Question

Ex. 13.4 The analysis of 10.00 mL 0.0100 M Ca²+ solution in water at pH 10.00 that was titrated using 0.0050 M EDTA. The conditional formation constant for this reaction under these conditions is 1.75X1010. What values of pCa would be expected (a) at the beginning of the titration; (b) 50% of the way through this titration; (c) at the equivalence point; (d) after an excess of 10.00 mL titrant has been added to this sample?

Answer 1

ANS Page 1 1. Initial point [co] = 0.0100M pla = -log (0.0106) pla "=2 2. At 50% Mole Ca2+ - 10 me x 0.0100 = 0.1 mmoles Mole EDTA = 5 mx 0.0050 = 0.025 mmoles Reaction table. - Ca²+ + EDTA - Ca- EDTA Before neaction: 1 0 .025 0 +0.025 o neacion! -0.025 -0.025 Aften reaction: 0.075 0 0.025 So [Ca 2+] = 0.075 - La 10+5 - = 5x10-3 i pea 2.306 3. At equédalence point moles Ca 2+ = 10ml x 0.0100 M = 0. Immone. moes EDTA = 20ml x 0.0050 M = 0.1 mmole. Reaction table - Ca²+ + EDTA Ca-EDTA Beforme neaction 0.1 0.1 0 reaction : -01 - oil toile After reaction! o o tool Now [ca2+] = 0 and pealt og Now take a look at the equetébmum reaction [CAEDTA ] [Ca] [EDTA from the reaction table, Ca-EDTA] = 0.1 = 3.33*10-3 m [ca] (EDTA] = 2, for back neaclion [Ca-Edral (3:3x10-3-2) Ki = (3.3610-3_2) nation constant = 1.75*1010 Прохорео g Kf = » 1.75×10 10 - 8.3x10-3_x an

> 175x100x = 3.3 X 10-3-X Page 2 → 75x100x + X-3.3 * 10 3 0 2 - 19- 4x1.75 X 100x33x10-3) - 2x1.75 81010 2 -1 +15198 68 - 54.84 2x10-7- [ca] 3.58 1010 Pca21 = -log (4.34 x 10-7) = 6.36 4. Aften equivalence point we continue with the same logic, since the only ca2+ come from the equilibrium, we need to use equilibrium constant. Addition of excess 10 me of EDTA: moles of Ca2+ = 10 ml x 0.0100 yyOilm moles moles of EDTA = 30mexo.0050M = 0.15 mmoles. Ca ²+ + EDTA - Ca-EDTA] Before reaction on 0.15 reaction -01 - 0il to! Aften neaction +0.05 tool So oun momenal concentration ane, [Ca ²+] =0, [EDTA = 0.0510+20 = 1.25x10-3 m [Ca-EDTA] = 0./10430 = 2.5810-3 m Now lets throw the back reaction [Ca ²+] = oth [EDIA] = (1.25x10-3+2). [ca-EDTA] = .5X10-3-2) go to the equilibrium equation, ki = [Ca-EDTAJ [Ca] [EDTA] 1.7581010 - 2.5x 10-3_W 2 x (* 125 X10-342) And again assume that x is small before Simplify the equation (and avoid the quadanatics and

Page 3 2.5x 10-3 1.25 x 10-3 xa >> 175 x 1.25*10-3x1010 = 2.5x10-3 > x= 2.58 10-3 1.75% 1010 x 1.25*10-3 a = 1.143x10-10 = [Ca2+] - pealt - log (1.143 × 10-10) = 9.94