Can you please help post lab question.. DATA AND CALCULATIONS Absorbance Trial 1
ID: 949856 • Letter: C
Question
Can you please help post lab question..
DATA AND CALCULATIONS
Absorbance
Trial 1
0.039
Trial 2
0.045
Trial 3
0.063
Trial 4
0.086
Absorbance of standard (Trial 5)
0.792
Temperature
25°C
Kc expression Kc =
[Fe3+]i
Trial 1
Trial 2
Trial 3
Trial 4
[SCN–]i
[FeSCN2+]eq
[Fe3+]eq
[SCN–]eq
Kc value
Average of Kc values
Kc = ________________________ at ___25_____°C
beaker
Fe(NO3)3
(mL)
KSCN
(mL)
H2O
(mL)
1
5
2
93
2
5
3
92
3
5
4
91
4
5
5
90
PROCESSING THE DATA
1. Write the Kc expression for the reaction in the Data and Calculation table.
2. Calculate the initial concentration of Fe3+, based on the dilution that results from addingKSCN solution and water to the original 0.0020 M Fe(NO3)3 solution. See Step 2 of theprocedure for the volume of each substance used in Trials 1-4. Calculate [Fe3+]i using theequation:
[Fe3+]i =
Fe(NO3)3 mLtotal mL X (0.0020 M)
This should be the same for all four test tubes.
3. Calculate the initial concentration of SCN–, based on its dilution by Fe(NO3)3 and water:
[SCN–]i =
KSCN mLtotal mL X (0.0020 M)
In beaker 1, [SCN–]i = (2 mL / 100 mL)(.0020 M) = .000040 M. Calculate this for the otherthree beakers.
4. [FeSCN2+]eq is calculated using the formula:
[FeSCN2+]eq =
AeqAstd
X [FeSCN2+]std
where Aeq and Astd are the absorbance values for the equilibrium and standard solutions,
respectively, and [FeSCN2+]std = (1/100)(0.0020) = 0.000020 M. Calculate [FeSCN2+]eq foreach of the four trials.
5. [Fe3+]eq: Calculate the concentration of Fe3+ at equilibrium for Trials 1-4 using the equation:
[Fe3+]eq = [Fe3+]i – [FeSCN2+]eq
6. [SCN–]eq: Calculate the concentration of SCN- at equilibrium for Trials 1-4 using theequation:
[SCN–]eq = [SCN–]i – [FeSCN2+]eq
7. Calculate Kc for Trials 1-4. Be sure to show the Kc expression and the values substituted infor each of these calculations.
8. Using your four calculated Kc values, determine an average value for Kc.
Absorbance
Trial 1
0.039
Trial 2
0.045
Trial 3
0.063
Trial 4
0.086
Absorbance of standard (Trial 5)
0.792
Temperature
25°C
Explanation / Answer
DATA AND CALCULATIONS
Absorbance Trial 1 0.029 Trial 2 0.045 Trial 3 0.063 Trial 4 0.086
Absorbance of standard (Trial 5) C = 0.000020 M 0.0792
Temperature 25°C
1. Kc expression Kc = [Fe(SCN)2+] / ([Fe3+][SCN-])
[Fe3+]i Trial 1 (5mL/100mL)x0.0020M= 0.000100 M
Trial 2 0.000100 M
Trial 3 0.000100 M
Trial 4 0.000100 M
[SCN–]i 0.000040 M 0.000060 M 0.000080 M 0.000100 M
[FeSCN2+]eq 0.0000008 0.0000012 0.0000016 M 0.0000022M
[Fe3+]eq 0.000100 M 0.000099M 0.000098M 0.000098 M
[SCN–]eq 0.000039 0.000059 0.000079 0.000098
Kc value 205 205 206 229
Average of Kc values Kc = 211 at 25 °C
Processing Data:
1. Kc = [Fe(SCN)2+] / ([Fe3+][SCN-])
2. The formula you provided is incorrect. Correct to:
[Fe3+]i = ( VFe(NO3)3 mL / Vtotal mL ) X (0.0020 M)
3. [SCN–]i = (VKSCN mL / Vtotal mL ) X (0.0020 M)
see table
4. [FeSCN2+]eq =(Aeq / Astd )X [FeSCN2+]std
See table , etc