Question

Calculate pFe2 at each of the following points in the titration of 25.00 mL of 0.02183 M Fe2 by 0.03651 M EDTA at a 7.00 pH:

A. 10.00 ml of PFe2+

B. Equivalence point Ve

C. 20.00 ml Fe2+

Answer 1

Given Volume of Fe2 25.00 mL Molarity of Fe 0.02183 M Molarity of EDTA- 0.03651 M 10.00 mL of pFe: Moles of Fe Molarity x Volume =0.021 83×0.025 -5.457 x10 mol Moles of EDTA= 0.03651×0.010 -3.651x10 mol 1 mole of Fe reacts with 1 mole of EDTA Remaining moles of Fe2+-(5.457-3.651) x10-4 =1.806×10-4 mol s, molarity of remaining Fe*1.806x10+ (0.025+0.010) =51.6×10-4 Then pFe--log (51.6×10-4) - 2.29 Hence, required value of pFe is 2.29 At equivalence point Moles of Fe2+-moles of EDTA

Moles of EDTA = 0.025×0.02183 = 5.46×104 So 5.46x10" moles of Fer-formed. 0.02183 5-1495 mL 0.03651 Volume of0.02183 Molarity of Fey_ 5.46x104 =0,013 0.025+0.01495 Let the amount dissociated be x. Then, FeY2 ト」EDTA] (0.013-x) 4.80×10% (00137-3) 4.80x10+x-0.013- x= 1.64×10-6 Thus, [Fe2]= 1.64×10-6 M Therefore, pFe -log (Fe2) =-log (1.64×10-6) -5.78 Hence, required value is 5.78 20.0 mL Fe Moles of EDTA-0.020× 0.03651-7.302x10" mol

Moles of Fe0.025x 0.02183 5.457 x10mol 1 mole of Fe reacts with 1 mole of EDTA Remaining moles of EDTA-f 7.302-5.457)× 10" -1.845x104 mol 1.845x104 Molaritv of remaining EDTA- - 41.0x104 M (0.025+0.020) Thus, [Fe2+1-0020002183=00121 M -00121 м 0.045 Now FeY2 Fe [EDTA] 0.0121 1.01x1011- Fe (0.0041) [Fe2+] 2.92 x10" Therefore. pFe"--log (292 ×10-11) 10.53 Hence, the required value is 10.53