Question

Prelab: Kc determination Answers can only be entered/graded between 2019-10-03 and 2019-10-08 A mixture is prepared by combining 10.23 mL of 2.49 x 103 M Fe(NO3)3 with 10.00 mL of 1.89 x 103 M KSCN. The solution turns red due to the formation of FESCN2, The absorbance is measured and, using a calibration plot, the [FeSCN2'] at equilibrium is found to be 1.83 x 104M. Complete the following ICE table FESCN2 Fe3 SCN (C)0 (B)0.00001 (A) 0.00001 initial (M) (F) 0.00500 (E)-0.00500 (D)-0.00500 change (M) (1)0.00500 (H) 0.0015 (G) 0.0015 equilibrium (M) (A) correct 0.6/1 (sfig err) (B) incorrect 0/1 (C) correct 1/1 (D) incorrect 0/1 [E) incorrect 0/1 W FS 3 2 P T R W

Answer 1

Fe3+ 20.23 Initial (M) 1-2537105 -3 1.259*10 - x + Z W Total volume of the son = 10:23 +10.00-20.23 mL Now, 10:23 mLx 2.49 +10 °CM) = 20.23% 5 P or s=/10 23x2.49x103 20.23 M = 1.259x18 cm) Initial [Fe3+] = 1-259x17%(M) Similarly, initial [san] = 10x1189x10 cm) = 9.343x7d & SON Fe (SCN) + change (M) – x . 9.3434104 0 Equilibrium (M) (1.259x10x) (9.343x102x) +x)=x. But at equilibrium, [Fe(scn)]2+= x = 1.83x10^CM) . At equilibrium, [Fe3+] = (1.259x1024.83+10) (m) = 1.076x103 (M) and . . [SON ] = (9.343 x 10 7 _ 1.83x104) (M) = 7.513x104(M) A= 1.259x10 (M), B = 9.343 x164(M),C=0 D = -1.83x10^(m), E = -1.83x104CM), F= +1-83x16M) G= 1.076x10(M), H= 7.513x104(M), 1 = 1.83x10CM) Fe3+ 3 + SON [Fe (SCN)]2+ Initial (M) A = 1.259X10B = 9.343x10 | c= 0 Change (M) D = -1.838109 E =:83x104 F = +1.831104 Equilibriewin) G = 1.076X10 H = 7.513x104 I = 1.83x104 -4 - 3 - 4 L 3 (