Question

Chapter 15 COC Reference NO,Aume that Out forms complex ions with X as follows Destion 18 Destion 19 Question 20 A solution is formed by mixing 50.0 mL of 10.9 MNX with 500 ml of 8.6 x 10-M Out (g) + X() OXog K; -10 x 10 CuX(aq) + X() - Cux, (a) K; -1.0 x 109 CuX;" ) + X (a) Cux,- (aq) K; -1.0 x 100 with an overall reaction Cu* (eq) +3X" () - CuX; (aq) -10x10' Calculate the following concentrations at equilibrium a Concentration of Cux,-)-D h. Concentration of Cux, . Concentratice of Cu*] Question 21 Question 221 estion 24 esten 25

Answer 1

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No. of molus in some of 10.9 M Nax = 10.9.x 50 1000 = 0.505 mole of Nax No of moles in so me of 8.6810SM CUNO3 1. = 8.6810150 Y 100 = 0.03x10-3 mole at Cunoa Toral Solechon voluere = 50+50 = 100 ml. so new conc. of cut= 0.43x10 100 XI Tconcot vu?, 4.3110 PM so new conc. of t= I concotx = acconding to eq 0 . cut + x = Kia 1x102 0.54501000 100 5.45 M) !! cux K= [cux] [cut] [x] put me values in above ean

LLLLL PAGE-3 1x10²= [cux] 4,30103 x 5.45 » [c4x} = 1x10²x 4.3x10 3x5.45 [[cux] = 2.34 M 1 according ro en © Cux + x = cuxz K2= 1x104 k= [cuxz ] [Cux] [*] [cuxz'] = ixio x 2.34 x 5.45 T [Cuxz]= 12.753 x 10 ml aconding of eq ② cuxitx - cuxz²- [Lux3] = kx [(4x2] [x] = 107x 12.753 x10' x 5.45 [cuxz2-] = 69.50 x 107 m K3= 1X103