Question

A 45.0 mL sample of 0.0100 M Cu2+ buffered at pH 8.00 is titrated with 0.0250 M EDTA. Calculate pCu2+ at each of the added volumes of EDTA. For the Cu2+−EDTA complex, log⁡Kf is 18.78. The fraction of free EDTA in the Y4− form, αY4−, can be found in the table below.

0 mLpCu2+=

6.00 mLpCu2+=

18.0 mLpCu2+=

23.0 mLpCu2+=

pH	αY4-
0	1.3 x 10-23
1	1.4 x 10-18
2	2.6 x 10-14
3	2.1 x 10-11
4	3.0 x 10-9
5	2.9 x 10-7
6	1.8 x 10-5
7	3.8 x 10-4
8	4.2 x 10-3
9	0.041
10	0.30
11	0.81
12	0.98
13	1.00
14	1.00

Answer 1

Volume of Cut2 solution = 45.0 mL Conc. of Cu+2 solution = 0.0100 M Conc.of EDTA = 0.0250 M logKf = 18.78 Formation constant of Cu(EDTA)]-2 = 6.03x1018 At pH=8, the fraction of EDTA that exists Y-4 is av-4 = 4.2 x 10 So the Conditional Formation constant = K) = 4.2x10-²x 6.03x1018 = 2.5326 x 1016 (a) [Cu+2] = 0.0100 pCu+2 = -log(0.0100) = 2.00 b) Addition of 6.0 mL of EDTA mmols of Cu+2 = M*V = 0.01 M* 15.0 mL = 0.45 mmol mmols of EDTA = M*V = 0.0250 * 6.0 mL = 0.15 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 = 0.45-0.15 = 0.30 mmol 020 =0.005882M moles 0.30 mmol [Cu*?] =- Total Volume (45.0+6.0) mL- pCu*? =- log(0.005882)= 2.23

c) Addition of 18.0 mL of EDTA: mmols of Cu+2 = M*V = 0.01 M* 15.0 mL = 0.45 mmol mmols of EDTA = M*V = 0.0250 * 18.0 mL = 0.45 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-] complex. So the concentration of Cu+2 is determined by the dissociation of (CuY-2] complex. initial moles of Cu+2 Total Volume 0.45 mmol = 0.007143M (45.0+18.0) mL y4 K' A Cur-2 Initial Cu+2 0 + x + 0 + x 0.007143 Change -X Equilibrium x x 0.007143 - X

0.007143 . (: x is small) CuY-21 [Cu ][Y-+] 2 0.007143 0.007143 - X X.X 0.007143 3x = -K 2. 82x10-19 2.53 x 1016 x=5.31x10-10 [Cu*?]=5.31x10-10 - pCut? = -log(Cu+27=- log(5.31x10-10)= 9.27 d) Addition of 23.0 mL of EDTA mmols of Cu+2 = M*V = 0.01 M* 15.0 mL = 0.45 mmol mmols of EDTA = M*V = 0.0250 * 23.0 mL = 0.575 mmol At this point EDTA present in excess. Net moles of EDTA = 0.575 -0.45 = 0.125 mmol the concentration of Cu+2 is determined by the dissociation of CuY-2] complex. 2 initial moles of Cu* [Cuy?]=? Total Volume 0.45 mmol = 0.006618M (45.0+23.0) mL no.of moles of EDTA -no. of moles of Cu*? [EDTA] = 10 Total Volume 0.125 mmol -= 0.001838M (45.0+18.0)mL

K = [Cuy?) -=1.42 x 10-16 ^4 *[Cu*?][Y+] 10+27 [CuY-?] =[Cu*] = 0.006618 Kyx[Y+] 2.53x10'6 x 0.001838 [Cu*?]=1.42x10-16 = pCu*P = -log[Cu*?]=-log(1.42x10-16) = 15.85 pCu+2 of the solution after the addition of 23.0 mL of EDTA = 15.85