Chemistry Postlab 16
Calculations
1)
a)
In order to calculate the concentrations of FeSCN
2+
of the standard solutions, the
molarity of HSCN of each dilution was obtained by using the formula Ci x Vi=Cf
x Vf. Since the equilibrium shifted to the right due to the high concentration of
Fe
3+
in the solution, the moles of FeSCN
2+
equaled to the moles of HSCN.
Example:
Dilution of solution A
Ci x Vi = Cf x Vf
Cf = 0.002 M x 1.00 mL / 100 mL = 2.00E5.0 M HSCN
The values of the standard solution are shown in Appendix A.
b)
In order to calculate the concentration of FeSCN
2+
of the equilibrium solutions,
the εl, calculated by taking the slope of the straight line passing through the values
of the standard solutions, was used.
The slope of the straight line is m = (Af – Ai)/(Cf Ci)
Since A = Cεl, (Af – Ci) can be expressed as CfεlCiεl or εl(Cf – Ci)
Therefore, the slope of the line is m= εl(Cf – Ci)/(Cf – Ci) = εl
The concentration of FeSCN
2+
can be calculated by diluting the absorbance by the
slope of the straight line of the standard solution graph (εl ).
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C = A/εl
[FeSCN
2+
] = A/εl
For instance:
[FeSCN
2+
] of tube 2 first reading
= 2.33E01 / 3.97E+03 = 5.87E05 M
FeSCN
2+
Values are shown in Appendix A.
c)
The number of moles of FeSCN
2+
at equilibrium in each mixture was determined
by expressing the molarity in terms of (moles/mL), and multiplying this value by
10mL of total solution.
For example:
For tube 2 first reading
5.87E5 moles/ 1000 mL x 10 mL = 5.87E7 moles of FeSCN
2+
Values are shown in Appendix A.
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 Fall '11
 staff
 Chemistry, Equilibrium, Molarity, total solution

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