PLEASE TYPE OR WRITE IN PRINT 4. When the solutions are prepared for the Beer\'s

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Question

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4. When the solutions are prepared for the Beer's Law Plot, a very dilute sample of Fe3- solution is mixed with a very concentrated SCN solution. What assumption has to be made to determine the Fe(SCN)2* concentration in the mixture? Why is the assumption reasonable? Determine the concentration of Fe(SCN)2* for each of the samples in part I. Consider the answer to problem 2 above and the method used to determine the concentration of the diluted sample. 5. Volume VolumeVolume 0.1 MDI water Sample Volume 0.0002 MIM KSCN Fe(NO3)3 olumeF]Fe(SCN1 (mL) (mL)(mL) HN03 (mL) 4 4 (mL) 2A ?? 4A 5A 32 15 13 7 2 10

Explanation / Answer

4) When solutions are prepared for Berr’s law plot, a very dilute solution of Fe3+ is usually mixed with a very concentrated solution of SCN-. The reaction between Fe3+ and SCN- is given as

Fe3+ (aq) + SCN- (aq) ----------> [Fe(SCN)]2+ (aq) ……(1)

We assume that the reaction goes to completion (for constructing Beer’s law plot). As per the stoichiometry of the reaction,

1 mole Fe3+ = 1 mole SCN- = 1 mole [Fe(SCN)]2+

The volume of the solution is constant for all the three species.

The concentration of Fe3+ solution is chosen to be much lower than that of SCN- to ensure that Fe3+ is the limiting reactant and the above reaction goes to completion, i.e, Fe3+ is quantitatively converted to [Fe(SCN)]2+ and we have,

moles Fe3+ = moles [Fe(SCN)]2+.

5) Take sample 1A as an example. We have 1 mL of 0.0002 M Fe(NO3)3 and the final volume of the solution is (1 + 2 + 4 + 33) mL = 40 mL, .i.e, we have,

C1 = 0.0002 M; V1 = 1 mL and V2 = 40 mL.

Use the dilution equation

C1*V1 = C2*V2

to determine the concentration of Fe3+ in the final solution. Plug in values and obtain

(0.0002 M)*(1 mL) = C2*(40 mL)

=====> C2 = (0.0002 M)*(1 mL)/(40 mL) = 0.000005 M.

Again as the stoichiometry of the reaction in (1) above,

[Fe3+] = [Fe(SCN)2+] (volume stays constant); therefore, the concentration of [Fe(SCN)]2+ in sample 1A = 0.000005 M.

Fill in the table below.

Sample

Volume of 0.0002 M Fe(NO3)3 (mL)

Volume of 1 M KSCN (mL)

Volume of 0.1 M HNO3 (mL)

Volume DI water (mL)

Total Volume (mL)

[Fe3+] (M)

[Fe(SCN)2+] (M)

0.000005

0.000010

0.000020

0.000048

0.000100

6a) The Beer’s law is given as

A = ?*C*l ……(2)

where A = absorbance of a solution having concentration C and molar absorptivity constant ?. l is the path length of the solution through which light passes.

The linear plot between absorbance and the concentration of a solution has the form

y = mx + c ……(3)

where y denotes the absorbance of the solution and x is the concentration of the solution. c may be a zero or non-zero proportionality constant. m is the slope of the plot between y and x.

Comparing equations (2) and (3), we have

A = y

C = x

and ?l = m.

6b) The slope of the line is equal to m and denotes the rate of change of absorbance for unit change in concentration. When we compare the two equations, the slope of the plot m denotes the product of the molar absorptivity constant and the path length of the solution, i.e,

m = ?*l.

6c) The molar absorptivity of a substance is a unique property of the substance and hence, is constant. For a particular experiment, the path length of the solution is constant. Therefore, the slope of the plot m = ?*l is constant.

The absorbance of an unknown solution is recorded by a spectrophotometer, i.e, y is known. Using equation (3) above, we can simply substitute values and determine x, which is the concentration of the unknown in the solution.

7) The concentration of the ions in the dilute solution is given as

C1*V1 = C2*V2

where C1 is the concentration of the stock solution; V1 is the volume of the stock solution; C2 is the concentration of the species in the dilute solution and V2 is the volume of the dilute solution. Here, we have C1 = 0.00200 M for both Fe3+ and SCN-; V1 = 5 mL for both and V2 = (5 + 15) mL = 20 mL; therefore,

(0.00200 M)*(5 mL) = [Fe3+]ini*(20 mL)

====> [Fe3+]ini = (0.00200 M)*(5 mL)/(20 mL) = 0.00050 M.

(0.00200 M)*(5 mL) = [SCN-]ini*(20 mL)

====> [SCN-]ini = (0.00200 M)*(5 mL)/(20 mL) = 0.00050 M.

8) The chemical equation for the equilibrium reaction is given as

Fe3+ (aq) + SCN- (aq) <======> [Fe(SCN)2+] (aq)

initial (M) 0.00050 0.00050 -

change (M) -0.0002 -0.0002 +0.0002

equilibrium (M) 0.0003 0.0003 0.0002

The equilibrium constant for the reaction is given as

Kc = [Fe(SCN)2+]/[Fe3+][SCN-]

= (0.0002 M)/(0.0003 M)(0.0003 M)

= 2222.22 M-1 ? 2222.2 M-1

The equilibrium constant for a reaction is written as a dimensionless quantity, hence, report the equilibrium constant as 2222.2 (ans).