Question

Plot of absorbance vs concentration for calibration curve. Data sheet stapled to back of DRA. Solution Stock solution Dilution 1 (50.0%) Dilution 2 (25.0%) Dilution 3 (12.5%) (SCN) (FeSCNI. 2.0 x10M 2.0 X10M 1.0 x 10-4 M 1.0104M 5.0 x10-5M 5.0*10-5M 2.5 x 10-5Ma.5 x10 -5M 0.836 0.389 0.180 0.085 Calibration curve trendline equation_=43 15.1 % - 0.032 Rvalue of trendline 0.9993 Absorbance Sample Number 0.0020 M Fe(NO), (ml) 0.0020 M KSCN (ML) - - - - - 0. 129 28 0.250 0.323 0.397 0.463 0.568 0.408 4 0.363 0.247 90 - -

Answer 1

I Initial concentration is the conc. of solution after mixing but before any reaction. It can by be calculated by using Miv, = Ma van Mg = MIV Stock After mixing M.= stock concentration ų I v- stock volume V = vol. after mixing like 9+1=10,8+2=10 for all solutions final sum of all volume = I final volume = 10mL For Fe 3+ for FNO3)₂ ] Sample 12 M, = 0.0020 M, V = 9 mL g k = 9+1=10m Mo = 0.0020Mx9 mL i 80MX9 m2 = 10.0018 M TomL For SCN- (or KSCN), M, 20:0020 v= 1 mL b = 1 + 9 = 100 mL. so Mg = 0.002oX IML 0.00020 M_1200 x 10 M) 10mL How 3+ + SCN- FeSCNo using ICE. U table FeSONT I 0.0018 2.08104 Fest SCN- ta cx -re E 0.0018-2 20x10x

for sample 1 [fes CN&+] e = 3.73x105x so, [fest Jeg = 0.0018-x=0.0018 -3.73X103 -0.0017627 [SCN Jeg = 20x10 x 20 x 10 L 30734105 = 1.627 X10-4M. So. KE [fe ScNet] eq 3.73x10-5 3 [Fe3+Jeg {SCN-Jeg - 0.0017627X10627x104 ko 1.301x102 - 1130.10 Similarly for sample 2: kg = 6.53410-5 0.0015 347 X 000003347 Similarly for sample 3% - 80 22x10-5 k3 0.008178 %0.0005178 = 120.46 Similarly can be calculated for other samples 3347 = 11 2 7.13