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Investigation 10: What is the Formation Constant for FeSCN2+ Experimental Raw Data for Students: Assign the data below to four different groups. Table 1: Standard Solution Preparation (0.200 M F (NO3)3, 0.00200 M KSCN-Pipette volume readings Solution Initial Final Reading I Volume Total volume Reading (mL) (mL) delivered in beaker (mL) into the (mL) beaker 0.200 M 0.10 9.20 F3(NO3)2 9.1 mL 7.01 mL 0.00200 M 7.10 9.19 12.09 mL KSCN Table 2: Equilibrium Solutions (0.00200 M F3(NO3)3, 0.00200 M KSCN-Pipette volume readings. Beaker 1 Solution Initial Final Reading Total volume Reading (mL) (mL) delivered in beaker (mL) into the (mL) beaker 0.00200 M 1.74 5.03 mL F (NO3)3 0.00200 M 3.93 5.97 | 2.04 mL 10.08 mL KSCN Water 2.14 5.15 3.01 mL 6.77 Beaker 2 Solution Initial Reading (mL) Final Reading (mL) Volume delivered (mL) into the beaker Total volume in beaker (mL) 3.47 8.80 0.00200 M F:(NO3)3 0.00200 M KSCN Water 6.90 9.90 15.33 mL | 3.00 mL | 2.28 ml 10.61 mL 5-34 7.62 Table 3: Absorbance Values for Each Solution (Standard and Equilibrium Mixtures). Solution Standard Beaker 1 Beaker 2 Absorbance 0.954 0.199 0.283

Using the information provided, these questions needed to be answered. My group and I are struggling to find the average Kf using the absorbance of the solution. Please help we are struggling.

Answer 1

Solution Date of standard solution Volvine of 0.2 m fe(Nor)z = 9.1 mL. moles of 0.2 m Fe(NO3) = Molering x volume = 0.2 x 9.1 ml = 1:82 mmol. Volume of 0.002 m KSIN = 2.09 mL. moles of KSCN . .= malarity & Volume = 0.002 x 2.og me = 0.00418 in modo Totul volume of solution = 9., mL +2.0i mL 1. = 11.19 ml. initial a [rela (f?*), = moles of Fe 3+ total volume 1.82 mmed 11.19 mL 0.1626 m Tre 3+ ] = a initial rsoni 7 = - 0.00418 monel 1 lg me 0.0003735(mol) = Scanned with CamScanner

you can see the concentration of . scni is very very less than the concentration of fest im can assume that the deartion between scht is totally completed so you Fe 3+ and Fest & schi - > rescn2+ 0.1626 m 0.0007735 m ņH -0.0003735 -0.0003735 + 0.0003735 E 0.1622 o 0.0003735 Now, absorbance of Stanland solution is giren A = 0.954 Lambda Beer's Law - A= ęcil d=1cm From 0.954 = EX 0:0003735 X 1 ię = 2553.8899 Cance.com determination of equilibrium concentration of te schet complex for Beaker af and Beaken - 2 for beaker-1 ; A = 0.199 -s Scanned with A=&cl 0,199 = 2553.8999 XCX! CamScanner

C = 0.0000779 m For beake - 2 A = 0.293 A = ecl 0.283 = 2553.88.99 XCXI C = 0.0001108 M fe 37 and schi in concentration of Determination of initial beaker-I and Beaker 2 for Beaker-1 Volume of 0.002 m. Felt = 5.03ml. - Volume of 0.002 om say = 2.04 ml. 120 E 3.0 mL. Total volume = 7.07+ 3.01 = 10:08 ml. using formula, miri= M2 V₂ [Fe3+] = 0·002 X 5.03 10:08 (fe3+] = 0.000998m = 0.002 X 2.04 10.08 0.0004047 m [schi] = Scanned with CamScanner

for beaken - 2 fel = 0'CO2 X 5.33 10.61 1 Fe3+] = ooolm = 0.0oż X 3.00 (simi I 10.61 = 0.0005655 m Determination of k i . For beaker-1 . Fest & scni - Pesany?t 0.000498 0.0004047 0.000998-H 0.0009201 0.0004047- 0.0003268 2=0.0000479 0-0600779 0.0000779 0.0009201 X 0:0003268 0.0009201 X 0.0003268 ke = [festn't] 1f23+] [ scri] Kf = 658 259.07 for beaker-2 fet San" - 0.001 0.0005655 3+ fescn²+ 0 +* = 0.0001108 0.0005655- 0.001-* 0.0008892 0.0004547 0.0001108 cs Scanned with CamScanner

Kf = lesen?t (pe 34 ] [ sc] Kf = 0.0001108 0.0008892 X 0.0004547 kg = 274.04 average value kt arerage = 254.07 +23.4.04 2 lavag k = 266.56 Well done both kf values are very cloge. # our first aim was to determine the & value and hence equilibrium concentration of (Festnet Scanned with CamScanner