 10-3 xo.ll = 0.297 x 10 moles 2 Moles of Nal = = 0.11 X 0.23 0.0253 moles. rives Imoloca This , Imole AgNo will give Imole Agt and Imole Nach gives Imola Now, Ag+ + 2CN - Ag (CNN) - Imole 2 mole. Imole. we can see that I mole of Ag + reacts with 2 mol of CN Thus, 0.297X103 moles of Ag + will react with 2x 0.297810²3 = 0.594X10 3 moles of CN Total volume of solution = volume of Ag NO₃ + volume of Now - Oill +0.23 = 0.34L.

No. of males of Ag (Cw), = no. of moles of - 0.297 X103 moles; Agt Concentration of Aglcw) = moles of Ag (en) " Total volume of solution 0.297 Xlon mal 0.11L+ 0.23 L [Ag (CN)] = 0.87 x 103 model Moles of CN left after formation of Ag (C) = Initial moles of CN - moles of CN used 0.0253 - 0.594 X 103 0.02471 maler, Thus, concentration of CN after formation of Aglende ااح ها (CNJ = moles of Cw left Total volume of solution = 0.0247 0.34 » [CN-] = 0.0726 molo /L)

Now, ko for reaction ① will be ko = [Ag (ca.) [Ag+] [CN-]2 1.08181 = 0.878103 [Ag +] [0.072672 [Ag+] = 0.87 X 100 1.0010² x 5:27 X 200 →/ Ag t] = 0.165 x 1021 mole /L)
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