Question

A 140.0 mL sample of 0.080 M Ca2+ is titrated with 0.080 M EDTA at pH 9.00. The value of log?f for the Ca2+−EDTA complex is 10.65 and the fraction of free EDTA in the Y4− form, ?Y4− , is 0.041 at pH 9.00. What is ?′f , the conditional formation constant, for Ca2+ at pH 9.00? ?′f= What is the equivalence point volume, ?e , in milliliters? ?e= mL Calculate the concentration of Ca2+ at ?=12?e . [Ca2+]= M Calculate the concentration of Ca2+ at ?=?e . [Ca2+]= M Calculate the concentration of Ca2+ at ?=1.1?e . [Ca2+]=

Answer 1

a) According to the condition given - 10.65 y = 4.47 x 100 Conditional formation constant as follows. Kp dytl4 = 0.041 14:47 x 100 Tkale = 1.83x10? by Mcate = 0.080 MEOTA = 0.080 M Veat2 = 140me MEDTA VEDTA - Mca +2 'catz UM X VEDTA 10.080 M X 140ml. VEDTA = 0.080 M X 140 ml 0.080M VEDTA = luome c) Now let us consider the reaction cate + y4--cay? For cat & M = 0.080M v = 14ome. V2 - V - 70 me. concentration of cath at V₂ is as follows

My = My x fonction de maining x diution factor EMIK ) lor in) : 0.080M X 140 md - 70ml 140ml + tome. s 0.080M X 70 ml 2iome. = 0.080 M X 0.333 = 2.6 x 10²m. The required cat2 concentration is 2266 x 10°°M. D) At Equivalence point potential cats iš Converted into cay - 2 140.0ml 2 [cay-3 = 0. 080 H x heo.0 +140.0 | =0.080 MX0.5 = 0.040 M ca +2 + EDTA - - 2 CAY? 0.040. initial 0.040-2 equilibriun conditional equitatum constant is as • Zcay”] [cat?] [EDTA] 118311090.040 -> x = 6.62 x 10 m [cat2 = 6,62x100m

€) we have velile = "1140me) =154me [cay-2) +0.08 M ( 1004 se me] Iko -0.08 x 160 294 50.08 X 0.476) = 0.038 M (EOTA) = 0.08M 154 me - luomen 154 ml + 140m = 0.08 x 14 294 = 0.08 x 6,04761 = 0.0038 M. [cay? [cat?] (EDTA] [cat?] = 0.038 1183 x 109 x 0.0038M 0,038 6954 000 = 5.46x10PM. E The value of [cat?} = 5.46x10 M.