Question

Zinc(II) ions forms a strong complex in the presence of EDTA. Calculate the equilibrium concentrations of Zn2+ and Zn(EDTA)2- after 0.600 mol of Zn(NO3)2 has been dissolved in 1.00 L of 2.60 M EDTA solution. The Kf of Zn(EDTA)2- is 3.8 x 1016.

Answer 1

Answer 8- Page No -o Given that, Zinc (11) ions forms a strong complex in the Presence of EATA. The ionic equation for formation of complex would be Int.com) + EDTA4%20qd) = 2n1 EOTA) .609) Dritial concentration of Zn+69)0.600 mollic = 0-600 M Initial concentration of EOTA" 600) - 2.64 Setting up the ICE table to find equilibrium concentration Zrt.co) EOTA"69) ZOLETA)? Can Initial Concentration_L0600.M. | 2:6M L O change in concentration ___x_ 1-x_ tx Equilibrium concentration 6.600.-x)M_|(16)-x) Mtum Kp for EnLEDTA)% = [Zn(ETA) C0q)] / [zinica)] [ESTA“Coq)] = * |(0-500 -*) (2.6 - x) = x1(1156 - 3.2x +x) 3.8 x100 x1 (1156 - 3:2+x2) (3.8*10'6) x®( 11216 x16) x + (5.988 x106) Do

- Page No - on Solving tiis quadratic equation, the possible values for are ab and 2.6. x=26 will give a ove value of cauilibrium Concentration of ant, which is not possible, . Accepted value of x is x = 0.6 Substituting for me to get quilibrium concentrating of int and in EOTA) [zn2+] = 0.600 - = 0.600 -0.6 - OM (EnCERTA)] = = 0.6H.