Question

A 30.0 mL sample of 0.0200M Cu2t buffered at pH 9.00 is titrated with 0.0300 M EDTA. Calculate pCut at each of the added volumes of EDTA. For the Cu2+-EDTA complex, log Kr is 18.78. The fraction of free EDTA in the Y form, ay can be found in this table. O mL pCu2+ 6.67 mL pCu2+ 20.0 mL pCu2

Answer 1

Conc.of EDTA = 0.0300 M logKf 18.78 Formation constant of Cu (EDTA)]2 = 6.03x1018 At pH -9, the fraction of EDTA that exists Y4 is -0.041 So the Conditional Formation constant= K 0.041x 6.03x1018 2.47 x1017 (a) [Cu+2] 0.0200 pCu*2 =log(0.0200) 1.69897 1.70 b) Addition of 2.50 mL of EDTA mmols of Cu*2 = M*V = 0.0200 M* 30.0 mL 0.6 mmol mmols of EDTA = M*V = 0.0300 * 6.67 mL 0.2001 mmol This is Before the equivalence point, as Cu*2 is present in excess. pCu*2 is calculated by the concentration of unreacted Cu*2 Net moles of Cu*2 0.6-0.2001 0.3999 mmol moles 0.3999 mmol [Cu1= 0.010905M (30.0 6.67mL Total Volume pCu log (0.010905) = 1.96236 1.96

c) Addition of 20.00 mL of EDTA: mmols of Cut2 = M*V 0.0200 M* 30.0 mL 0.6 mmol mmols of EDTA = M*V = 0.0300 * 20.0 mL 0.6 mmol Number of moles of Cu*2 Number of moles of EDTA. This is the equivalence point. At equivalence point, all Cu*2 is converted to [CuY-2] complex. So the concentration of Cu 2 is determined by the dissociation of [CuY2] complex. initial moles of Cu 0.6 mmol [CuY= =0.012M (30.0 20.0)mL Total Volume к Y-4 CuY2 Cut2 Initial 0 0.012 Change + X +X - X 0.012 x Equilibrium х х [CuY2 к, [Cu Y 0.012-x 0.012 (x is small) 2 х.х 0.012 0.012 4.86 x 10-20 2.47 x1017 K x= 2.20 x 10-10 [Cu 2.20x10-10 -log (2.20x 10-10) =9.656801 = 9.66 - -log Cu pCu =_ 1