Question

HOME F12 END, POD did in class 11. In Chem VO1 BL, we will make tetro-ammine copper (W) by mixing Cu" (og) (- [Cu(H20).Xa) with NH3(aq) [Cu(H20).laq) + 4NH3(aq) = [Cu(H20),(NH2).]?aq) + 4H200 For which, k, = 1.20 x 10' Q. What are the two ligands (Lewis Bases) in the above equation? b. if 100.0 mL of 3.5 x 10' M Cu(NO3)2 is mixed with 150.0 mL of 0.20 M NH3. What is the [Cu] at equilibrium? [Cu(H20).]?aq) (NH3(aq)] ICH [Cu(H2O)2(NH3)4]? 49)

Answer 1

(a) Ligands are the electron rich species which donate at least one pair of electrons to metal in a complex In the above complex [Cu(H2O)2(NH3)]2, Two ligands are H20 and NH3 (b)

Number of moles of Cu*2 =[Cu (H,O)]+2 = 3.5 x 10-3 M 100.0 mL 0.35 mmol Number of moles of NH2 = 0.20 M * 150.0 mL 30 mmol The balanced equation for the formation of the complex [Ag(NH3)2]* is [Cu(H20)l*4 NH3 ---> [Cu(H20)2(NH3)4]*2 Since the concentration of Cu*2 is less than that of NH3, assume that all Cu 2 is converted to the complex. [Cu(H20)J*2 [Cu(H20)2(NH3)al2 mmol 4 NH3 Initial 30 0.35 -1.4 Change -0.35 0.35 Final 0 28.6 0.35 Concentration of the complex= moles/total volume 0.35 mmol/(100+150) mL = 0.0014 M The conc. of NH3remaining the solution = 28.6 mmol/(100+150)mL 0.1144 M At this point, there is no Ag* left in the solution. The concentration of Cut2 is determined by the dissociation of the complex [Cu(H20)2(NH3)a12 Cu(H20)l*2 conc 4 NH3 + Initial 0 0.1144 0.0014 +4x Change + X -X Final 0.0014 x 0.1144+4x X [Cu(H,O), (NH,), K 0.0014 x =1.20x1013 4 [CuH,ONH,] x.(0.1144+4x) x is small, so 0.0014-x= 0.0014 and 0.1144 + 4x = 0.1144 0.0014 r = 1.20 x1013 x.(0.1144) 0.0014 = 6.8115 x10-13 1.20x10 x(0.1144) Concentration of Cu*2 6.81 x10-13 M