TE But to get the equilibrium concentration of FeSCN2+ in any sar use the measur

Question

TE But to get the equilibrium concentration of FeSCN2+ in any sar use the measured absorbance given on page B-5. So these absorbances ar ation of FeSCN2+ in any sample solution, you must e the measured absorbance of FeSCN2+ in the solution and the Beer's Law equation page B-5. So these absorbances are added into the spreadsheet as shown below Table B-2. 2.00E-03 2.00E-03 Sample solutions (Initial)Abs Sample solutions (Equilibrium) V(KSCN, L VFeNO3), LM(KSCN) M(Fe(NO3)3) M(FeSCN 2+) M (SCN-) M(Fe3+) Keq 0.00200 0.01000 0.004000.01000 0.00600 0.01000 0.00800 0.01000 0.010000.01000 0.0995 0.2060 0.2872 0.4306 0.5575 Avg = The highlighted cells in the spreadsheet shown are the ones in which you must en- ter formulas that allow calculation of the specified results. Calculation of K for each sample solution is done using the equation (as on page 4-2 of lab manual): You must turn in your copies of Figure B-3, Table B-1, and Table B-2 with values in all highlighted cells. The average Keq should be somewhere around 2 × 102, You should also turn in the set of calculations by hand for one sample solution. This should con- vince you that the formulas you entered into the spreadsheet were correct.

Explanation / Answer

5 X 10^-4

The total volume used in each trial = 0.02 L

Initial conc of KSCN = 2 X 10^-3

Volume of KSCN Used = 0.002 L

Moles of KSCN = Molarity X volume = 0.002 X 2X10^-3 = 4 X 10^-6 moles

Molariy of KSCN in solution = moles / total volume = 4 X 10^-6 moles / 0.02 L = 2 X 10^-4 M

Volume of KSCN Used = 0.004 L

Moles of KSCN = Molarity X volume = 0.004 X 2X10^-3 = 8 X 10^-6 moles

Molariy of KSCN in solution = moles / total volume = 8 X 10^-6 moles / 0.02 L = 4 X 10^-4 M

Volume of KSCN Used = 0.006 L

Moles of KSCN = Molarity X volume = 0.006 X 2X10^-3 = 12 X 10^-6 moles

Molariy of KSCN in solution = moles / total volume = 12 X 10^-6 moles / 0.02 L = 6 X 10^-4 M

Volume of KSCN Used = 0.008 L

Moles of KSCN = Molarity X volume = 0.008 X 2X10^-3 = 16 X 10^-6 moles

Molariy of KSCN in solution = moles / total volume = 16 X 10^-6 moles / 0.02 L = 8 X 10^-4 M

Volume of KSCN Used = 0.01 L

Moles of KSCN = Molarity X volume = 0.01 X 2X10^-3 = 20 X 10^-6 moles

Molariy of KSCN in solution = moles / total volume = 20 X 10^-6 moles / 0.02 L = 10 X 10^-4 M

For Fe(NO3)3

Initial conc of Fe(NO3)3 = 2 X 10^-3

Volume of KSCN Used = 0.001 L

Moles of KSCN = Molarity X volume = 0.001 X 2X10^-3 = 2 X 10^-6 moles

Molariy of KSCN in solution = moles / total volume = 2 X 10^-6 moles / 0.02 L = 1 X 10^-4 M

M(KSCN) M(Fe(NO3)3 V (total) = 0.02 L Sample solution V(KSCN) L V(Fe(NO3)3 L M(KSCN) M(Fe(NO3)3 0.002 0.01 2 X 10^-4 1 X 10^-4 0.004 0.01 4X10^-4 2 X 10^-4 0.006 0.01 6X10^-4 3 X 10^-4 0.008 0.01 8X10^-4 4 X 10^-4 0.01 0.01 10X10^-4

5 X 10^-4

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