4. A student determining the equilibrium constant for the formation of the FeNcs

Question

4. A student determining the equilibrium constant for the formation of the FeNcs21 ion followed the procedure in this experiment. In Part I, 2.00 x 10-1M Fe and 2.00 x 10 M SCN ion stock solutions were used. All standard solutions were prepared in 100-mL volumetric flasks. Complete the following enter and the data in Table 3 on the next page. (1) Calculate the initial molar SCN ion concentration in standard solutions S3 and S5 (2) Calculate the molar FeNCS2+ ion concentration in standard solutions S3 and S5 (3) Calculate the absorbances for standard solutions S3 and S5 (4) Prepare a Beer's Law plot, on the graph paper earlier in this experiment, for the FeNCs ion, using data for standard solutions si-S6 In Part II, 2.00 x 10 3M Fe ion and 2.000 x 10-3M SCN ion stock solutions were used to prepare the equilibrium mixtures indicated in Table 4 on the next page. The %T of the equilibrium solutions (E2-EA) are listed in Table 5 on the next page. Make the following calculations in the spaces provided, and enter the results in Table 5. (5) Calculate the initial Fes+ ion concentration in equilibrium mixtures E2-E4 (6) Calculate the initial SCN ion concentration in equilibrium mixtures E2-E4. (7) Express the T readings for solutions E2-EA as equivalent absorbances. solutions (8) Use your Beer's Law plot to determine the equilibrium FeNcs2+ ion concentration in E2-E4. (9) Determine the equilibrium molar concentration in solutions E2-E4. Fe ion 10) Determine the equilibrium molar SCN ion concentration in solutions E2-E4. using the data 01 calculate Keq for the formation of FeNcs2+ ion at the experimental temperature, for solutions E2-E4. mean Keg for the formation FeNOS ion at (12) Use your calculated Kegs fro (11) calculate the of to the experimental temperature.

Explanation / Answer

concentration of Fe^3+ = 0.002 M

concentration of SCN^- = 0.002 M

[SCN] in standard solution in S3 = 4mL * 0.002M/100mL

                                                = 8*10^-5 M

[SCN] in standard solution in S5 = 8mL * 0.002M/100mL

                                                =1.6*10^-4 M

Fe(SCN)^2+ concentration in S3 = 8*10^-5 M

Fe(SCN)^2+ concentration in S5 = 1.6*10^-4 M

Absorbance of S3 = log (100/T%)

                           = log (100/65.8)

                            = 0.182

Absorbance of S5 =log(100/45.4)

                           =0.343

                           

                                                 

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